Basic Differentiation: Hamilton Education Guides Manual 3 - Over 560 Solved Problems 0964995484, 9780964995482

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Basic Differentiation

Manual Title: Basic Differentiation Author: Dan Hamilton Editor: John Hamilton Cover design by: John Hamilton Copyright  2000 All rights reserved. Printed in the United States of America. No part of this manual may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. Request for permission or further information should be addressed to Hamilton Education Guides via [email protected]. First published in 2000 Library of Congress Catalog Card Number 99-97605 Library of Congress Cataloging-in-Publication Data ISBN 0-9649954-8-4

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Hamilton Education Guides Manual Series

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eManual versions available in the Amazon Kindle Store

Hamilton Education Guides Manual Series

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Contents Basic Differentiation

Quick Reference to Problems 1 1.1

The Difference Quotient Method .............................................................................. 2

1.2

Differentiation Rules Using the Prime Notation ..................................................... 7

1.3

Differentiation Rules Using the

1.4

The Chain Rule .......................................................................................................... 30

1.5

Implicit Differentiation .............................................................................................. 45

1.6

The Derivative of Functions with Fractional Exponents........................................ 50

1.7

The Derivative of Radical Functions ........................................................................ 57

1.8

Higher Order Derivatives.......................................................................................... 72

d Notation .......................................................... 19 dx

Appendix – Exercise Solutions.............................................................................................. 87 Section 1.1 87 Section 1.4 98 Section 1.7 106

Hamilton Education Guides

Section 1.2 89 Section 1.5 103 Section 1.8 109

Section 1.3 93 Section 1.6 104

i

Acknowledgments The primary motivating factor in writing the Hamilton Education Guides manual series is to provide students with specific subjects on mathematics. I am grateful to John Hamilton for his editorial comments, cover design, and suggestions on easier presentation of the topics. I would also like to acknowledge the original contributors of the Hamilton Education Guides books for their editorial reviews. Finally, I would like to thank my family for their understanding and patience in allowing me to prepare this manual.

Hamilton Education Guides

ii

Introduction and Overview It is my belief that the key to learning mathematics is through positive motivation. Students can be greatly motivated if subjects are presented concisely and the problems are solved in a detailed step by step approach. This keeps students motivated and provides a great deal of encouragement in wanting to learn the next subject or to solve the next problem. During my teaching career, I found this method to be an effective way of teaching. I hope by presenting equations in this format, more students will become interested in the subject of mathematics. This manual is a chapter from my Mastering Algebra – Advanced Level book with the primary focus on the subject of basic differentiation. The scope of this manual is intended for educational levels ranging from the 11th grade to adult. The manual can also be used by students in home study programs, parents, teachers, special education programs, preparatory schools, and adult educational programs including colleges and universities as a supplementary manual. A fundamental understanding of basic mathematical operations such as addition, subtraction, multiplication, and division is required. This manual addresses basic derivatives and how they are simplified and mathematically operated. Students learn topics that include the computation of derivatives using the Difference Quotient method and Chain Rule; differentiation rules using the prime notation; how to find the derivative of exponential and radical expressions; and the computation of higher order derivatives. Detailed solutions to the exercises are provided in the Appendix. Students are encouraged to solve each problem in the same detail and step by step format as shown in the text. It is my hope that all Hamilton Education Guides books and manuals stand apart in their understandable treatment of the presented subjects and for their clarity and special attention to detail. I hope readers of this manual will find it useful. With best wishes, Dan Hamilton

Hamilton Education Guides

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Basic Differentiation

Quick Reference to Problems

Basic Differentiation Quick Reference to Problems 1.1

The Difference Quotient Method .............................................................................. 2 lim h→0

1.2

1 x + h +1 + x +1

1

(θ + 1)

Differentiation Rules Using the

3

h

=

2 x

 2 + x3   = ;   x +1 

f ( x) =

2 x 2 + 3x + 1 x2 +1

=

d Notation .......................................................... 19 dx

[

)] = ;

(

d  u u2    = + du  1 − u 1 + u 

d ( x + 1) x 2 − 3 dx

2

  t3 + t 2  h( t ) =  4   t −1

(

 f ( x) =  x 3 + 2x 

=;

)

2

 − x2 

4

=

= ; xy + x 2 y 2 + y 3 = 10 x = ; 3x 3 y 3 + 2 y 2 = y + 1 =

The Derivative of Functions with Fractional Exponents........................................ 50

( )

1

2 3

+ (2 x + 1)

3 5

=;

y = ( x + 1)

1 2

( x + 3) 2

1 3

1

=;

y=

x 3 (x − 1) 1 + x2

=

The Derivative of Radical Functions ........................................................................ 57 r (θ ) = θ 2 +

1.8

x+3

Implicit Differentiation .............................................................................................. 45

y= x

1.7

=;

=;

x 2 y 2 + y = 3y 3 − 1

1.6

(x + h ) + 3 −

= ; lim h→0

The Chain Rule .......................................................................................................... 30  1  f ( x ) =  2 + x x 

1.5

f ( x) =

=;

3

d  2x   5x + 2  dx  x + 1

1.4

4(x + h )2 − 4 x 2 h

Differentiation Rules Using the Prime Notation ..................................................... 7 r (θ ) = θ 2 +

1.3

= ; lim h→0

1

(θ + 1) 3 = ;

r (θ ) =

θ3 +1 θ 2 +1

= ; h(t ) =

3 2

t +1 t3

=

Higher Order Derivatives.......................................................................................... 72 f (u) =

Hamilton Education Guides

u3 − 1 =; u +1

r (θ ) = θ 2 +

1

(θ + 1) 3

=;

f (t) =

t 3 + t 2 +1 = 10

1

Basic Differentiation The objective of this manual is to improve the student’s ability to solve problems involving derivatives. In Section 1.1 derivatives are computed using the Difference Quotient method. Various differentiation rules using the prime and d notation are introduced in Sections 1.2 and dx

1.3, respectively. The use of the Chain Rule in finding the derivative of functions that are being added, subtracted, multiplied, and divided are addressed in Section 1.4. Implicit differentiation is discussed in Section 1.5. Finding the derivative of exponential and radical expressions is addressed in Sections 1.6 and 1.7, respectively. Finally, computation of second, third, fourth, or higher order derivatives are discussed in Section 1.8. Each section is concluded by solving examples with practice problems to further enhance the student’s ability.

1.1

The Difference Quotient Method

In this section students learn how to differentiate functions using the difference quotient equation as the limit h approaches zero. The expression

f ( x + h) − f ( x ) h

is referred to as the difference

quotient equation. A function f ( x ) is said to be differentiable at x if and only if lim h→0 f ( x + h) − f ( x ) h

exists. If the limit exists, then the result is referred to as the derivative of f ( x ) at

which is denoted by f ′( x ) . It should be noted that this approach is rather long and time consuming and is merely presented in order to show the usefulness of the differentiation rules which are addressed in the subsequent sections. The following examples show the steps in finding derivatives of functions using the Difference Quotient method: x

Example 1.1-1: Use the Difference Quotient method to find the derivative of the following functions. a. f ( x ) = 3x + 1

b. f ( x ) = 4 x 2

c. f ( x ) = x + 3

d. f ( x ) = (2 x − 7) 2

e. f ( x ) = − x + 1

f. f ( x ) =

Solutions:

1 x +1

a. To find the derivative of the function f (x ) = 3x + 1 First - Substitute f (x + h ) = 3(x + h ) + 1 = 3x + 3h + 1 and f (x ) = 3x + 1 into the difference quotient equation, i.e.,

f (x + h ) − f (x ) h

=

(3x + 3h + 1) − (3x + 1)

3h/ 3/ x + 3h + 1/ − 3/ x/ − 1/ = / = = 3 h

h

h/

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= lim h→0 3 = 3

b. To find the derivative of the function f (x ) = 4x 2

(

)

First - Substitute f (x + h ) = 4(x + h )2 = 4 x 2 + h 2 + 2 xh = 4 x 2 + 4h 2 + 8 xh and f (x ) = 4x 2 into the Hamilton Education Guides

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Basic Differentiation

1.1 The Difference Quotient Method

f (x + h ) − f (x ) h

difference quotient equation, i.e., =

4h/ (h + 2 x ) h/

4(h + 2 x ) 1

=

=

( 4x

2

)

+ 4h 2 + 8 xh − 4 x 2 h

=

4/ x/ 2/ + 4h 2 + 8 xh − 4 x/ 2/ h

= 4h + 8 x

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

f (x + h ) − f (x ) h

= lim h→0

= lim h→0 (4h + 8 x ) = (4 ⋅ 0) + 8 x = 0 + 8 x = 8 x

c. To find the derivative of the function f ( x ) = x + 3 First - Substitute f (x + h ) = (x + h ) + 3 = f (x + h ) − f (x ) h

equation, i.e.,

x+h+3

and f (x ) = x + 3 into the difference quotient

x+h+3 − x+3 h

=

To remove the radical from the numerator multiply both the numerator and the denominator by x + h + 3 + x + 3 to obtain the following: f (x + h ) − f (x ) h

=

x+h+3 − x+3 h

=

x+h+3 − x+3 x+h+3 + x+3 ⋅ h x+h+3 + x+3

1

=

x+h+3 + x+3

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= lim h→0

1 x+h+3 + x+3

=

1 x+3 + x+3

=

1 2 x+3

d. To find the derivative of the function f (x ) = (2 x − 7 )2 First - Substitute f (x + h ) = [ 2 (x + h ) − 7 ] 2 = 4 (x + h ) 2 + 49 − 28 (x + h ) = 4 x 2 + 4h 2 + 8 xh + 49 − 28 x − 28h and f (x ) = (2 x − 7 ) 2 = 4 x 2 + 49 − 28 x into the difference quotient equation, i.e., = =

(4 x

2

) (

+ 4h 2 + 8 xh + 49 − 28 x − 28h − 4 x 2 + 49 − 28 x h

4h 2 + 8 xh − 28h h

=

h/ (4h + 8 x − 28) h/

)=

f (x + h ) − f (x ) h

4/ x/ 2/ + 4h 2 + 8 xh + 4/ 9/ − 2/ 8/ x/ − 28h − 4/ x/ 2/ − 4/ 9/ + 2/ 8/ x/ h

= 4h + 8 x − 28

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= lim h→0 (4h + 8 x − 28) = (4 ⋅ 0) + 8 x − 28 = 0 + 8 x − 28 = 8 x − 28

e. To find the derivative of the function f (x ) = − x + 1 First - Substitute f (x + h ) = − (x + h ) + 1 = − x + h + 1 and f (x ) = − x + 1 into the difference Hamilton Education Guides

3

Basic Differentiation

1.1 The Difference Quotient Method

quotient equation, i.e.,

f (x + h ) − f (x ) h

x + h +1 + x +1 h

= −

To remove the radical from the numerator multiply both the numerator and the denominator by x + h + 1 − x + 1 to obtain the following: f (x + h ) − f (x ) h

x + h +1 + x +1 h

= −

x + h +1 + x +1 x + h +1 − x +1 ⋅ h x + h +1 − x +1

= −

1

= −

x + h +1 + x +1

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= lim h→0

f (x + h ) − f (x ) h

= − lim h→0

1 x + h +1 + x +1

f. To find the derivative of the function f (x ) = 1

First - Substitute f (x + h ) =

f (x + h ) − f (x ) h

i.e.,

x + h +1

1

=

x + h +1 h

−

= −

1

= −

x +1 + x +1

1 2 x +1

1 x +1

and f (x ) =

1 x +1

into the difference quotient equation,

1 x +1

=

x +1 − x + h +1 h x +1 ⋅ x + h +1

To remove the radical from the numerator multiply both the numerator and the denominator by x + 1 + x + h + 1 to obtain the following: f (x + h ) − f (x ) h

=

=

x +1 − x + h +1 h x +1 ⋅ x + h +1

⋅

x +1 + x + h +1 x +1 + x + h +1

=

(x + 1) − (x + h + 1) h (x + 1) x + h + 1 + h (x + h + 1)

x/ + 1/ − x/ − h − 1/ −h/ = h (x + 1) x + h + 1 + h (x + h + 1) x + 1 h/ (x + 1) x + h + 1 + (x + h + 1) x + 1

[

]

=

(x + 1)

x +1 −1

x + h + 1 + (x + h + 1) x + 1

Second - Compute f ′(x ) as the lim h→0 in the difference quotient equation, i.e., f ′( x )

= −

= lim h→0

f (x + h ) − f (x ) h

= lim h→0

(x + 1)

−1

x + h + 1 + (x + h + 1) x + 1

=

(x + 1)

−1

x + 1 + (x + 1)

x +1

1

2 ( x + 1)

x +1

Example 1.1-2: Given the derivative of the functions in example 1.1-1, find: a. f ′(2)

b. f ′(3)

c. f ′(1)

d. f ′(0)

e. f ′(15)

f. f ′(0)

Solutions: a. Given f ′(x ) = 3 then, f ′(2) = 3 Hamilton Education Guides

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Basic Differentiation

1.1 The Difference Quotient Method

Note that since the derivative is constant f ′(x ) is independent of the x value. f ′(2) can also f (x + h ) − f (x ) h

be calculated directly by using f ′(x ) = lim h→0

[ 3 (2 + h ) + 1] − (6 + 1)

and by replacing x with 2 , i.e., f ′(2) = lim h→0 = lim h→0

3h/ h/

= lim h→0

h

[ 3 (x + h ) + 1] − (3x + 1) h

= lim h→0

6 + 3h + 1 − 7 h

= lim h→0 3 = 3

b. Given f ′(x ) = 8 x then, f ′(3) = 8⋅ 3 = 24 4 (x + h ) 2 − 4 x 2 f (x + h ) − f (x ) = lim h→0 h h

f ′(3) can also be calculated directly by using f ′(x ) = lim h→0

and by replacing x with 3 , i.e., f ′(3) = lim h→0

= lim h→0

(

)

4 9 + h 2 + 6h − 36 4 ( 3 + h) 2 − 4 ⋅ 32 = lim h→0 h h

h/ (4h + 24 ) 36 + 4h 2 + 24h − 36 = lim h→0 = lim h→0 4h + 24 = 0 + 24 = 24 h h/

c. Given f ′(x ) =

1 2 x+3

then, f ′( 1) =

1

=

2 1+ 3

1 2 4

=

1 2⋅2

=

1 4

Again, f ′( 1) can also be calculated directly by using the equation f ′(x ) = lim h→0 = lim h→0 = lim h→0

=

(x + h ) + 3 −

x+3

h

h+4 − 4 h

1 0+4 +2

=

1 4 +2

and by replacing x with 1 , i.e., f ′( 1) = lim h→0 h+4 − 4 h+4 + 4 ⋅ h h+4 + 4

= lim h→0

=

1 2+2

=

= lim h→0

h

(

h h+4 +2

)

f (x + h ) − f (x ) h

(1 + h ) + 3 −

1+ 3

h

= lim h→0

1 h+4 +2

1 4

d. Given f ′(x ) = 8 x − 28 then, f ′(0) = (8 × 0) − 28 = −28 e. Given f ′(x ) = − f. Given f ′(x ) = −

1 2 x +1

then, f ′(15) = −

1 2

(x + 1)3

1 2 15 + 1

then, f ′(0) = −

= −

1 2 (0 + 1)3

1 2 16

= −

= −

1 2 ⋅1

1 2⋅4

= −

= −

1 8

1 2

In problems 1.1-2 d, e, and f students may want to practice finding f ′(x ) for the specific values of

Hamilton Education Guides

5

Basic Differentiation

x

1.1 The Difference Quotient Method

by using the general equation f ′(x ) = lim h→0

above stated solutions.

f (x + h ) − f (x ) . h

The answers should agree with the

Finally, it should be noted that every differentiable function is continuous. However, not every continuous function is differentiable. The proof of this statement is beyond the scope of this manual and can be found in a calculus book. In the following section we will learn simpler methods of finding derivative of functions using various differentiation rules. Section 1.1 Practice Problems – The Difference Quotient Method 1. Find the derivative of the following functions by using the Difference Quotient method. a. f (x ) = x 2 − 1

b. f (x ) = x 3 + 2 x − 1

c. f (x ) =

x x −1

e. f (x ) = 20 x 2 − 3

f. f (x ) = x 3

g. f (x ) =

10 x−5

d. f (x ) = − h. f (x ) =

1 x2

ax + b cx

2. Compute f ′(x ) for the specified values by using the difference quotient equation as the lim h→0 . a. f (x ) = x 3 at x = 1

b. f (x ) = 1+ 2 x at x = 0

c. f (x ) = x 3 + 1 at x = −1

d. f (x ) = x 2 (x + 2) at x = 2

e. f (x ) = x −2 + x −1 + 1 at x = 1

f. f (x ) = x + 2 at x = 10

Hamilton Education Guides

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Basic Differentiation

1.2

1.2 Differentiation Rules Using the Prime Notation

Differentiation Rules Using the Prime Notation

In the previous section we differentiated several functions by writing the difference quotient equation and taking the limit as h approaches to zero. This process, however - as was mentioned earlier, is rather long and time consuming. Instead, we can establish some general rules that make the calculation of derivatives simpler. These rules are as follows: Rule No. 1 - The derivative of a constant function is equal to zero, i.e., if f (x ) = k ,

then f ′( x ) = 0

For example, the derivative of the functions f ( x ) = 10 , g( x ) = −100 , and s( x ) = 250 is equal to f ′( x ) = 0 , g ′( x ) = 0 , and s ′( x ) = 0 , respectively. Rule No. 2 - The derivative of the identity function is equal to one, i.e., if f ( x ) = x ,

then f ′( x ) = 1

For example, the derivative of the functions f ( x ) = 5x , g( x ) = −10 x , and s( x ) = 5x is equal to f ′(x ) = 5 ⋅1 = 5 , g ′(x ) = −10 ⋅1 = −10 ,

and s ′(x ) = 5 ⋅1 = 5 , respectively.

Rule No. 3A - The derivative of the function f ( x ) = x n is equal to f ′( x ) = n x n −1 , where n is a positive integer. For example, the derivative of the functions f ( x ) = x , f ( x ) = x 2 , f ( x ) = x 3 , and f ( x ) = x 4 is equal to f ′( x ) = 1 ⋅ x1−1 = x 0 = 1 , f ′( x ) = 2 ⋅ x 2−1 = 2 x1 = 2 x , f ′( x ) = 3 ⋅ x 3−1 = 3x 2 , and f ′(x ) = 4 ⋅ x 4−1 = 4x 3 , respectively. Rule No. 3B - The derivative of the function f ( x ) = x n is equal to f ′( x ) = n x n −1 , where n is a negative integer. For example, the derivative of the functions f ( x ) = x −1 , f ( x ) = x −2 , f ( x ) = x −3 , f ( x ) = x −4 , and −

1

f ( x) = x 8

is equal to f ′( x ) = −1 ⋅ x −1−1 = − x −2 , f ′( x ) = −2 ⋅ x −2−1 = −2 x −3 , f ′(x ) = − 3 ⋅ x −3−1 = −3x −4 , 1 8

and f ′( x ) = − ⋅ x

1 − −1 8

9

1 − =− x 8, 8

respectively.

Note that this rule can also be used to obtain the derivative of functions that are in the form of 1 f ( x) = n x

by rewriting the function in its equivalent form of f ( x ) = x − n . For example, the

derivative of the functions f (x ) = is equal to f ′(x ) = −8 ⋅ x

−8−1

f ′( x ) = −1 ⋅ x = −8 x

−9

−1−1

= −x

2 1 1 = −2 x −3 , = x − 2 , f (x ) = − = x −1 , f (x ) = 3 2 x x x −2

,

f ′( x ) = −2 ⋅ x

−2−1

= −2 x

−3

,

and f (x ) =

f ′( x ) = ( −2 ⋅ −3) ⋅ x

−3−1

= 6x

1 x

8

−4

= x −8

, and

, respectively.

Rule No. 4 - If the function f ( x ) is differentiable at x , then a constant k multiplied by f ( x ) is also differentiable at x , i.e., ( k f ) ′ x = k f ′( x )

Hamilton Education Guides

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Basic Differentiation

1.2 Differentiation Rules Using the Prime Notation

Note that this rule is referred to as the scalar rule. For example, the derivative of the functions f ( x ) = 5x , g( x ) = −10 x , and s( x ) = 5x is equal to f ′( x ) = 5 , g ′( x ) = −10 , and s ′( x ) = 5 , respectively. Rule No. 5 - If the function f ( x ) and g( x ) are differentiable at x , then their sum is also differentiable at x , i.e.,

( f + g )′ (x ) =

f ′( x ) + g ′( x )

In other words, the derivative of the sum of two differentiable functions, ( f + g )' (x ) , is equal to the derivative of the first function, f ' (x ) , plus the derivative of the second function, g ' (x ) . Note that this rule is referred to as the summation rule.

)

(

For example, the derivative of the functions h( x ) = (5x − 3) + 2 x 2 − 1 and s( x ) = 6 x 3 + (3x + 2) is        f ( x)

f ( x)

g( x )

    g( x )

equal to h ′( x ) = f ′( x ) + g ′( x ) = 5 + 4 x and s ′( x ) = f ′( x ) + g ′( x ) = 18 x 2 + 3 . Rule No. 6 - If the function f ( x ) and g( x ) are differentiable at x , then their product is also differentiable at x , i.e., ( f ⋅ g )′ (x ) = f ′( x ) g( x ) + g ′( x ) f ( x ) In other words, the derivative of the product of two differentiable functions, ( f ⋅ g )' (x ) , is equal to the derivative of the first function multiplied by the second function, f ' (x ) ⋅ g (x ) , plus the derivative of the second function multiplied by the first function, g ' (x ) ⋅ f (x ) . Note that this rule is referred to as the product rule.

(

)

For example, the derivative of the functions f ( x ) = (3x − 5)(6 x + 1) and g( x ) = −10 x 2 5x 3 − 2 is equal to f ′( x ) = [ 3 ⋅ (6 x + 1) ] + [ 6 ⋅ (3x − 5) ] = 18 x + 3 + 18 x − 5 = (18 x + 18 x ) + (− 5 + 3) = 36 x − 2 and

[

)] [

(

]

g ′( x ) = − 20 x ⋅ 5 x 3 − 2 + 15 x 2 ⋅ −10 x 2 = −100 x 4 + 40 x − 150 x 4 = −250 x 4 + 40 x

Rule No. 7 - Using the rules 1 , 4 , 5 , and 6 we can write the formula for differentiating polynomials, i.e., if f ( x ) = a n x n + a n−1 x n−1 + a n−2 x n−2 +  + a3 x 3 + a2 x 2 + a1 x1 + a0 , then f ′( x )

= n a n x n−1 + ( n − 1) a n−1 x n−2 + ( n − 2) a n−2 x n−3 +  + 3a3 x 2 + 2a2 x + a1

For example, the derivative of the polynomials f ( x ) = 6 x 4 + 5x 3 − 3 , g( x ) = 2 x 5 − 3x 2 − 4 x , and h( x ) =

1 2 x − 2x + 5 3

is equal to f ′( x ) = (6 ⋅ 4) x 4−1 + (5 ⋅ 3) x 3−1 = 24 x 3 + 15x 2 , g ′( x ) = (5 ⋅ 2) x 4 − (2 ⋅ 3) x − 4

2 1 = 10x 4 − 6 x − 4 , and h′( x ) =  2 ⋅  x − 2 = x − 2 , respectively. 

3

3

Rule No. 8 - If the function f ( x ) and g( x ) are differentiable at x , then their quotient is also differentiable at x , i.e.,

Hamilton Education Guides

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Basic Differentiation

1.2 Differentiation Rules Using the Prime Notation

′  f   ( x)  g

f ′(x )g (x ) − g ′(x ) f (x )

=

[ g (x ) ] 2

'

 f    ( x ) , g

In other words, the derivative of the quotient of two differentiable functions,

is equal to

the derivative of the function in the numerator multiplied by the function in the denominator, f ' ( x ) ⋅ g ( x ) , minus the derivative of the function in the denominator multiplied by the function in the numerator, g ' (x ) ⋅ f (x ) , all divided by the square of the denominator, [ g (x ) ] 2 . Note that this rule is referred to as the quotient rule. For example, the derivative of the functions f ( x ) =

[ 3 ⋅ (1 + x ) ] − [ 1⋅ (1 + 3x ) ] = = (1 + x )2 =

6 x 4 + 6 x − 9 x 4 − 15x 2

( x + 1) 3

2

=

3 + 3x − 1 − 3x

(1 + x ) 2

=

1 + 3x 1+ x

and g( x ) =

3x 2 + 5 x3 +1

is equal to f ′(x )

[ 6x ⋅ (x + 1) ]− [ 3x ⋅ (3x + 5) ] and g ′(x ) = (x + 1) 3

2

(1 + x ) 2

2

3

2

2

−3x 4 − 9 x 2

( x + 1) 3

2

In the following examples the above rules are used in order to find the derivative of various functions: Example 1.2-1: Differentiate the following functions. a. f ( x ) = 5 + x

b. f ( x ) = x 3 + 3x − 1

d. f ( x ) = 10 x 3 + 5x 2 + 5

e. f (x ) = 3 x 2 + 2 x

g. f ( x ) = ax 2 + b

h. f ( x ) =

1 1 − x x3

j. f ( x ) =

m. f ( x ) =

x4 1− x

p. f ( x ) =

2 x

k.

 2 + x3     x +1 

v. f ( x ) =

x 

f ( x ) =x −5 + 3 x −3 − 2 x −1 + 10

x 4 + 10 2

x +1

(

)

q. f ( x ) = x 2 + 1 ( x + 5)

x 

2 x 2 + 3x + 1

5+ x

 3x 2 + 5   ( x + 1)  x 

t. f ( x ) =  w. f ( x ) =

x2 +1

Solutions: a. Given f ( x )=

)

x5 x4 x3 x2 − + − 10 4 6 4

n. f ( x ) =

1  1  s. f (x ) = 1 +  1 − 3  

(

then f ′( x ) =

Hamilton Education Guides

0 + x1−1

=

x0

=

ax 2 + bx + c ax 2 − b

c. f ( x ) = 2 − x f. f ( x ) = ax 3 + bx 2 + c i. f ( x ) = l. f ( x ) = o. f ( x ) =

10 x2

x2 + 2 x4

3+ x

x3 − 5

r. f ( x ) = ( x + 1)( x + 2) u. f ( x ) = x. f ( x ) =

1+ x2 1− x2 3− x 1 −5 x

1

9

Basic Differentiation

1.2 Differentiation Rules Using the Prime Notation

then f ′( x ) =

b. Given

f ( x ) = x3 + 3x − 1

c. Given

f ( x )= 2 − x

d. Given

f ( x ) = 10 x 3 + 5 x 2 + 5

then

(

e. Given f (x ) = 3 x 2 + 2 x

f ( x ) = ax 3 + bx 2 + c

g. Given

f (= x ) ax 2 + b

h. Given

f ( x) =

=

=

(5x

then f ′( x ) =

⋅ 10 ) − ( 0 ⋅ x 5 )

50 x 4 16 x 3 18 x 2 8 x − + − 100 16 36 16

−

=

( 4x

4 −1

j. Given

f ( x )=

1 1 − x x3

k. Given

f ( x ) =x −5 + 3 x −3 − 2 x −1 + 10

then f ′( x ) =

=

)

=

30x 2 + 10x

(

= 3 2x + 2 ⋅ x 0 =

⋅ 4) − ( 0 ⋅ x4 )

+

( 3x

(0 ⋅ x ) − ( 2x (x ) 2

2 −1

3 −1

⋅ 10 )

2 2

⋅ 6 ) − ( 0 ⋅ x3 )

−

( 2x

2 −1

6x + 6

⋅ 4) − ( 0 ⋅ x2 ) 42

=

0 − ( 2 x ⋅ 10 ) x4

−20x/ x 4/ = 3

=

(x )

then f ′( x ) =

or, we can rewrite

f ( x)

3

as f ( x ) =

-

1 3x 2/ + x 2 x 6/ =4

1

3 2 + 3 − + 10 x x x 5

2

3 2

−

=

20 x3

=

-

1 3 + x 2 x4

−5 x −5 −1 + 3 ⋅ −3 x −3−1 − 2 ⋅ −1x −1−1 + 0

( 0 ⋅ x ) − ( 5 x ⋅1) + ( 0 ⋅ x ) − ( 3x ⋅ 3) − ( 0 ⋅ x ) − (1 ⋅ 2 ) + 0 = x (x ) (x )

Hamilton Education Guides

= 3 (2 x + 2 ⋅1) =

2a x

62

x

5 2

)

3ax 2 + 2bx

( 0 ⋅ x ) − (1 ⋅ 1) − ( 0 ⋅ x3 ) − ( 3x3−1 ⋅ 1) then f ′( x ) = = 2 3 2

4

3x 2 + 3

1 1 1 1 ⋅ 5 x 5 −1 − ⋅ 4 x 4 −1 + ⋅ 3 x 3−1 − ⋅ 2 x 2 −1 10 4 6 4

the quotient rule. 5

=

x4 x 3 x 2 x - + 2 1 2 2

10 x2

f ′( x ) =

3x 2 + 3 ⋅ 1

or, using the quotient rule we obtain

42

f ( x) =

-5x -6 - 9x -4 + 2x -2

a ⋅ 2 x 2 −1 + 0

then f ′( x ) =

i. Given

=

10 ⋅ 3 x 3−1 + 5 ⋅ 2 x 2 −1 + 0

a ⋅ 3 x 3−1 + b ⋅ 2 x 2 −1 + 0

x4 x 3 x 2 x - + 2 1 2 2

=

102

=

=

3x 2 + 3 ⋅ x0

= −1

− x0

then f ′( x ) = 3 2 x 2−1 + 2 ⋅ x1−1

x5 x 4 x3 x 2 − + − 10 4 6 4

5 −1

f ′( x)

then f ′( x ) =

5 4 4 3 3 2 2 x − x + x − x 10 4 6 4

f ′( x ) =

then

=

0 − x1−1

(

)

f. Given

=

f ′( x)

=

3 x 3−1 + 3 ⋅ x1−1 − 0

2

−

5x 4 x

10

−

9x 2 x

6

+

and then find f ′( x ) using

2 x

2

=

5 x 4/ 9 x 2/ 2 −= − 6/ 4 + 2 // 6 = 10 x x x

10

Basic Differentiation

=

−

5 9 2 − + x6 x4 x2

2 x5 − 4 x5 − 8 x3 x8

m. Given =

4 x3 − 3x 4

(1 − x )

then

(x

2

x 4 + 10 x2 + 1

+ 1)

2

x3 − 5 − 9 x 2 − 3x3 3

− 5)

f ( x) =

)]

[ 2x ⋅ x ]− [ 4x ⋅ ( x = 4

4 −1

=

3

2

+2

)]

x8

4 2

8/ =5

[ 4x

(

(

− 2 x2 + 4 x5

)

][

⋅ (1 − x ) − (0 − 1) ⋅ x 4

]

=

(1 − x ) 2

4 x 3 (1 − x ) + x 4

(1 − x )

2

4 x3 − 4 x 4 + x 4

=

(1 − x )

2

=

2

[ ( 4x

then f ′( x ) =

=

( x + 1) 2

4 −1

) ] [ ( 2x (x + 1)

)(

+ 0 x 2 +1 −

2

=

(

2 x x 4 + 2 x 2 − 10

(x

2

+1

)

2

[ (0 + 1) ( x − 5) ]− [ ( 3x f ′( x ) = ( x − 5) 3

3−1

3

2 −1

)(

+ 0 x 4 + 10

)]

2

2

2 x 5 + 4 x 3 − 20 x

then

=

(

) ( ( x + 1)

4 x 3 x 2 + 1 − 2 x x 4 + 10 2

2

) )

− 0 ( 3 + x)

2

]

=

(x

3

− 5) − 3x 2 ( 3 + x )

(x

3

− 5)

2

− 2x 3 − 9x 2 − 5

2  2 + x3    x  x +1 

 3 x 2 ( x + 1)  − ( 2 + x 3 )      2 x 1) + (  

q. Given

x

) ][ (x )

+ 0 ⋅ x 4 − 4 x 4−1 ⋅ x 2 + 2

−2 x 3/ ( x 2 + 4 )

=

then f ′( x ) =

(1 - x )

3+ x f ( x) = 3 x −5

(x

2 −1

4x 3 - 3x 4

=

2

[ ( 2x f ′( x ) =

−2 x 5 − 8 x 3 x8

=

4 x 5 + 4 x 3 − 2 x 5 − 20 x

p. Given

=

2

f ( x) =

o. Given

=

-5x -6 - 9x -4 + 2x -2

x4 1− x

f ( x) =

n. Given

=

=

x2 + 2 f ( x) = x4

l. Given

=

1.2 Differentiation Rules Using the Prime Notation

(x

−5

)

2

3 −1 3  ( 0 ⋅ x ) − (1 ⋅ 2 )   2 + x 3   ( 0 + 3 x ) ( x + 1)  − (1 + 0 ) ( 2 + x )  2  +      2 x2 ( x + 1)    x + 1     x 

then f ′( x ) = 

=

f ( x ) =( x 2 + 1) ( x + 5 )

Hamilton Education Guides

3

−

2 x2

 2 + x3  2  3x3 + 3x 2 − x3 − 2    +  2  ( x + 1)  x + 1  x 

= −

2  2 + x 3 x 2  x + 1

 2  2 x 3 + 3x 2 − 2  +    x  (x + 1)2   

then f ′( x ) = ( 2 x 2 −1 + 0 ) ( x + 5) + (1 + 0 ) ( x 2 + 1) =

2 x ( x + 5 ) + ( x 2 + 1)

11

)

Basic Differentiation

=

1.2 Differentiation Rules Using the Prime Notation

=

2 x 2 + 10 x + x 2 + 1

3x 2 + 10x + 1

A second method would be to multiply the binomial terms by one another, using the FOIL method, and then taking the derivative of f (x ) as follows: f (x )

(

)

= x 2 + 1 (x + 5) = x 3 + 5 x 2 + x + 5 then f ′( x ) = 3x 3−1 + (5 ⋅ 2) x 2−1 + x1−1 + 0 = 3 x 2 + 10 x + 1

r. Given

f ( x) = ( x + 1)( x + 2 )

then f ′( x ) = (1 + 0 )( x + 2 ) + (1 + 0 )( x + 1) = ( x + 2 ) + ( x + 1) =

2x + 3

An alternative way is by multiplying the terms in parenthesis together and then taking the derivative of the product, i.e., f ( x)

= (x + 1)(x + 2) =

x2 + 2x + x + 2

then

f ′( x)

1   1 s. Given f (x ) = 1 +  1 − 3  then f ′( x ) = x



x 

=

−

1 1  3x 2  1  1 − 3  + 6 1 +  2  x  x  x  x

=

−

1 1 3 3 + + + x 2 x5 x 4 x5

=

=

−

=

=

2 x 2 −1 + 2 ⋅ 1 + 1 + 0

2x + 2 + 1

=

2x + 3

 ( 0 ⋅ x ) − (1 ⋅ 1)  1 − 1  + 0 − ( 0 ⋅ x3 ) − ( 3x3−1 ⋅ 1)  1 + 1  0 +     x2 x3   x6 x     

1 1 3x 2 3x 2 + 5+ 6 + 7 2 x x x x

=

−

1 1 3 x 2/ 3 x 2/ + 5= + + 7/ 5 2 x x x 6/ 4 = x

4 3 1 + x5 x4 x 2

A perhaps simpler way is to write f (x ) in the following form: f (x )

1 



= 1 +  1 −  x 

[

1   x3 

)] [

(

(

)(

(

)]

= 1 + x −1 1 − x −3

= − x − 2 ⋅ 1 − x −3 + 3x −4 ⋅ 1 + x −1 = 4 x −5 + 3x −4 − x −2 =

t.

)

[ ( 0 − x )⋅ (1 − x ) ]+ [ ( 0 + 3x )⋅ (1 + x ) ]

then f ′( x ) =

(

−1−1

)(

= − x −2 + x −2−3 + 3x −4 + 3x −4−1

−3

)

−3−1

−1

= − x −2 + x −5 + 3x −4 + 3x −5

4 3 1 + x5 x4 x 2

 3x 2 + 5  = f ( x)  Given  ( x + 1) a  x 

[(

)][ (

 2−1 + 0 x − 1⋅ 3x 2 + 5  3 ⋅ 2x f x ′ then ( ) =  x2  

) ]  (x + 1) + 1⋅  3x   

 

+ 5   x  

2

 2   2  6 x 2 − 3 x 2 − 5    2    (x + 1) +  3 x + 5  =  3 x − 5 ( x + 1) +  3 x + 5  =  2 2       x  x   x   x 



u. Given

=

f ( x) =

1 + x2 1 − x2

 

then f ′( x ) =

2x − 2x 3 + 2x + 2x 3 4

2

x − 2x + 1

Hamilton Education Guides

=







 



[ ( 0 + 2x )(1 − x ) ]− [ ( 0 − 2x )(1 + x ) ] (1 − x ) 2 −1

2

2 −1

2 2

(2 x + 2 x ) − 2/ x 3/ + 2/ x 3/

(1 − x )

2 2

=

4x

(1 − x ) 2

2

=

2 x (1 − x 2 ) + 2 x (1 + x 2 )

(1 − x )

2 2

2

12

Basic Differentiation

v. Given

f ( x) =

1.2 Differentiation Rules Using the Prime Notation

2 x 2 + 3x + 1 x2 + 1

then f ′( x ) =

[ (4x + 3) (x + 1) ]− [ 2x ( 2x = (x + 1) 2

ax 2 + bx + c a ax 2 − b

f ( x) =

2

− abx 2 − 2abx − 2acx − b 2

x. Given

2

−b

)

2

3− x f ( x) = a 1 −5 x

=

1 3− x − + 2 +5 x x 2 1   − 5 x 

=

5 x3 − 2 x 2 + 3x 25 x 3 − 10 x 2 + x

=

=

=

+ bx + c

=

)]

2−1

(x + 1) 2

2−1

2

2

)] [

)(

)]

)(

+ 0 2 x 2 + 3x + 1

=

(

− 3x 2 + 2x + 3

(x

2

+ b ax 2 − b − 2ax 2−1 ax 2 + bx + c

( ax

2

−b

)

+1

)

2

)]

2

2/ a/ 2/ x/ 3/ − 2abx + abx 2 − b 2 − 2/ a/ 2/ x/ 3/ − 2abx 2 − 2acx

=

( ax

2

−b

)

2

− abx 2 − 2a (b + c ) x − b 2

( ax

then f ′( x ) =

1 3 1 − + 2 − +5 x x x 2 1   − 5 x 

x/ ( 5 x 2 − 2 x + 3)

=

x/ ( 25 x − 10 x + 1) 2

) ] [ ( 2x (x + 1)

)(

+ 3 + 0 x 2 +1 −

4/ x/ 3/ + 4 x + 3 x 2 + 3 − 4/ x/ 3/ − 6 x 2 − 2 x

[ ( 2ax

2

2

( ax

2

2−1

2

then f ′( x ) =

[ (2ax + b) (ax − b) ] − [ 2ax ( ax = ( ax − b) =

)]

+ 3x + 1

2

2

w. Given

2

[ ( 4x

2

−b

)

2

  1     ( 0 ⋅ x ) − (1 ⋅ 1)   ( 3 − x )  −1 ⋅  x − 5   −   x2         1   − 5 x  3 2 − +5 x2 x 1 10 − + 25 x2 x

=

=

2

3x − 2 x 2 5 + x3 1 x − 10 x 2 25 + x3 1

=

=

 1  1  − + 5  + 2 (3 − x )  x  x 2 1   − 5 x 

3x − 2 x 2 + 5 x3 x3 x − 10 x 2 + 25 x 3 x3

5x 2 - 2x + 3 25x 2 - 10x + 1

Example 1.2-2: Find f ′( 0) , f ′( 1) , and f ′(−2) for the following functions. a. f (x ) = (x + 5) x 2

b. f ( x ) = 3x 2 + 1

d. f ( x ) = x −1 ( x + 2)

e. f ( x ) = x 3 + 2 x +

Hamilton Education Guides

c. f ( x ) = x −5 − 2 x −4 − 3x −2 + 1 1 x

f. f ( x ) = x 2 ( x + 1)

13

Basic Differentiation

1.2 Differentiation Rules Using the Prime Notation

x2 + 4

g. f ( x ) =

(

3x + 1

)

i. f (x ) = x 2 + 3 (x − 1)

h. f ( x ) = x 5 − 2 x 2 + 3x + 10

2

Solutions: a. Given f (x ) = (x + 5) x 2 , then f ′( x)

= (1 + 0 ) ⋅ x 2  +  2 x 2 −1 ⋅ ( x + 5) = x 2 + 2 x (x + 5) =

f ′( 0)

= ( 3 ⋅ 02 ) + (10 ⋅ 0 ) =

f ′ (1)

= ( 3 ⋅ 12 ) + (10 ⋅ 1) =

f ′ ( −2 )

=

x 2 + 2 x 2 + 10 x

=

3 x 2 + 10 x

and

0

=

3 + 10

3 ⋅ ( −2 )2  + (10 ⋅ −2 )  

13

= ( 3 ⋅ 4 ) − 20 =

=

-8

6 ⋅1

=

12 − 20

b. Given f ( x) = 3x 2 + 1 , then f ′( x)

= (3 ⋅ 2) x 2−1 + 0 =

f ′( 0)

=

6⋅0

=

and

6x

f ′ (1)

0

=

f ′ ( −2 )

6

=

6 ⋅ −2

=

-12

c. Given f ( x ) = x −5 − 2 x −4 − 3x −2 + 1 , then f ′( x)

= − 5 x −5−1 + (− 2 ⋅ −4) x −4−1 + (− 3 ⋅ −2) x −2−1 + 0 =

f ′( 0)

=

−

f ′ (1)

=

−

f ′ ( −2 )

5 8 6 + + 16 15 13

=

5 0

5 8 6 + + 0 6 05 03

−

5

( −2 )

6

+

8 0

6 0

5 8 6 − + + 1 1 1

=

= − + + =

8

( −2 )

5

+

6

( −2 )

=

3

−

f ′(0)

−5 x −6 + 8 x −5 + 6 x −3

=

−

5 8 6 + + x6 x5 x3

and

is undefined due to division by zero

9 5 8 6 + + 64 −32 −8

=

=

−0.078 − 0.25 − 0.75

-1.078

d. Given f ( x ) = x −1 ( x + 2) , then f ′( x)

=

 − x −1−1 ( x + 2 )  + (1 ⋅ x −1 )

f ′( 0)

=

−

f ′ (1)

=

−

2 02

= −

2 12

=

−

2 0

2 1

f ′(0)

=

Hamilton Education Guides

=

−2 −1  − x ( x + 2 )  + x

=

 x + 2 1 − x2  + x  

=

1 2 1 − − 2+ x x x

=

−

2 x2

and

is undefined due to division by zero

-2

14

Basic Differentiation

=

f ′ ( −2 )

−

2

( −2 )

2

1.2 Differentiation Rules Using the Prime Notation

=

−

2 4

=

-

1 2

1 x

e. Given f ( x ) = x 3 + 2 x + = x 3 + 2 x + x −1 , then f ′( x)

=

f ′( 0)

= 3x 2 + 2 − x −2 = 3x 2 + 2 −

1 x2

= 3 ⋅ 02 + 2 −

1 02

= 2−

f ′ (1)

=

1 x2

=

1 12

=

f ′ ( −2 )

=

3 x 3−1 + 2 x1−1 − x −1−1

3 x 2 + 2 − x −2

=

=

3x 2 + 2 −

=

3 x 2 + 2 − x −2

3 x 2 + 2 − x −2

and

3 ⋅ 12 + 2 −

1

3 ⋅ ( −2 ) + 2 − 2

( −2 )

2

=

1 0

3+ 2−

3⋅ 4 + 2 −

1 4

=

f ′(0) is undefined due to division by zero 1 1

=

3 + 2 −1

=

4

12 + 2 − 0.25

=

13.75

f. Given f ( x ) = x 2 ( x + 1) , then  2 x 2 −1 ( x + 1)  + (1 ⋅ x 2 )

f ′( x)

=

f ′( 0)

= ( 3 ⋅ 02 ) + ( 2 ⋅ 0 ) =

0

f ′ (1)

= ( 3 ⋅ 12 ) + ( 2 ⋅ 1) =

3+ 2

f ′ ( −2 )

=

3 ⋅ ( −2 )2  + ( 2 ⋅ −2 )  

g. Given f ( x ) =

x2 + 4 3x 2 + 1

=

2  2 x ( x + 1)  + x

=

5

= (3 ⋅ 4) − 4 =

=

( 3 x 2 + 1) ⋅ 2 x  − ( x 2 + 4 ) ⋅ 6 x      2 2 x + 3 1 ( )

f ′( 0)

=

−

f ′ (1)

=

−

f ′ ( −2 )

=

22 ⋅ 0

2

22 ⋅ 1

(3 ⋅1

2

−

+ 1)

2

=

−

=

−

22 ⋅ −2 3 ⋅ ( −2 ) + 1   2

Hamilton Education Guides

12 − 4

2x2 + 2x + x2

=

=

3x 2 + 2 x

and

=

6 x 3 + 2 x − 6 x 3 − 24 x

8

, then

f ′( x)

( 3 ⋅ 02 + 1)

=

2

0

( 0 + 1)

2

22

( 3 + 1)

=

−

2

= =

=

−44

(12 + 1)

2

3

+ 2 x ) − ( 6 x 3 + 24 x )

( 3x + 1) 2

0 12

=

−

0 1

22 42

=

−

22 16

=

=

44 169

−

−

(6x

=

44 132

=

2

( 3x 2 + 1)

2

=

−

22 x

( 3x 2 + 1)

2

0

-1.375

=

0.26

15

Basic Differentiation

1.2 Differentiation Rules Using the Prime Notation

h. Given f ( x ) = x 5 − 2 x 2 + 3x + 10 , then f ′( x)

=

5 x 5 −1 − 2 ⋅ 2 x 2 −1 + 3 + 0

f ′( 0)

=

5 ⋅ 04 − 4 ⋅ 0 + 3

f ′ (1)

=

5 ⋅ 14 − 4 ⋅ 1 + 3

f ′ ( −2 )

=

= =

=

=

0−0+3

=

5−4+3

3

4

= (5 ⋅16) + 8 + 3 = 80 + 11 =

5 ⋅ ( −2 ) + ( −4 ⋅ −2 ) + 3 4

(

5x4 − 4x + 3

91

)

i. Given f (x ) = x 2 + 3 (x − 1) , then

[ (

)]

f ′( x)

= [ 2 x ⋅ (x − 1) ] + 1 ⋅ x 2 + 3

f ′( 0)

=

3 ⋅ 02 − 2 ⋅ 0 + 3

f ′ (1)

=

3 ⋅ 12 − 2 ⋅ 1 + 3

f ′ ( −2 )

=

= =

0−0+3

= =

2x2 − 2x + x2 + 3

=

3x 2 − 2 x + 3

3

3−2+3 = 4

=

3 ⋅ ( −2 ) + ( −2 ⋅ −2 ) + 3 2

Example 1.2-3: Given g( x ) =

3 ⋅ 4 + 4 + 3 = 12 + 7

=

19

1 + 1 and h( x ) = x , find f (x ) , f ′( x ) and f ′(0) . x x h( x )

a. f ( x ) = x g( x )

b. f ( x ) = 2 x 2 − 5h( x )

c. f ( x ) = g( x ) +

d. h( x ) = 3x f ( x )

e. h( x ) = 1 − f ( x )

f. 3h( x ) = 2 x f ( x ) − 1

Solutions: a. Given g( x ) =

1 +1 x

and f ( x ) = x g( x ) , then 1

f ( x ) = x g (x ) = x ⋅  + 1 = 1 + x therefore f ′ ( x ) = 1 and f ′ ( 0 ) = 1 x 

b. Given h( x ) = x and f ( x ) = 2 x 2 − 5h( x ) , then f ( x ) = 2 x 2 − 5h(x ) = 2 x 2 − 5 ⋅ x = 2 x 2 − 5 x therefore f ′ ( x ) = (2 ⋅ 2 ) x 2−1 − 5 = 4x - 5 and

= (4 ⋅ 0) − 5 =

f ′( 0)

c. Given g( x ) = f ( x)

=

= g (x ) +

− x −2

=

-

1 x2

-5

1 + 1 , h( x ) = x , x x h( x )

=

and

f ′( 0)

and f ( x ) = g( x ) +

1  x  + 1 + x  x

=

1   + 1 + 1 x 

1 02

=

-

=

−

1 0

=

x , h( x )

then

1 +2 x

= x −1 + 2 therefore

f ′( x)

=

− x −1−1 + 0

which is undefined due to division by zero.

d. Given h( x ) = x and h( x ) = 3x f ( x ) , then

Hamilton Education Guides

16

Basic Differentiation

h( x ) 3x

f ( x) =

1.2 Differentiation Rules Using the Prime Notation

x 3x

=

1 3

=

therefore,

f ′( x)

=

0

and

f ′( 0)

=

0

-1

and

f ′( 0)

=

-1

=

6/ x − 6/ x − 2 [ 2 x ⋅ 3] −  2 ⋅ ( 3x + 1) = / 2/ = 2 ( 2x) 4x

e. Given h( x ) = x and h( x ) = 1 − f ( x ) , then f ( x)

= 1 − h(x ) = 1 − x therefore,

f ′( x)

=

f. Given h( x ) = x and 3h( x ) = 2 x f ( x ) − 1 , then f ( x)

=

-

3h(x ) + 1 2x

=

1 2x 2

and

=

f ′( 0)

=

3x + 1 2x −

1 2 ⋅ 02

therefore, =

-

1 0

f ′( x)

−

2 4x 2

which is undefined due to division by zero.

Section 1.2 Practice Problems - Differentiation Rules Using the Prime Notation 1. Find the derivative of the following functions. Compare your answers with Practice Problem number 1 in Section 1.1. a. f (x ) = x 2 − 1

b. f (x ) = x 3 + 2 x − 1

c. f (x ) =

x x −1

e. f (x ) = 20 x 2 − 3

f. f (x ) = x 3

g. f (x ) =

10

2. Differentiate the following functions: a. f (x ) = x 2 + 10 x + 1

(

d. f (x ) = 2 x 5 + 10 x 4 + 5 x

(

)(

g. f (x ) = x 3 + 1 x 2 − 5 j. f (x ) =

)

)

(

e. f (x ) = a 2 x 3 + b 2 x + c 2

f. f (x ) = x 2 ( x − 1) + 3x

(

)

 2x 3 + 5   p. f (x ) = x 2 − 1   

)

x 5 + 2x 2 −1 3x 2

n. f (x ) = (x + 1) ⋅

x



x



x2 ax + b f (x ) = cx

c. f (x ) = 3x 4 − 2 x 2 + 5

k. f (x ) =

1 m. f (x ) = x 2  2 + 

h.

1

b. f (x ) = x 8 + 3x 2 − 1

h. f (x ) = 3x 2 + x − 1 (x − 1)

x3 +1 x

x−5

d. f (x ) = −

q. f (x ) =

2x x −1

3x 4 + x 2 + 2 x −1

(

)

i. f (x ) = x x 3 + 5 x 2 − 4 x l. f (x ) = o. f (x ) =

x2 (x − 1) + 3x x 3 + 3x − 1 x4

r. f (x ) = x −1 +

1 x −2

3. Compute f ′(x ) at the specified value of x . Compare your answers with the practice problem number 2 in Section 2.1. a. f (x ) = x 3 at x = 1

b. f (x ) = 1+ 2 x at x = 0

c. f (x ) = x 3 + 1 at x = −1

d. f (x ) = x 2 (x + 2) at x = 2

e. f (x ) = x −2 + x −1 + 1 at x = 1

f. f (x ) = x + 2 at x = 10

Hamilton Education Guides

17

Basic Differentiation

1.2 Differentiation Rules Using the Prime Notation

4. Find f ′(0) and f ′(2) for the following functions:

(

)

(

)

a. f (x ) = x 3 − 3x 2 + 5

b. f (x ) = x 3 + 1 (x − 1)

c. f (x ) = x x 2 + 1

d. f (x ) = 2 x 5 + 10 x 4 − 4 x

e. f (x ) = 2 x −2 − 3x −1 + 5 x

f. f (x ) = x −2 x 5 − x 3 + x

g. f (x ) =

x 1+ x 2

1 x

h. f (x ) = + x 3

i. f (x ) =

(

)

ax 2 + bx cx − d

5. Given f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , find h(x ) and h ′(x ) . a. h(x ) = x 3 f (x )

b. f (x ) = 3 + h(x )

c. 2 g (x ) = h(x ) − 1

d. 3h(x ) = 2 x g (x ) − 1

e. 3 [ f (x ) ] 2 − 2h(x ) = 1

f. h(x ) = g (x ) ⋅ 3 f (x )

g. 3h(x ) − f (x ) = 0

h. 2 g (x ) + h(x ) = f (x )

i. f (x ) = x 3 + 5 x 2 + h(x )

j. h(x ) =

k. h(x ) = 2 f (x ) + g (x )

l. [ h(x ) ] 2 − f (x ) = 10

m.

n.

x3 +1 − f (x ) x 2 g (x ) f (x ) = h( x )

Hamilton Education Guides

3 f (x ) 1 = h( x ) x

o. f (x ) =

1 h( x ) + 4

18

Basic Differentiation

1.3

1.3 Differentiation Rules Using the

Differentiation Rules Using the

d dx

d Notation dx

Notation

In the previous section the prime notation was used as a means to show the derivative of a function. For example, derivative of the functions y = f ( x ) = x 2 + 3x + 1 was represented as y ′ = f ′(x ) = 2 x + 3 . However, derivatives can also be represented by what is referred to as the “double-d” notation. For example, the derivative of the function y = f (x ) = x 2 + 3x + 1 can be shown as form:

dy d = f (x ) = 2 x + 3 . dx dx

Following are the differentiation rules in the double-d notation

Rule No. 1 - The derivative of a constant function is equal to zero, i.e., if f ( x ) = k ,

then

d f ( x) = 0 dx

Rule No. 2 - The derivative of the identity function is equal to one, i.e., if f ( x ) = x ,

then

d f ( x) = 1 dx

Rule No. 3 - The derivative of the function f ( x ) = x n is equal to positive or negative integer.

d f ( x ) = n x n −1 , dx

where n is a

Rule No. 4 (scalar rule) - If the function f ( x ) is differentiable at x , then a constant k multiplied by f ( x ) is also differentiable at x , i.e., d d ( k f ) x = k  dx f ( x ) dx  

[

]

Rule No. 5 (summation rule) - If the function f ( x ) and g( x ) are differentiable at x , then their sum is also differentiable at x , i.e., d dx

[( f + g ) x ] =

d d g( x ) f ( x) + dx dx

Rule No. 6 (product rule) - If the function f ( x ) and g( x ) are differentiable at x , then their product is also differentiable at x , i.e., d   dx ( f ⋅ g ) (x )  

d d =  f ( x ) g( x ) +  g( x ) f ( x )  dx



 dx



Rule No. 7 - Using the rules 1 , 4 , 5 , and 6 we can write the formula for differentiating polynomials, i.e., if f ( x ) = a n x n + a n−1 x n−1 + a n−2 x n−2 +  + a3 x 3 + a2 x 2 + a1 x1 + a0 then, d f ( x ) = n a n x n −1 + ( n − 1) a n −1 x n −2 + ( n − 2) a n −2 x n −3 +  + 3a3 x 2 + 2a 2 x + a1 dx

Rule No. 8 (quotient rule) - If the function f ( x ) and g( x ) are differentiable at x , then their quotient is also differentiable at x , i.e.,

Hamilton Education Guides

19

Basic Differentiation

1.3 Differentiation Rules Using the

 d  f      ( x)  dx  g  

=

d Notation dx

d  d   dx f ( x ) g( x ) −  dx g( x ) f ( x )    

[ g( x )]2

Note 1 - depending on the letter used to express the terms of a function, the double-d notation of a derivative is then shown as

( where a is equal to the letter used in the left hand side of the equation) . ( where b is equal to the letter used in the right hand side of the equation)

da db

For example, •

if the function y is represented by f (x ) , i.e., y = f ( x ) = x 2 + 2 x , then its derivative is shown as

(

)

dy d d 2 f ( x) = x + 2x = 2x + 2 . = dx dx dx

•

if the function y is represented by f (t ) , i.e., y = f (t ) = t 3 + 2t 2 + 4 , then its derivative is shown as

•

(

)

dy d d 3 = f (t) = t + 2t 2 + 4 = 3t 2 + 4t . dt dt dt

if the function u is represented by f (v ) , i.e., u = f ( v ) = v 3 + 3v , then its derivative is shown as

(

)

d du d = v 3 + 3v = 3v 2 + 3 . f (v ) = dv dv dv

•

if the function p is represented by f (r ) , i.e., p = f ( r ) = 2r 3 − 2r 2 − 3 , then its derivative is shown as

•

)

if the function y is represented by f (z ) , i.e., y = f ( z ) = z 5 + 3z 2 + 1 , then its derivative is shown as

•

(

dp d d f (r) = = 2r 3 − 2r 2 − 3 = 3r 2 − 4r . dr dr dr

(

)

dy d d 5 = z + 3z 2 + 1 = 5z 4 + 6z . f ( z) = dz dz dz

if the function v is represented by f (x ) , i.e., v = f ( x ) = x 8 + 4 , then its derivative is shown as

(

)

dv d d 8 f ( x) = x + 4 = 8 x 7 , etc. = dx dx dx

In the following examples the above rules are used in order to find the derivative of various functions: Example 1.3-1: Find

dy for the following functions. dx

a. y = x 3 − 2 x 2 + 5 d. y =

3x 2 1+ x

e. y = 5x +

(

g. y = (x + 1) x 2 − 3

)

=

d 3 ( x − 2 x 2 + 5) dx

Hamilton Education Guides

x +1

i. y = 5 +

x − 3   5 

=

d 3 ( x ) + dxd ( −2 x 2 ) + dxd ( 5) dx

=

3 x 2 + ( −2 ⋅ 2 ) x + 0

1 x

)

f. y = x 3 x 2 + 1

k. y = x 2 

Solutions: dy dx

(

2x

2

h. y = 5 x (x + 1)

j. y = x (x + 1) (x − 2)

a.

c. y = x 2 +

b. y = 4 x 5 − 3x 2 − 1

1− x x

l. y = (x + 1) (x − 1)−2

=

3x 2 - 4x

20

Basic Differentiation

1.3 Differentiation Rules Using the

b.

dy dx

=

d ( 4 x5 − 3x 2 − 1) dx

c.

dy dx

=

d  2 1 x +  dx  x

d.

dy dx

=

d  3x 2    dx  1 + x 

e.

dy dx

=

d  2x   5x + 2  dx  x +1

=

( x 2 + 1) ⋅ 2  − [ 2 x ⋅ 2 x ]  5+  2 2 + x ( 1)

dy dx

=

d 3 2  x ( x + 1)   dx 

= ( x 2 + 1) ( x3 ) +  x3 ( x 2 + 1) = ( x 2 + 1) ⋅ 3x 2 + x3 ⋅ 2 x = dx dx

dy dx

=

d 3 2  x ( x + 1)   dx 

=

dy dx

=

d ( x + 1) ( x 2 − 3)   dx 

=

x2 − 3 + 2x2 + 2x

dy dx

=

d ( x + 1) ( x 2 − 3)   dx 

dy dx

=

d [5 x (x + 1)] dx

= ( x + 1) ( 5 x ) + ( 5 x ) ( x + 1) = ( x + 1) ⋅ 5 + [5 x ⋅ 1] =

dy dx

=

d [5 x (x + 1)] dx

=

f.

g.

h.

i.

dy dx

=

(

d x 2 + x −1 dx

=

)

=

d d 2x ( 5 x ) +  2  dx dx  x + 1 

=

=

5+

(x

2



2

=

5+



)=

=

1 x2

2x -

[ ( 1 + x )⋅ 6 x] − [ ( 3x 2 )⋅1] 3x + 6x = 2 (1 + x ) ( 1 + x) 2

2

-2x 2 + 2

(x

2

+ 1)

2

5x 4 + 3x 2



d 5 d 3 x x + dx dx

d

or,

= 5 x 5−1 + 3x 3−1 = 5 x 4 + 3 x 2 d





3x 2 + 2x - 3



(

d dx

(

d 5x 2 + 5x dx



or,

d x 3 − 3x + x 2 − 3 dx

=



Hamilton Education Guides

2x − x −2

= ( x 2 − 3) ( x + 1) + ( x + 1) ( x 2 − 3) = ( x 2 − 3) ⋅ 1 + ( x + 1) ⋅ 2 x  dx dx

=

=

=

=

d



(

=

+ 1)

2x + ( − x −2 )

=

20x 4 - 6x

d  2   d 2  ( x + 1) dx ( 2 x )  −  2 x dx ( x + 1)  d (5x ) + 2 dx ( x 2 + 1)

2x2 + 2 − 4x2

d

d x5 + x3 dx

= (4 ⋅ 5) x 4 + (− 3 ⋅ 2) x + 0 =

d 2 ( x ) + dxd ( x −1 ) dx

d   2 d  2  (1 + x ) dx ( 3 x )  − ( 3 x ) dx (1 + x )  2 (1 + x )

=

d  1− x  5 +  dx  x 

d d d 4 x 5 ) + ( −3 x 2 ) + ( −1) ( dx dx dx

=

d Notation dx



)=

d d 1− x  ( 5) +   dx dx  x 

d dx



)=

d 3 d 2 d (− 3x ) + d (− 3) x + x + dx dx dx dx



d d 5x 2 + 5x dx dx

=

5x + 5 + 5x

= (5 ⋅ 2) x 2−1 + 5 x1−1 =

d  d     x dx (1 − x )  − (1 − x ) dx ( x )  d ( 5) + dx x2

=

=

3x 2 + 2x - 3

=

10x + 5

or,

10x + 5

0+

[ x ⋅ −1] − (1 − x ) ⋅ 1 x2

21

Basic Differentiation

j.

dy dx

1.3 Differentiation Rules Using the 1 x2

=

−x −1 + x x2

=

d [x (x + 1) (x − 2)] dx

=

-

d ( x 2 + x ) ( x − 2 )   dx 

=

= ( x − 2 ) ⋅ ( 2 x + 1) + ( x 2 + x ) ⋅ 1 = dy dx

k.

dy dx

d d 3 d (− 2 x ) x + − x2 + dx dx dx

=

( )

d  2  x − 3  x   dx   5  

=

 2x2 − 6x   2 1   +x ⋅  5 5   

=

d  2  x − 3  x   dx   5  

=

5 ( 3 x 2 − 6 x )  − ( x 3 − 3 x 2 ) ⋅ 0      52

=

d  −2 ( x + 1)( x − 1)  dx 

=

=



] = dxd ( x

3

=

2 x2 − 6 x x2 + 5 5

d  x3 − 3x 2    dx  5 

=

= ( x − 1)−2 

d dx



=



or,

3x 2 - 2x - 2

− 2x 2 + x 2 − 2x



) = dxd ( x

= 3x 3−1 + −2 x 2−1 − 2 x1−1 = 3x 2 − 2 x − 2 x 0 =

 x − 3  d 2   2 d  x − 3    5  dx ( x )  + ( x ) dx  5        

=

= ( x − 1)−2 − 2 ( x + 1)( x − 1)−3 =

dy dx

)

=

dy dx

l.

[(

d  x ( x + 1)( x − 2 )  dx 

d dx

= ( x − 2 ) ( x 2 + x ) + ( x 2 + x ) ( x − 2 )

2x2 + x − 4x − 2 + x2 + x

d x 2 + x (x − 2 ) dx

=

dy dx

d Notation dx

=

2x2 − 6x + x2 5

5 ( 3x 2 − 6 x ) − 0

)

3x 2 - 2x - 2

 x − 3    2  1 ⋅ 5 − 0   5  ⋅ 2 x  + ( x ) ⋅  52        

= =

3x 2 − 6 x 5

5

2

=

5/ ( 3 x 2 − 6 x ) 5

2/ =1

d d −2 ( x + 1) + ( x + 1) ( x − 1)  dx dx   

1

d  x +1    dx  ( x − 1)2 

− x 2 − 2x

=

3 x ( x − 2) 5

or,

 d 3  3  2  2 d 5 dx ( x − 3 x )  − ( x − 3 x ) dx ( 5 )  52

=

( x − 1)

3

2

−

2 ( x + 1)

( x − 1) 3

=

3x 2 − 6 x 5

=

3 x ( x − 2) 5

= ( x − 1)−2 ⋅ 1 + ( x + 1) ⋅ −2 ( x − 1)−3 

or,

d 2 d 2    ( x − 1) dx ( x + 1)  − ( x + 1) dx ( x − 1)  4 ( x − 1)

[ (x −1) ⋅1] = 2

=

d  −2 ( x + 1)( x − 1)  dx 

−

[ (x + 1)⋅ 2 (x − 1) ] 2 ( x + 1) 1 (x − 1)2 − 2(x + 1) (x − 1) (x − 1)2/ − 2(x/ + 1/ ) (x − 1) − = = = 2 ( x − 1) ( x − 1) 3 (x − 1)4/ =2 (x − 1)4/ =3 (x − 1)4 (x − 1)4

Hamilton Education Guides

=

=

(x − 1)4

22

Basic Differentiation

1.3 Differentiation Rules Using the

d Notation dx

Example 1.3-2: Find the derivative of the following functions.

(

)

(

)

a.

d 3x 2 + 5 x − 1 = dx

b.

d 8 x 4 + 3x 2 + x = dx

d.

d  2t 2 + 3t + 1   dt  t3 

e.

g.

d dt

j.

d  3  1   w + 1   dw   w 

m.

 3  − 2t   t    4 

(

=

=

)

=

[( )

d 3 2 t t + 1 ( 3t − 1) dt

]=

[(

]

)

c.

d u3 + 5 (u + 1) = du

d 1 − t 2 (1 + t ) + t = dt

]

f.

d  t 2 + 1   dt  t 2 − 1

h.

u2  d  u   = + du  1 − u 1 + u 

i.

d  s 3 + 3s 2 + 1   =  ds  s3 

k.

d  2x    = dx  1 + 2 x 

l.

d  s2   =  ds  1+ s 2 

n.

d  4u 3 + 2    = du  u 2 

o.

d dx

[( )

=

 x3   2  =  x +1 

Solutions: a.

d ( 3x 2 + 5 x − 1) dx

b.

d (8 x 4 + 3x 2 + x ) dx

c.

d ( u 3 + 5 ) ( u + 1)   du 

= ( u + 1)

=

=

d.

( 4t

=

)(

(

t

6x + 5

32x 3 + 6x + 1

= ( u + 1) ⋅ 3u 2  + ( u 3 + 5) ⋅ 1

4u 3 + 3u 2 + 5

d 3  3d   2 2 t dt ( 2t + 3t + 1)  − ( 2t + 3t + 1) dt ( t )  t6

t 2/ 2t 2 + 6t + 3

d (1 − t 2 ) (1 + t ) + t   dt 

=



=

6 x + 5x0

= (8 ⋅ 4 ) x 4 −1 + ( 3 ⋅ 2 ) x 2 −1 + x1−1 =

d 3 ( u + 5) + ( u 3 + 5) dud ( u + 1) du

6

6/ = 4

= ( 3 ⋅ 2 ) x 2 −1 + 5 x1−1 + 0 =

d (8 x 4 ) + dxd ( 3x 2 ) + dxd ( x ) dx

+ 3t 3 − 6t 4 + 9t 3 + 3t 2 t

= − e.

4

d d d 3 x 2 ) + ( 5 x ) + ( −1) ( dx dx dx

=

3u 3 + 3u 2 + u 3 + 5

d  2t 2 + 3t + 1    dt  t3 

=

=

) = − 2t d dt

=

2

)=

4t 4 + 3t 3 − 6t 4 − 9t 3 − 3t 2 t6

Hamilton Education Guides

=

t 3 ⋅ ( 4t + 3)  − ( 2t 2 + 3t + 1) ⋅ 3t 2    t6

−2t 4 − 6t 3 − 3t 2 t6

=

(

− t 2 2t 2 + 6t + 3 t

6

)

+ 6t + 3 t4

[ (1 − t ) (1 + t ) ]+ dtd t =  (1 + t ) dtd (1 − t ) + ( 1 − t ) dtd ( 1 + t )  + 1 2

2

{(1 + t ) ⋅ −2t  + (1 − t ) ⋅1} + 1 = ( −2t − 2t 2

=

2

+1− t2 ) +1

=

−2t − 2t 2 + 1 − t 2 + 1

2

=

-3t 2 - 2t + 2

23

Basic Differentiation

f.

d  t2 +1   dt  t 2 − 1 

=

g.

d 2 d 2  2   2  ( t − 1) dt ( t + 1)  − ( t + 1) dt ( t − 1)  2 ( t 2 − 1)

=

2t 3 − 2t − 2t 3 − 2t

(t

2

− 1)

= −

=

2

d  3  2t  t  −  dt   4 

(t

[ ( t − 1)⋅ 2t ]− [ ( t + 1)⋅ 2t ] ( t − 1)

=

  d d  4 (− 2t ) − (− 2t ) (4)   − 2t  2   3  dt dt   ⋅ 3t  + t   42  4        

2

2

- 1)

h.

d  2t 4 − dt  4

=

d  u u2  +   du  1 − u 1 + u 

= −

6t 3  3  − 8  + t   4   16 

   

=

 / 4 d  2t − dt  4/  2

d  u  d  u2    +  du  1 − u  du  1 + u 

=

[(1 − u )⋅1] − [u ⋅ −1] + [(1 + u )⋅ 2u ] − [ u = (1 − u )2 (1 + u )2 i.

d  s 3 + 3s 2 + 1    ds  s3 

=

( 3s

5

=

)(

= −

+ 6 s 4 − 3s 5 + 9 s 4 + 3s 2 s

(

)

j.

d  3  1   w + 1   dw   w 

k.

d  2x    dx  1 + 2 x 

=

2

=

]

⋅1

    

= −

1 d 4 t 2 dt

=

=

1− u + u

(1 − u )

− 2t ) − ( 2t 3 + 2t )

(t

2

− 1)

2

5

d ( w3 + 1) w−1   dw 

=

6t 3 t 3 − 12t 3 − 4t 3 16t 3 − = = − = − 2t 3 4 2 8 8  3  2t  t  −    4 

d dt

as follows:

1 2

= − ⋅ 4t 4−1 = −

2

+

2u + u 2 − u 2

(1 + u )

4

5

4

6

d ( w3−1 + w−1 ) dw

d d     (1 + 2 x ) dx ( 2 x )  − ( 2 x ) dx (1 + 2 x )  2 (1 + 2 x )

=

=

2

=

[ s ⋅ ( 3s = 3

) = 3s + 6s − 3s − 9s − 3s s

Hamilton Education Guides

3

4t 3 = − 2t 3 2

d   d d 2  2 d     (1 − u ) du u  − u du (1 − u )  (1 + u ) du u  − u du (1 + u )  + 2 2 (1 + u ) (1 − u )

d 3  3 d 3   3 2 2  s ds ( s + 3s + 1)  − ( s + 3s + 1) ds s  s6

6

2

2

( 2t

2

6t 3  3  [4 ⋅ −2] − [(− 2t ) ⋅ 0]  + t   16 4   

 3  2t  t  −    4 

=

-4t 2

 − 2t  d 3   3 d  − 2t   t  + t     4  dt   dt  4 

=

=

A second way of solving this problem is to simplify d dt

d Notation dx

1.3 Differentiation Rules Using the

2

=

2

1

(1 - u )

2

+

2u

(1 + u )

2

)] [(

) ]

+ 6 s − s 3 + 3s 2 + 1 ⋅ 3 s 2 s6

−3s 4 − 3s 2 s6

d 2 d −1 w + w dw dw

=

=

−3s 2 ( s 2 + 1) s

6/ =4

2w − w−2

=

=

2w -

(

− 3 s2 +1 s

)

1 w2

[(1 + 2 x )⋅ 2] − [2 x ⋅ 2] 2 2 + 4x − 4x = = 2 1 + ( 2x ) (1 + 2 x ) (1 + 2 x ) 2

4

2

24

Basic Differentiation

l. m.

d  s2    ds  1 + s 2 

1.3 Differentiation Rules Using the

  2 d 2 d 2 2  (1 + s ) ds s  −  s ds (1 + s )  2 (1 + s 2 )

=

d 3 2 t ( t + 1) ( 3t − 1)   dt 

d ( t 5 + t 3 ) ( 3t − 1)   dt 

=

(1 + s 2 ) ⋅ 2 s  −  s 2 ⋅ 2 s      2 2 (1 + s )

=

=

2s + 2s 3 − 2s 3

=

(1 + s )

2 2

d

d Notation dx

2s

(1 + s )

2 2

d

= ( 3t − 1) ( t 5 + t 3 ) + ( t 5 + t 3 ) ( 3t − 1) dt dt 







= ( 3t − 1) ⋅ ( 5t 4 + 3t 2 ) + ( t 5 + t 3 ) ⋅ 3 = (15t 5 + 9t 3 − 5t 4 − 3t 2 ) + ( 3t 5 + 3t 3 ) = 18t 5 − 5t 4 + 12t 3 − 3t 2 Another way of solving this problem is by multiplication of the binomial terms using the FOIL method prior to taking the derivative of the function as follows. d 3 2 t ( t + 1) ( 3t − 1)   dt 

n.

d  4u 3 + 2    du  u 2 

=

4u 4 − 4u u4

)]

4u/ ( u 3 − 1) u

=

4/ = 3

(

4 ( u 3 − 1) u

3

)

(

d 3t 6 − t 5 + 3t 4 − t 3 dt

=

d 2  2 d   3 3 u du ( 4u + 2 )  − ( 4u + 2 ) du u  u4

= =

[ (

d 3 3 2 t 3t − t + 3t − 1 dt

=

=

=

x 4 + 3x 2

( x + 1) 2

2

=

(

x2 x2 + 3

(x

2

+1

)

2

⋅ 12u 2 ) − ( 4u 3 + 2 ) ⋅ 2u u

= 4

u

)

d 3  3 d  2  x +1 x  − x x 2 +1  3    du   du d x   o. =   = 2 dx  x 2 +1  2 x +1

(

2

 u3

 u 3 −1   3   u 

= 4

(

(u

)

) = 18t

3

−

=

4

1  u 3 



= 4  1 − 

[ ( x + 1)⋅ 3x ]− ( x ( x + 1) 2

2

2

3

5

− 5t 4 + 12t 3 − 3t 2

12u 4 − 8u 4 − 4u u4

1   u3 

⋅ 2x

)

3x 4 + 3x 2 − 2 x 4

=

2

( x + 1) 2

2

)

Example 1.3-3: Find the derivative of the following functions at the specified value. a.

d (t + 1)(t − 2) + 3t dt

[

 x −1 d  c.   dx  ( x + 1)(2 x + 1) 

e.

d  s 2 + 3s    ds  s 2 + 1 

]

d  u 2 + 1   du  u3 − 1 

at t = 1

b.

at x = 0

 3  d  x + 1 ( x − 1)  d.  dx  2x 2  

(

at s = −1

f.

at u = 2

)

d  z 2 + 3z − 5    z dz  

at x = 2

at z = 2

Solutions: a.

d ( t + 1)( t − 2 ) + 3t  dt 

=

Hamilton Education Guides

d d ( t + 1)( t − 2 )  + ( 3t ) dt  dt

d dt

d dt

d dt

= ( t − 2 ) ( t + 1) + ( t + 1) ( t − 2 ) + ( 3t ) 



25

Basic Differentiation

1.3 Differentiation Rules Using the

= ( t − 2 ) ⋅ 1 + ( t + 1) ⋅ 1 + 3 = ( t − 2 ) + ( t + 1) + 3 = Therefore, at t = 1

b.

d  u2 + 1    du  u 3 − 1 

=

(u

=

− 1)

2

Therefore, at u = 2

c.

 d  x −1   dx  ( x + 1)( 2 x + 1) 

=

=

=

(u

3

− 1)

d  u2 + 1    du  u 3 − 1 

2

2t + 2

= ( 2 ⋅ 1) + 2 =

+ 3 x + 1)

2

3 d  ( x + 1) ( x − 1)    dx  2x2  

=

( ) ( u − 1)

(2

− 1)

3

2

2 ⋅ 16 72

= −

= −

32 49

= -0.653

( 2x

2

+ 3 x + 1) − ( 4 x 2 + 3 x − 4 x − 3)

( 2x

2

+ 3 x + 1)

=

2

( 2x

2

+ 3 x + 1) − ( 4 x 2 − x − 3)

( 2x

2

+ 3 x + 1)

2

-2x 2 + 4x + 4

( 2x

2

+ 3x + 1)

 d  x −1   dx  ( x + 1)( 2 x + 1) 

2

=

 2 x 2 ( 4 x 3 − 3 x 2 + 1)  − ( x 4 − x 3 + x − 1) ⋅ 4 x      4x4

=

8 x5 − 6 x 4 + 2 x 2 − 4 x5 + 4 x 4 − 4 x 2 + 4 x 4x4

=

−2 x 2 + 4 x + 4

( 2x

2

+ 3 x + 1)

=

2

=

( −2 ⋅ 0 ) + ( 4 ⋅ 0 ) + 4 = ( 2 ⋅ 0 ) + [3 ⋅ 0] + 1   2

2

2

4 1

=

4

d  2 d 4   4 3 3 2  ( 2 x ) dx ( x − x + x − 1)  − ( x − x + x − 1) dx ( 2 x )  2 ( 2x2 )

(8 x

5

− 6 x 4 + 2 x 2 ) − ( 4 x5 − 4 x 4 + 4 x 2 − 4 x ) 4x4

4 x5 − 2 x 4 − 2 x 2 + 4 x 4x4

3 d  ( x + 1) ( x − 1)    2 x2 dx   

Hamilton Education Guides

2 ⋅  23 + ( 3 ⋅ 2 ) + 2 

d d  2    2 ( 2 x + 3 x + 1) dx ( x − 1)  − ( x − 1) dx ( 2 x + 3 x + 1)  2 ( 2 x 2 + 3x + 1)

=

=

)

= −

2

3

=

Therefore, at x = 2

( u − 1)

2

2

3

2

3

4

2

3

u u 3 + 3u + 2

= −

d  x 4 − x3 + x − 1   dx  2x2 

=

(

=

2+2

[ ( u − 1)⋅ 2u ]− [ ( u + 1)⋅ 3u ] = ( u − 1)

u u 3 + 3u + 2

= −

2

d  x −1  dx  2 x 2 + 3 x + 1 

2 x 2 + 3x + 1 − 4 x 2 + x + 3

Therefore, at x = 0

d.

−u 4 − 3u 2 − 2u

=

( 2 x 2 + 3 x + 1) ⋅ 1 − ( x − 1) ⋅ ( 4 x + 3)    2 2 ( 2 x + 3x + 1)

( 2x

=

= 2t + 2

t − 2 + t +1+ 3

d 2 d 3  3   2  ( u − 1) du ( u + 1)  − ( u + 1) du ( u − 1)  2 ( u 3 − 1)

2u 4 − 2u − 3u 4 − 3u 2 3

d ( t + 1)( t − 2 ) + 3t  dt 

d Notation dx

=

2 x 4 − x3 − x + 2 2 x3

2 x/ ( 2 x 4 − x 3 − x + 2 ) 4x

=

4/ = 3

(2 ⋅ 2 ) − 2 4

3

2⋅2

3

−2+2

=

2 x 4 − x3 − x + 2 2 x3

=

24 16

=

1.5

26

Basic Differentiation

e.

d  s 2 + 3s    ds  s 2 + 1 

=

( 2s

3

1.3 Differentiation Rules Using the

d 2 d 2  2   2  ( s + 1) ds ( s + 3s )  − ( s + 3s ) ds ( s + 1)  2 ( s 2 + 1)

=

+ 3s 2 + 2 s + 3) − ( 2 s 3 + 6 s 2 )

(s

+ 1)

2

2

d  s 2 + 3s    ds  s 2 + 1 

Therefore, at s = −1

f.

d  z 2 + 3z − 5    dz  z 

=

=

=

2

+ 1)

−3s 2 + 2 s + 3

=

(s

2

+ 1)

2

=

=

(s

2

+ 1)

2

 −3 ⋅ ( −1)2  + ( 2 ⋅ −1) + 3   2 ( −12 ) + 1  

=

−3 − 2 + 3 22

=

 z ⋅ ( 2 z + 3)  − ( z 2 + 3 z − 5 ) ⋅ 1 z2

=

9 4

=

-

1 2

z2 + 5 z2

d  z 2 + 3z − 5    dz  z 

Therefore, at z = 2

−3s 2 + 2 s + 3

=

2

d  d 2   2   z dz ( z + 3 z − 5 )  − ( z + 3 z − 5 ) dz ( z )  z2

2 z 2 + 3z − z 2 − 3z + 5 z2

 2   2  ( s + 1) ( 2 s + 3)  − ( s + 3s ) ⋅ 2 s  2 ( s 2 + 1)

=

2 s 3 + 3s 2 + 2 s + 3 − 2 s 3 − 6 s 2

(s

d Notation dx

=

z2 + 5 z2

=

22 + 5 22

= 2.25

Example 1.3-4: Given the functions below, find their derivatives at the specified value. x 2 + 2x − 1

a.

dy dx

c.

dv , given v = x 3 + 1 3x 2 + 5 dx

, given y =

x3

(

u2 dv , given v = at u = 4 du 1− u du 3x d. , given u = at x = 2 dx ( x − 1) 2

at x = 2

)(

)

b.

at x = 5

Solutions: a.

dy dx

=

=

d  x2 + 2x − 1    dx  x3 

2 x 4 + 2 x3 − ( 3x 4 + 6 x3 − 3x 2 ) x

6

Therefore, at x = 2

b.

=

d 3   3 d 2   2  x dx ( x + 2 x − 1)  − ( x + 2 x − 1) dx ( x )  2 ( x3 )

dv du

=

d  u2    du  1 − u 

=

Therefore, at u = 4

dy dx

=

=

2 x 4 + 2 x3 − 3x 4 − 6 x3 + 3x 2 x6

d  x2 + 2x − 1    dx  x3 

=

Hamilton Education Guides

=

d  u2    du  1 − u 

=

−u 2 + 2u

(1 − u )

2

 x 3 ( 2 x + 2 )  − ( x 2 + 2 x − 1)( 3 x 2 )    x6

− x 4 − 4 x3 + 3x 2 x6

− x 4 − 4 x3 + 3x 2 x6

d 2   2 d   (1 − u ) du ( u )  − u du (1 − u )  2 (1 − u ) dv du

=

=

−24 − 4 ⋅ 23 + 3 ⋅ 22 26

=

[(1 − u )⋅ 2u ] − [u 2 ⋅ −1] = = (1 − u )2 =

− (4 )2 + (2 ⋅ 4 )

(1 − 4)

2

=

−16 + 8

(− 3)

2

=

−36 64

= −0.56

2u − 2u 2 + u 2

(1 − u )

=

−8 9

2

=

−u 2 + 2u

(1 − u )

2

= −0.889

27

Basic Differentiation

c.

dv dx

=

d ( x 3 + 1)( 3 x 2 + 5 )   dx 

=

du dx

=

=

9 x 4 + 15 x 2 + 6 x 4 + 6 x dv dx

Thus, at x = 5

d.

1.3 Differentiation Rules Using the

d  3x    dx  ( x − 1)2 

=

(

=

=

dv

= ( 3x 2 + 5) ( x3 + 1) + ( x3 + 1) ( 3x 2 + 5) = ( 3x 2 + 5) ⋅ 3x 2 + ( x3 + 1) ⋅ 6 x dx dx

15 x 4 + 15 x 2 + 6 x

d  x2 + 2x − 1    dx  x3 

=

)

(x − 1) 4 du dx

=

=

=

15 x 4 + 15 x 2 + 6 x

2 d 2    d ( x − 1) du ( 3 x )  − 3 x du ( x − 1)  4 ( x − 1)

3 x 2 − 2x +1 − 6x 2 + 6x

Therefore, at x = 2

dv

3x 2 − 6 x + 3 − 6 x 2 + 6 x

( x − 1) d  3x    dx  ( x − 1)2 

=

− 3x 2 + 3

(x − 1)

4

dy dx

x2

3 ( x − 1)2 − 6 x ( x − 1)

=

( x − 1)4

(x − 1)4

=

e. y = 4 x 2 +

1− x3

(

)

(x − 1)(x + 3) x

(− 3 ⋅ 2 )+ 3 2

(2 − 1)

4

=

−12 + 3 1

d dx

Notation

= −

9 1

= −9

c. y = x 3 −

1 x −1

f. y =

h. y = x 2 (x + 3)(x − 1) x −1    3 

k. y = x

2

x  x −3   1 +  x  5 

n. y = x 3 1 +

ax 2 + bx + c bx

q. y =

m. y =  p. y =

= 9780

− 3x 2 + 3

b. y = 3x 2 + 5

g. y = x 3 x 2 + 5 x − 2 j. y =

9375 + 375 + 30

for the following functions:

a. y = x 5 + 3x 2 + 1 d. y =

=

( x − 1)2 ⋅ 3 − 3 x ⋅ 2 ( x − 1)      4 ( x − 1)

=

=

4

15 ⋅ 54 + 15 ⋅ 52 + 6 ⋅ 5

Section 1.3 - Differentiation Rules Using the 1. Find

d Notation dx



1 x

x 2 + 2x x3 +1

i. y = 5 x −

1 x3

l. y = x 2 (x + 3)−1

1   x −1 

x3 − 2

1 2x −1  o. y =   x  3x + 1 

r. y =

4

x −3

5x

(1 + x ) 2

2. Find the derivative of the following functions: a.

d 3t 2 + 5t dt

(

)=

b.

d 6 x 3 + 5x − 2 dx

d.

d  t 2 + 2t  = dt  5 

e.

d  s 3 + 3s − 1  =  ds  s2 

g.

d 2 t (t + 1) t 2 − 3 dt

h.

d (x + 1) x 2 + 5 dx

[

(

)] =

Hamilton Education Guides

(

[

(

)=

)] =

(

)=

c.

d u 3 + 2u 2 + 5 du

f.

d  3 w 2  = w + dw  1 + w 

i.

d  u 2 u  = − du  1 − u 1 + u 

28

Basic Differentiation

j.

1.3 Differentiation Rules Using the

d  3r 3 − 2r 2 + 1  =  r dr  

k.

d  3s 2 1 − ds  s 3 + 1 s 2

  =  

l.

d Notation dx

d  u 3 u + 1  = − du  1 − u u 2 

3. Find the derivative of the following functions at the specified value.

(

)

[ ( x + 1) ( x − 1) ]

b.

d dx

d  t 2 + 1  at t = −1 dt  t − 1 

e.

d  u3    du  ( u + 1) 2 

[ ( v + 1)v ] at

h.

d  x 3  at x = 0 dx  x 2 +1 

a.

d 3 x + 3x 2 + 1 dx

d. g.

d dv

2

at x = 2

3

v = −2

2

at x = 1

at u = 1

[

] at s = 0  w ( w + 1)  at w = 2 

c.

d 3s 2 (s − 1) ds

f.

d dw 

i.

d  3  u 2   u du   1 − u   

2

3w 2



at u = 0

4. Given the functions below, find their derivatives at the specified value.

(

)

a.

ds , given s = t 2 − 1 + (3t + 2) 2 dt

c.

2 dw , given w = x 2 + 1 + 3x dx

(

Hamilton Education Guides

)

at t = 2

b.

t 3 + 3t 2 + 1 dy , given y = dt 2t

at x = −1

d.

dy , given y = x 2 x 3 + 2 x + 1 dx

(

at t = 1

)

2

+ 3x

at x = 0

29

Basic Differentiation

1.4

1.4 The Chain Rule

The Chain Rule

The chain rule is used for finding the derivative of the composition of functions. In general, the chain rule for two and three differentiable functions are defined in the following way: a. The chain rule for two differentiable functions f ( x ) and g( x ) is defined as:

( f  g )′ (x ) =

d { f [g (x )]} dx

= f ′[ g( x ) ] ⋅ g ′( x )

b. The chain rule for three differentiable functions f ( x ) , g( x ) and h( x ) is defined as:

( f  g  h) ′ ( x ) =

d [ f {g [h(x )]}] dx

= f ′{g [h(x )]}⋅g ′[h(x )]⋅ h ′(x )

The derivative of four or higher differentiable functions using the chain rule involves addition of additional link(s) to the chain. Note that the pattern in finding the derivative of higher order functions is similar to obtaining the derivative of two or three functions, given that they are differentiable. One of the most common applications of the chain rule is in taking the derivative of functions that are raised to a power. In general, the derivative of a function to the power of n is defined as: d n n−1 d f ( x) = n f ( x) ⋅ f ( x) = n f ( x) dx dx

[

]

[

]

[

]n−1 ⋅ f ′( x)

[ ]

n

which means, the derivative of a function raised to an exponent, f ' (x ) , is equal to the exponent times the function raised to the exponent reduced by one, n [ f (x )] n −1 , multiplied by the derivative of the function, f ' (x ) , i.e., n [ f (x )] n −1 ⋅ f ' (x ) . Note that the key in using the chain rule is that we always take the derivative of the functions by working our way from outside toward inside. The following examples show in detail the use of chain rule in differentiating different types of functions. Students are encouraged to spend adequate time working these examples. Example 1.4-1: Find the derivative of the following functions. (It is not necessary to simplify the answer to its lowest level. The objective is to learn how to differentiate using the chain rule.)

(

d. f ( x ) = x 3 − x 5

(

g. f ( x ) = x + x 3 j.

 1  f (t ) =   1+ t2 

(

)

)

 1  + x  x2 

8

e. f ( x ) = 

 1   1+ x2 

4

h. f ( x ) = 

3

m. g(u) = u3 + 3u 2

(

b. f ( x ) = (1 + x ) 6

a. f ( x ) = (3 − 5x ) −2

 θ2   k. r(θ ) =  1+ θ 

)

3

Hamilton Education Guides

 t3

c. f ( x ) = 1 + 2 x 2 3



2

x

 a x + b  cx −d

 r2 + r  l. p(r ) =   1+ r  2

(



o. s(t ) =  1 + t 3 

3

3

i. f ( x ) = 

3

 + t 2   t 4 −1 

n. h( t ) = 

1 f. f ( x ) = 1 + 

)

)

2

3

−1  3

 

30

Basic Differentiation

 x2

p. f ( x ) = 

 3

−

1.4 The Chain Rule −1

2x   5 

( 

)

2

( 

)

3



 θ   1+ θ 2 

q. r(θ ) =  

s. f ( x ) =  x 3 + 2 x − x 2  

4

(  



5

(  

w. f ( x ) =  2 − x −1



Solutions: −2 ( 3 − 5 x ) then

b. Given

f ( x= )

6 (1 + x ) then

c. Given

f ( x= )

(1 + 2 x ) then f ′ ( x ) =

d. Given

f (= x)

(x

e. Given

 1  f (= x)  2 + x  x 

f. Given

1 f ( x= ) 1 +  x 

g. Given

f ( x= )

h. Given

 1  f ( x) =  2  1+ x 

=

− x5 )

8

3

3

then

f ′( x)

(

3 1 + 2x 2

3−1

(

(1 + x )⋅ (1 + x )

2 2

=

f ′( x)

8−1

(

−4 x

(1 + x )

2 1+ 2

  2  1+ x  

1

=

(

) ⋅ (− 2 x 3−1

− 2 −1

3

(

= 12 x 1 + 2 x 2

− x5

) ( 3x 7

) = 3 (x

+1

−2

) ⋅ ( 0 − x ) = -3x 3−1

−1−1

-2

2

+x

− 5x 4

)

)

) (− 2 x 2

(1 + x )

-1 2

2

−3

+1

)

a

) (1 + 3 x ) 3

2

)]

  2  0 ⋅ 1 + x − [2 x ⋅1] ⋅  2   1+ x 2

(

)

−2 x  2  ⋅  2 2   1 + x  (1 + x 2 )

=

(1 + x )

2 3

  2  1+ x  

Hamilton Education Guides

[ (

) = 8( x

⋅ 4x

-4x

A second method is to rewrite f (x ) =  f ′( x)

− 5 x 5−1

(

4 x + x3

2 −1

2 2

3 1 + x −1

−1 3

= 2 

3−1

= 3 x −2 + x

(1 + x ) then f ′ ( x ) =

f ′( x)

derivative of f ( x ) . Hence,

−3

−1

2 −1

) ⋅ ( 3x

3 4

then

)

)

( x + x ) then f ′ ( x ) = ( f  g )′ (x ) = 2

 + 2x 3  

(

u. f ( x ) = 2 − x −1

) ⋅ ( 0 + 4 x ) = 3 (1 + 2 x )

= 8 x3 − x5

3

f ( x= )

3

)

(

= ( x −2 + x ) then

=

−3

 + x 

= 6 ( 1 + x ) 6−1 ⋅ 0 + x1−1 = 6 ( 1 + x ) 5 ⋅1 = 6 ( 1 + x ) 5

2 3

2 ⋅ −2 x

2

f ′( x)

2

(

f ( x= )

f ′( x)

)

)

4

= − 2 ( 3 − 5 x )−2−1 ⋅ 0 − 5 x1−1 = − 2 ( 3 − 5 x )−3 ⋅ (− 5) = 10 ( 3 − 5 x )−3

a. Given

3

 t3   1+ t2 

r. f (t ) = 

t. f ( x ) =  x −1 + x −3



v. f ( x ) =  1 + 2 x 2 − x −2 

−1

(

1

= − 2 1+ x 2

)

2

= (1 + x 2 )

−2 −1

−1 2

⋅ ( 0 + 2x)

 

= (1 + x 2 ) and then take the −2

(

= − 2 1+ x 2

)

−3

⋅ ( 2x)

=

-4x

(1 + x 2 )

3

31

Basic Differentiation

1.4 The Chain Rule

ax+b f ( x) =   cx−d 

i. Given

2

then

 ax + b 

 = 2   cx − d 

f ′( x)

 a x + b   (acx − ad ) − (acx + bc )    (cx − d )2

2 (a x + b )(− ad − bc )

(c x − d )(cx − d )

 1  f (t ) =  2  1+ t 

j. Given

=

3

−2t

⋅

(1 + t ) (1 + t ) 2 2

=

2

2 2

3

 a x + b   acx − ad − acx − bc   

− 2 (a x + b )(ad + bc )

(c x − d )

then

 1 3  1+ t 2

=

3 ⋅ −2t

=

(1 + t ) ⋅ (1 + t ) 2 2

2 2

  

(cx − d )3

3−1

(

(1 + t )



  2  1+ t 

derivative of f ( t ) . Hence,

k. Given

=

3θ 4

⋅

θ 2 + 2θ

(1 + θ )2 (1 + θ )2

l. Given

=

 θ2  r (θ ) =   1+θ 

3

=

 r2 + r  p(r ) =    1+ r 

 r2 +r   3  (1+ r)  

2

then

3

f ′(t )

r ′ (θ )

1

(

= − 3 1+ t 2

=

 θ2 3  1+θ 

p′ ( r )

 r2 +r   3  1+ r   

=

 2r + 2r 2 + 1 + r − r 2 − r      ( 1 + r )2

Hamilton Education Guides

=

3

)

−3−1

3−1

)

(1 + θ )2 ⋅ (1 + θ )2 then

   

(

3θ 4 ⋅ θ 2 + 2θ

(

2 2+ 2

A second method is to rewrite f (t ) = 

)

  0 ⋅ 1 + t 2 − 2t ⋅1  ⋅ 2   1+ t 2  

−6t

=

 a x + b   − ad − bc 

  = 2    c x − d   (cx − d )2 

− 2(ax + b ) (ad + bc )

=

1+ 2

f ′(t )

 [a ⋅ (cx − d )] − [c ⋅ (ax + b )] ⋅  (cx − d )2  

/ //   / / / = 2  (cx − d )2  c x − d  

  = 2   c x − d  

=

2 −1

=

=

)

−1 3

 

= (1 + t 2 ) and then take the −3

(

= − 3 1+ t 2

)

−4



3θ 4+1 (θ + 2 )

(1 + θ )4

[(

(1 + t )

2 4

2

 2θ + 2θ 2 − θ 2     (1 + θ )2 

3θ 5 (θ + 2 )

(1 + θ )4

 [( 2r + 1) ( 1 + r )] − 1 ⋅ r 2 + r ⋅  ( 1 + r )2

2  r (1/ + r )   r 2 + 2r + 1 

2

  1+θ   

=

-6t

=

2t

[ ] = 3  θ

(1 + θ )2+ 2

)

2 4

⋅ ( 0 + 2t )

=

(

  2 

(1 + t )

= (1 + t 2 )

3θ 4 ⋅θ (θ + 2 )

2   − 2t    1 + t 2

-6t

 [2θ ⋅ (1 + θ )] − 1 ⋅θ 2 ⋅  (1 + θ )2

3−1

 1 3  1+ t 2

)] 

 (1/ + r )2/ 

/ /  = 3r 2  = 3  =    (1/ + r/ )2/   (1/ + r/ )   ( 1 + r )2 

3r 2 ⋅ 1

=

3r 2

32

Basic Differentiation

1.4 The Chain Rule

A simpler way is to note that p(r ) = u ) ( u 3 + 3u 2 ) then m. Given g ( = 3

 t3  + t2   4 1 t −  

h (t ) n. Given=

2

then

 r2 +r     1+ r   

3

=

(

 r (1 + r )     (1 + r ) 

) ⋅ [ 3u 3−1

g′(u )

= 3 u 3 + 3u 2

h′ ( t )

= 2  4 + t 2   t −1 

 t3



2 −1

2

3

 r (1/ + r/ )     (1/ + r/ ) 

=

+ ( 3 ⋅ 2) u

[(

3

=

] = 3(u

3

r3

then

+ 3u 2

) ][ ( ( )

p′ ( r )

) ( 3u 2

2

=

+ 6u

3r 2

)

)]

 4  2 3 3  t − 1 ⋅ 3t − t ⋅ 4t − 0  ⋅ + 2t  2 4  t −1  



  − t 6 − 3t 2  t3   3t 6 − 3t 2 − 4t 6  t3 + 2t  +t2   + 2t  = 2  = 2  4 + t 2   2  4  t4 −1 2    4 t − 1    t −1   t −1 

(



( 

o. Given s(t ) =  1 + t 3 =

−9t 2 (1 + t 3 )

−4

=

-

)

−1 

−1

−2

 t3  f (t ) =  2  1+ t 

⋅

3t 2 + t 4

(1 + t ) (1 + t ) 2 3

2 2

(

then s′ ( t ) = − 3 1 + t 3

)

(

−3−1

⋅ 0 + 3t 2

) = −3(1 + t )

3 −4

⋅ 3t 2

3 4

1−θ 2  θ  − ⋅ 2 2   1 + θ  (1 + θ 2 )

4t 9

−3

(1 + t )

θ  2  1+θ 

=

)

)

9t 2

q. Given r (θ ) = 

r. Given

(

3

(



3  = 1 + t

 x2 2x  f (= x)  −  5   3

p. Given

=

)

4

=

−1

then

then

=

=

r ′ (θ )

 x2 2x  − −  5   3

 θ  − 2  1+θ 

2

1+θ 2  1−θ 2 −  ⋅ 2  θ  (1 + θ 2 )

then

4t 9

f ′(t )

⋅

=

2 3

=

 t3  4 2  1+ t 

t 2 (3 + t 2 )

(1 + t ) (1 + t )

Hamilton Education Guides

=

f ′( x)

2 2

=

4 −1

−1−1

−1−1

2 2 ⋅ x −  5 3

1 ⋅ (1 + θ 2 ) − 2θ ⋅ θ   ⋅ 2 2   1 θ + ( )  

(1 + θ )

2 2/

−

−2

 x2 2x   2 2 − −   x −  3 5 3 5    

=

θ

2

⋅

1−θ 2

=

(1 + θ )

2 2/

 3t 2 ⋅ (1 + t 2 ) − 2t ⋅ t 3   ⋅ 2 2   + 1 t ) (  

(

4t 9+ 2 3 + t 2

(1 + t )

2 3+ 2

)=

=

=

−

−2  2 2  θ  1 + θ − 2θ  − 2 2   1 + θ   (1 + θ 2 )   

1−θ 2

θ

2

=

 t3  4 2  1+ t 

3

θ2 - 1 θ2  2 4 4  3t + 3t − 2t   (1 + t 2 )2   

4t11 ( 3 + t 2 )

(1 + t )

2 5

33

Basic Differentiation

f ( x ) = ( x 3 + 2 x ) − x 2   

s. Given

2

(

 = 4  x 3 + 2x 

)

2

(

 = 3  x −1 + x − 3 

u. Given -

f ( x= )

)

2

 + x 

2

4

then

{[ 2( x

3

f ′( x)

(

= 4  x 3 + 2 x 

 − x2  

4−1

(

 ⋅  2 x 3 + 2 x 

) ⋅ ( 3x

2

) (− x

−2

2−1

)

  + 2  − 2 x  

)] }

)(

(

then f ′ ( x ) = 3  x −1 + x −3 

{[2 ( x

2

+ 2x 3x 2 + 2 − 2x



3

)

−1

+ x −3

) (− x

−2

 + x 

3−1

(

 −1 −3  2 x + x  

2 −1

)

  − 3 x − 4  + 1  

)] }

− 3 2 − x −1

−1 −3

2

− 3 x −4 + 1

(

( 2 − x ) then f ′ ( x ) =

)

)

−3−1

(

⋅ 0 + x −2

) = −3(2 − x )

−1 −4 − 2

x

=

−3 x −2

(2 − x )

−1 4

3

x ( 2 - x -1 ) 2

4

(1 + 2 x 2 ) − x −2  f ( x) =  

v. Given

3

(

 = 5  1 + 2x 2 

)

3

5

then

f ′( x)

(

( 

= 5  1 + 2 x 2

)

3

 − x −2  

5−1

(

)

3−1   ⋅ 3 1 + 2 x 2 ⋅ 4 x + 2 x − 2−1   

)

4 2    − x − 2  12 x 1 + 2 x 2 + 2 x − 3    

( 2 − x −1 )−3 + 2 x 3  f ( x) =  

w. Given =

3

 − x2 ⋅ 

2 f ( x ) = ( x −1 + x −3 ) + x   

t. Given

=

1.4 The Chain Rule

−1

(

 then f ′ ( x ) = −  2 − x −1 

)

−3

 + 2 x3  

−1−1

(

)

−3−1 − 2   ⋅ − 3 2 − x −1 ⋅ x + 6 x2   

-2

-3 -4 - ( 2 - x -1 ) + 2x 3  -3x -2 ( 2 - x -1 ) + 6x 2     

Example 1.4-2: Find the derivative at x = 0 , x = −1 , and x = 1 in example 1.4-1 for problems a - g. Solutions: a. Given f ′( x) = 10 ( 3 − 5x ) −3 , then f ′( 0)

f ′ ( −1)

=

10 3 − ( 5 ⋅ 0 ) 

=

−3

10 3 − ( 5 ⋅ −1) 

Hamilton Education Guides

=

10 [3 − 0]

=

−3

=

−3

−3

10 [3 + 5]

10 ⋅ 3−3

=

=

10 ⋅8−3

10 ⋅

=

1 33

10 ⋅

= 1 83

10 27

=

=

10 512

0.37

=

0.019

34

Basic Differentiation

1.4 The Chain Rule

f ′(1) = 10 3 − ( 5 ⋅ 1) 

−3

=

=

10 [3 − 5]

−3

10 ⋅ ( −2 )

−3

=

10 ⋅

1

( −2 )

=

3

10 −8

=

−

10 8

=

-1.25

b. Given f ′( x ) = 6 (1 + x ) 5 , then = 6 (1 + 0 )5 =

f ′( 0) f ′ ( −1) f ′ (1)

=

6 ⋅ 15

= 6 (1 − 1)5 =

= 6 (1 + 1)5 =

=

6 ⋅ 05

=

6 ⋅ 25

=

6⋅1

6

6⋅ 0

=

0

6 ⋅ 32

=

192

( ) f ′ ( 0 ) = ( 12 ⋅ 0 ) [ 1 + ( 2 ⋅ 0 ) ] = 0 ⋅ ( 1 + 0 ) 2

c. Given f ′( x ) = 12 x 1 + 2 x 2 , then 2

2

f ′ ( −1)

[

= ( 12 ⋅ −1) 1 + 2 ⋅ (− 1)2

]

2

2

= 0 ⋅ 12 = 0 ⋅ 1 =

0

= − 12 ( 1 + 2) 2 = − 12⋅ 3 2 = −12⋅ 9 = −108

[ ( ) ] = 12 (1 + 2) = 12⋅ 3 = 12⋅ 9 = 108 d. Given f ′( x ) = 8 ( x − x ) ( 3x − 5 x ) , then f ′( 0) = 8 ( 0 − 0 ) [ 3 ⋅ 0 − 5 ⋅ 0 ] = 8 ⋅ 0 ⋅ 0 = 8 ⋅ 0 ⋅ 0 = 0 f ′ (1)

2

= ( 12 ⋅1) 1 + 2 ⋅12 3

5 7

3

f ′ ( −1)

f ′ (1)

[

(

) [ 3 ⋅1 7

5 7

2

2

4

] [ 3 ⋅ (− 1) 7

= 8 (− 1)3 − (− 1)5

= 8 13 − 15

2

(

− 5 ⋅ 14

e. Given f ′( x ) = 3 x −2 + x =

f ′( 0)

f ′ ( −1)

f ′ (1)

2

2

4

7

2

− 5 ⋅ (− 1)4

] = 8 ( 1 − 1)

) (−2 x 2

−3

] = 8 [− 1 + 1]

7

( 3 ⋅1 − 5 ⋅1) =

2

2

 



2

f ′(0)

− 1  − + 1 = 3 (1 − 1) 2 (2 + 1) = = 3  2 3     (− 1)   (− 1) 

 1

 + 1 2 1 

= 3 

2

8 ⋅ 07 ( 3 − 5 )

8 ⋅ 07 ⋅ −2

=

= 8 ⋅ 0 ⋅ −2 =

8 ⋅ 0 ⋅ −2 = 0 0

2

)

2

1

[ 3 ⋅1 − 5 ⋅1] =

 1   2  + x  − + 1 , then + 1 = 3 2 3 x   x 

 1   2    2  1 3  + 0   − + 1 = 3  + 0   − + 1  2 3 0 0     0   0  

7

is undefined due to division by zero.

3 ⋅ 02 ⋅ 3

= 0

 2   − + 1 = 3 (1 + 1) 2 (− 2 + 1) = 3 ⋅ 22 ⋅ −1 = 3 ⋅ 4 ⋅ −1 = -12 3  1 

(

f. Given f ′( x ) = −3x −2 1 + x −1

Hamilton Education Guides

)

2

= −

(

3 1+ x

)

−1 2

x2

= −

 1 3 1 +   x x2

2

, then

35

Basic Differentiation

f ′ ( −1)

f ′ (1)

1   3 1 +   −1 

= −

= −

2

(

= 4 0 + 03

f ′ ( −1)

f ′ (1)

is undefined due to division by zero.

= −

3⋅ 02 0 = − = 0 1 1

3

3

2

=

−

3 ⋅ 22 1

= −

3⋅ 4 1

=

-12

) (1 + 3x ), then ) (1 + 3 ⋅ 0 ) = 4 ⋅ 0 ⋅ 1 = 4 ⋅ 0 ⋅ 1 = 0

[

(

3 (1 − 1) 2 1

3 (1 + 1) 2 1

(

= 4 − 1 + (− 1)3

= 4 1 + 13

f ′(0 )

2

 1 3 1 +   1

g. Given f ′( x ) = 4 x + x 3 f ′( 0)

2

2

(− 1)2

1

1  3 1 +  0  − 0

=

02

= −

= −

2

 1 3 1 +   0

= −

f ′( 0)

1.4 The Chain Rule

2

3

] [1 + 3 ⋅ (− 1) ] 3

2

= 4 ⋅ (− 1 − 1) 3 ⋅ ( 1 + 3) = 4 ⋅ (− 2) 3 ⋅ 4 = 4 ⋅ (−8) ⋅ 4 = −128

) (1 + 3 ⋅1 ) = 4 ⋅ (1 + 1) ⋅ (1 + 3) = 4 ⋅ 2 ⋅ 4 = 4 ⋅ 8 ⋅ 4 = 128 3

2

3

3

Example 1.4-3: Use the chain rule to differentiate the following functions. Do not simplify the answer to its lowest term.

(

)

5

a.

d 2 x +3 dx

d.

 2 d  t +3  dt  t − 1 

g.

d  2 u +4 du 

j.

d  x + 2   dx  x − 2 

m.

(

)

(

)(

2

   

6

=

5 d 2 ( x + 3) dx

)

 u −1   3

=

= 4

=

(

= 5 x2 + 3

Hamilton Education Guides

)

5−1

(

d x2 + 3 dx

(

)

4

b.

3 d  2  x + 5 + 1  dx  

e.

3  3 d  θ + 2θ    dθ  (θ + 1) 2   

h.

3 d  3 t + 2t 2 + 1 t 2 + t + 1   dt  

k.

3 d  ( 2 x + 5)    dx  (1 + x ) 2   

n.

3 4 d  3 2x + 1 x 2 + 1   dx  

5

d  2 1   x + 3 dx  x + 5

Solutions: a.

=

(

)

(

) = 5(x

=

)(

(

2

=

=

)(

)

+3

4

⋅ 2x

)

)

=

=

)

(

c.

3 d  2  u + 1 (u + 5)  du  

f.

4 d  2 2 r r +3   dr  

(

)

i.

d  x 3 + 3x    dx  1 − x 2 

8

l.

2 d  (1 − x )    dx  x 3 + 2 x   

=

=

=

=

=

10x ( x 2 + 3 )

4

36

Basic Differentiation

b.

4 3 d  2 x + 5 ) + 1 (  dx 

(

)

1.4 The Chain Rule

(

)

3 = 4  x 2 + 5 + 1  

(

3

3

)

4 −1

(

)

(

)

3 d  2  x + 5 + 1  dx  

2

(

)

(

)

3

3

= 4  x 2 + 5 + 1 ⋅ 3 x 2 + 5 ⋅ 2 x  = 4  x 2 + 5 + 1 ⋅ 6 x x 2 + 5       c.

3 d  2 u + 1) ( u + 5 )  (    du

(

)

3   2 +  u + 1 ⋅1  

d.

5  2  d  ( t + 3)  dt  t − 1   

=

e.

=

g.

)

(





)

4

(

)

 ⋅ 2t  − t 2 + 3 



5

=

3

(

 2 3  3 ( θ + 1) θ + 2θ

) ( 3θ 2

2

= ( r 2 + 3) 

4

6 d  2 ( u + 4 ) ( u 3 − 1) du 

6  2 2 +  u + 4 ⋅ 3u   

)

=

(

)

3

3 −1

3

(x

2

+5

)

2

d 2  u + 1)   ( du 

6u ( u + 5 ) ( u 2 + 1) + ( u 2 + 1) 2

=

)

(

d  x 2 + 5 + 0 dx 

3

=

5

2

=

(

(

)

)

3−1 d   2 3 θ 3 + 2θ  ⋅ ( θ + 1) ⋅ 3 θ + 2θ d θ  

( θ + 1) 4

(θ + 1)

(

)

3

4

( θ + 1) 

4

= ( r 2 + 3)

2 r ( r 2 + 3 ) + 8r 3 ( r 2 + 3 )

d

= ( u 3 − 1) ( u 2 + 4 ) du 

(

3−1

(θ + 1)2 ⋅ 3 (θ 3 + 2θ )2 ⋅ ( 3θ 2 + 2 )  − (θ 3 + 2θ )3 ⋅ 2 (θ + 1) ⋅ 1    

4 d 2  2 d 2 r  + r r + 3)  ( dr   dr 

4 3 2r ( r 2 + 3) +  4r 2 ( r 2 + 3) ⋅ 2r   

)

5 −1 d 5    2 2 2  ( t − 1) ⋅ 5 ( t + 3) ⋅ dt ( t + 3)  − ( t + 3) ⋅ 1 2 ( t − 1)

( t - 1)

  + 2  −  2 θ 3 + 2θ  

( θ + 1)





4

=

(

  = 24 x  x 2 + 5 + 1  



10t ( t - 1) ( t 2 + 3 ) - ( t 2 + 3 )

=

2

= ( u + 5) ⋅ 3 ( u 2 + 1)

3 3 d 2 d 2   3 3 (θ + 1) dθ (θ + 2θ )  − (θ + 2θ ) dθ (θ + 1)  4 (θ + 1)

3 d 2 −1  − (θ 3 + 2θ ) ⋅ 2 (θ + 1) ⋅ (θ + 1) dθ   4 (θ + 1)

)

)

(

5 5 d d 2   2  ( t − 1) dt ( t + 3)  − ( t + 3) dt ( t − 1)  2 ( t − 1)

( t − 1)2

(



2

= (u + 5) ⋅ 3 u 2 + 1 ⋅ 2u   + u 2 + 1  

(

4 d  2 2 r ( r + 3)    dr 

=



 2 ( t − 1) ⋅ 5 t + 3

3  3  d  (θ + 2θ )  dθ  (θ + 1)2   

=

f.

=

3 3 d d 2  u + 1)  + ( u 2 + 1) ( u + 5) ( du du   

= ( u + 5)

3

3

 = 4  x 2 + 5 + 1 ⋅ 3 x 2 + 5   

) (

4

6

=

3

4

4 −1 d   ⋅ 2r  +  r 2 ⋅ 4 ( r 2 + 3) ⋅ ( r 2 + 3)    dr 

2r ( r 2 + 3 ) ( r 2 + 3 ) + 4r 2  3

6 d   2  3  + ( u + 4 ) du ( u − 1)    

)

(

)

(

) (

 =  u 3 −1 ⋅ 6 u 2 + 4 

(

)(

)

6 −1

⋅

(

)

d  u2 + 4  du 

)

(

)

5 6 5 6 =  u 3 − 1 ⋅ 6 u 2 + 4 ⋅ 2u  + 3u 2 u 2 + 4  = 12u u 3 − 1 u 2 + 4  + 3u 2 u 2 + 4         

Hamilton Education Guides

37

Basic Differentiation

(

= 3u u 2 + 4 h.

1.4 The Chain Rule

) [ 4 ( u − 1) + u ( u 5

3

(

)(

(

) (

2

+4

)

)]

(

) dtd ( t

3 d  3  t + 2t 2 + 1 t 2 + t + 1  =  t 2 + t + 1  dt   

) (

3

3

) (

) (

) (

)

3 d 2   + 2t 2 + 1  +  t 3 + 2t 2 + 1 t + t +1  dt   

)

(

(

)

) (

)

3 3 3−1 d   =  t 2 + t + 1 ⋅ 3t 2 + 4t  +  t 3 + 2t 2 + 1 ⋅ 3 t 2 + t + 1 ⋅ t 2 + t + 1  =  t 2 + t + 1 ⋅ 3t 2 + 4t  dt      

(

) (

(

)

2  3  2 2 +  t + 2t + 1 ⋅ 3 t + t + 1 ⋅ (2t + 1)  

i.

d  x + 3x    dx  1 − x 2  3

8

=

 x 3 + 3x   8  1− x 2    7

[(

)(

8−1

= t2 + t +1

d  x 3 + 3 x  ⋅  dx  1 − x 2 

=

)] [( ( )

)

) {[ t ( t 2

j.

=

k.

 x+2 2   x−2

=

−

2 −1

d  x+2 ⋅   dx  x − 2 

 x + 2  ( x − 2 ) ⋅ 1 − ( x + 2 ) ⋅ 1 2 ⋅ 2  x−2 ( x − 2)

3 d  ( 2 x + 5)    2 dx  (1 + x ) 

−

l.

2

=

(

]

(1 + x )4

) (

)

) ( ( )

7

[(

1− x2  x 3 + 3x    ⋅ = 8 2   1− x 

]}

) (

)

)( 3 x

)] [ ( (1 − x )

2

+ 3 + 2x 2 x 2 + 3

)]

2 2

d d  ( x − 2 ) ( x + 2 ) − ( x + 2 ) ( x − 2 ) dx dx  x + 2      2 ⋅ 2  x−2 − x 2 ( )

x + 2  x/ − 2 − x/ − 2 ⋅  x − 2  ( x − 2 )2

 = 2

2 d 3 3 d 2   (1 + x ) dx ( 2 x + 5 )  − ( 2 x + 5 ) dx (1 + x )  4 (1 + x )

 3 2 −1 d (1 + x ) (2 x + 5) ⋅ 2(1 + x ) dx  

[ 2 ( 2 x + 5)

=

][ (

)

+ t + 1 ( 3 t + 4 ) + 3 t 3 + 2 t 2 + 1 ( 2 t + 1)

d d     7 x 3 + 3x  −  x 3 + 3x 1− x 2 1− x 2   x 3 + 3 x   dx dx     ⋅ 8 2  1− x 2  2   1− x

1 − x 2 ⋅ 3 x 2 + 3 − x 3 + 3 x ⋅ (− 2 x )  x 3 + 3x    ⋅ = 8 2 2   1− x  1− x 2

d  x+2   dx  x − 2 

2

[( 1 + x) =

2

−4 x+2 ⋅ 2 − x  ( x − 2 )2 

 = 2

=

= −

8( x + 2 )

( x − 2 )3

 3−1 d 2 ( 2 x + 5) ( 1 + x ) ⋅ 3 ( 2 x + 5 ) dx  

( 1 + x) 4

] [ ( 2 x + 5)

⋅ 3 ( 2 x + 5) 2 ⋅ 2 −

( 1 + x) 4

3

⋅ 2 ( 1 + x ) ⋅1

]

[6 ( 1 + x) =

( 2 x + 5) 2 ] ( 1 + x) 4 2

( 1 + x)] 2(2 x + 5)2 ( x − 2) 2(2 x + 5)2 (3 + 3 x − 2 x − 5) 2(1/ + x/ ) (2 x + 5)2 [3(1 + x ) − (2 x + 5) ] = = = (1 + x )3 (1 + x )3 (1 + x )4/ =3 ( 1 + x) 4

2 d  (1 − x )   3  dx  x + 2 x 

3

=

d 2 2 d  3   3 ( x + 2 x ) dx (1 − x )  − (1 − x ) dx ( x + 2 x )  2 ( x3 + 2 x )

Hamilton Education Guides

=

2 −1 d  3  ( x + 2 x ) ⋅ 2 (1 − x ) ⋅ dx (1 − x )  2 ( x3 + 2 x )

38

Basic Differentiation

[ ( 1 − x ) ⋅ ( 3x + 2) ] 2

−

m.

(x

2

3

+ 2x

1  d  2 x + 3  dx  x +5

=

n.

1.4 The Chain Rule

)

4

2

=

[2 ( x =

1   4  x2 + 3  +5 x 

3

+ 2x

) ( 1 − x)⋅ −1]− [( 1 − x) ( 3x ( x + 2x) 2

⋅

1  d  2 x + 3  dx  x +5

=

 d  3   d 3  3 ( x + 5 ) dx (1)  − 1 ⋅ dx ( x + 5 )   1    2  4 x + 3  ⋅ 2 x + 2 x +5    x3 + 5) (  

3 4 d  3 ( 2 x + 1) ( x 2 + 1)  dx 

= ( x 2 + 1) 

4

+2

)]

=

2

3

4 −1

2

-2 ( x 3 + 2x ) (1 - x )  - (1 - x )2 ( 3x 2 + 2 )     

(x

3

+ 2x )

2

3

d  1  1  d 2  4  x2 + 3  ⋅  (x ) +  3  + x dx dx 5     x + 5 

=

3  1   3x 2   4  x2 + 3 2x  2 x +5   ( x3 + 5 )  

3 3 4 d ( 2 x3 + 1)  + ( 2 x3 + 1) dxd ( x 2 + 1)  dx

(

) (

)

4 2 4 3 −1 d 3 4 −1 d   = ( x 2 + 1) ⋅ 3 ( 2 x3 + 1) ⋅ ( 2 x3 + 1)  + ( 2 x3 + 1) ⋅ 4 ( x 2 + 1) ⋅ ( x 2 + 1)  =  x 2 + 1 ⋅ 3 2 x 3 + 1 ⋅ 6 x 2  dx dx    



(

) (

)

3 3  3  2 +  2 x + 1 ⋅ 4 x + 1 ⋅ 2 x  

=



18x 2 ( x 2 + 1)4 ( 2x 3 + 1)2  + 8x ( 2x 3 + 1)3 ( x 2 + 1)3     

In some instances students are asked to find the derivative of a function y , where y is a function of u and u is a function of x . We can solve this class of problems using one of two methods. The first method, and perhaps the easiest one, is performed by substituting u into the y equation and taking the derivative of y with respect to x . The second method is to find the derivative of y

by using the equation

dy dy du = ⋅ dx du dx

. This method is most often used in calculus books and can

be time consuming. For example, let’s find the derivative of the function y = u 2 + 1 where u = x + 1 using each of these methods. First Method: Given the function y = u 2 + 1 where u = x + 1 , substitute u with x + 1 in the function y and simplify, i.e., y = u 2 + 1 = (x + 1)2 + 1 = x 2 + 2 x + 1 + 1 = x 2 + 2 x + 2 . Next, take the derivative of y with respect to x , i.e.,

dy dx

=

(

dy 2 x + 2x + 2 dx

) = 2x

2 −1

+ 2 x1−1 + 0

Second Method: Given the function y = u 2 + 1 where u = x + 1 , find 2u 2−1 + 0

= 2u and

dy dy du = ⋅ dx du dx

we obtain

du dx

= x1−1 + 0 = x 0 = 1 . Next, substitute

to find the derivative dy dx

dy dx

, i.e.,

dy dx

dy du

= 2x + 2 .

dy du

and

and

du dx

du dx

, i.e.,

dy du

=

in the equation

= 2u ⋅1 = 2u . Substituting u = x + 1 in place of u

= 2u = 2 (x + 1) = 2 x + 2 .

Hamilton Education Guides

39

Basic Differentiation

1.4 The Chain Rule

The second method is generally beyond the scope of this book, therefore the first method is used in order to solve this class of problems. Examples 1.4-4 and 1.4-5 below provide additional examples as to how these types of problems are solved. Example 1.4-4: Find a. y = c. y = e. y =

1 1+ u u 1+ u

2

1+ u

dy dx

given:

and u = 3x + 1

b. y = u 2 + 2u + 1 and u = 3x + 2

and u = 5x + 1

d. y = u 2 + 1 and u =

(

Solutions: a. Given y = dy dx

=

1 1+ u

x2 −1

)

f. y = u3 + 1 and u = x 2 + 1

and u = x 2 + 1

u3

x +1

and u = 3x + 1 , then y =

d    d  ( 3 x + 2 ) dx (1)  − 1 ⋅ dx ( 3 x + 2 )  2 ( 3x + 2 )

=

1 1 + ( 3x + 1)

0−3

( 3x + 2 )

2

=

-

1 3x + 2

=

−1

and

3

( 3x + 2 )

2

b. Given y = u 2 + 2u + 1 and u = 3x + 2 , then y = ( 3x + 2) 2 + 2( 3x + 2) + 1 = ( 3x + 2) 2 + 6 x + 5 and dy dx

=

d d d 2 ( 3x + 2 ) + 6 x + 5 dx dx dx

c. Given y = dy dx

=

=

u 1+ u

2

and u = 5x + 1 , then y =

d ( 3x + 2 ) + 6 + 0 dx

[

][

5 1 + ( 5 x + 1) 2 − 10 ( 5 x + 1) 2

[1 + ( 5x + 1) ]

2 2

x +1 x2 −1

]

2

=

2 2

[1 + (5x + 1) ]

2 2

2

 x +1   +1  x 2 − 1

=

5 − 5 (5 x + 1) 2

[1 + ( 5 x + 1) ]

2 2

and

 2 ( x − 1) dxd ( x + 1) − ( x + 1) ⋅ dxd ( x 2 − 1)  x + 1   2 2 ⋅ 2  x −1 ( x 2 − 1)

=

=

2  x + 1  ( x − 1) ⋅ 1 − ( x + 1) ⋅ 2 x  ⋅ 2 2  2  x −1 ( x 2 − 1)

=

18 ( x + 1)

{[ 1 + ( 5x + 1) ]⋅ 5 }− [( 5x + 1)⋅ 2 ( 5x + 1)⋅ 5 ] [1 + ( 5x + 1) ]

5 + 5 (5 x + 1) 2 − 10 (5 x + 1) 2

, then y = 

  x + 1  2 −1 d  x + 1   2  2  ⋅  2  + 0 dx  x − 1     x − 1 

Hamilton Education Guides

=

= 2 ( 3x + 2 ) ⋅ 3 + 6 =

(5x + 1) and 2 1 + ( 5x + 1)

d  2 2      d   1 + ( 5 x + 1)  dx ( 5 x + 1)  − ( 5 x + 1) ⋅ dx 1 + ( 5 x + 1)   2 1 + ( 5 x + 1)2   

d. Given y = u 2 + 1 and u = dy dx

= 2 ( 3x + 2 )2 −1 ⋅

=

 2  2  x + 1   x − 1 − 2x − 2x  ⋅ 2 2  2  x −1  ( x 2 − 1)  

=

 2  x + 2x + 1   x +1  2 2  ×  2  x -1   ( x 2 -1)   

40

Basic Differentiation

 x + 1  ( x + 1) −2  2 ⋅ 2  x − 1  ( x 2 − 1) 2

=

1+ u

e. Given y =

dy dx

1.4 The Chain Rule

=

−2

( x + 1)( x + 1)

(x

− 1)( x 2 − 1)

2

and u = x + 1 , then y =

u3

=

2

 2 x ( x 2 + 1)3  − ( x 2 + 2 ) ⋅ 3 ( x 2 + 1)2 ⋅ 2 x     

=

(x

2

+ 1)

6

(

)

(

)

f. Given y = u3 + 1 and u = x 2 + 1

(

)

 = − 3 x 2 + 1 

−3−1

⋅

−1

( x + 1)

−2

(x

(

( x + 1) 2

)

3

 x +1  -2  2   x -1

=

3

3

x2 + 2

=

( x + 1) 2

3

( x 2 + 1)3 ⋅ 2 x  − ( x 2 + 2 ) ⋅ 3 ( x 2 + 1)2 d ( x 2 + 1)      dx

=

(x

2

+ 1)

6

 2x ( x 2 + 1)3  - 6x ( x 2 + 2 )( x 2 + 1)2      6 2 ( x + 1)

=

( 

)



, then y =  x 2 + 1

d  x 2 +1  + 0 dx 

3

− 1)

2

1+ x 2 + 1

3 3 d 2 d 2  2   2 ( x + 1) ⋅ dx ( x + 2 )  − ( x + 2 ) ⋅ dx ( x + 1)  6 ( x 2 + 1)

=

dy dx

2

2

(

)

= − 3 x 2 +1

−4

−1  3

 +1 

⋅ 2x

(

)

(

)

= x2 +1

= − 6x x 2 +1

−3

+1

−4

=

-

(x

6x 2

+ 1)

4

Example 1.4-5: Find y ′ given that: 3u 2 and u = x 2 1+ u 1 − 5u d. y = 2 and u = 1 − x u

a. y = 3u3 − 1 and u = x 2 + 1 c. y = e. y = g. y =

b. y =

u +1 and u = x 2 + 3 u −1 u

and u = 2 x − 1

2

u +1

2u

(u − 1)

2

f. y = u +

1 and u = (2 x − 1) 4 4

h. y = u 4 − 1 and u =

and u = x + 2

1+ x 1− x

Solutions:

)

(

3

a. Given y = 3u3 − 1 and u = x 2 + 1 , then y = 3 x 2 + 1 − 1 and y′

(

b. Given y = y′

)

= 3⋅3 x 2 +1

=

3−1

⋅ 2x − 0

(

)

2

= 9 x 2 +1 ⋅ 2x =

18x ( x 2 + 1)

2

3u 2 3x 4 and u = x 2 , then y = and 1+ u 1+ x 2

4 −1 2 4     ( 3 ⋅ 4 x ) ⋅ (1 + x )  −  2 x ⋅ 3 x  2 (1 + x 2 )

Hamilton Education Guides

=

12 x 3 (1 + x 2 ) − 6 x 5

(1 + x )

2 2

=

12 x 3 + 6 x 5

(1 + x )

2 2

=

6x 3 ( 2 + x 2 )

(1 + x )

2 2

41

Basic Differentiation

c. Given

1.4 The Chain Rule

( x + 3) + 1 = x + 4 and and u = x + 3 , then y = ( x + 3) − 1 x + 2 [ 2x ( x + 2) ]− [2x ( x + 4) ] ⋅ ( x + 2 ) ]− [ 2 x ⋅ ( x + 4) ] = ( x + 2) ( x + 2)

u +1 y= u −1

[ 2x y′ =

2 −1

d. Given y =

2

1 − 5u

2

2

2

2

2−1

1 − 5(1 − x )

(1 − x )

]

⋅ −1 ⋅ ( 5 x − 4)

=

( 1 − x) 4

5 ( x 2 − 2 x + 1) + 2 ( 5 x − 4 − 5 x 2 + 4 x )

(1 − x )

e. Given y =

y′

=

=

u

u +1

2 ( 2 x − 1) + 2 − 4 ( 2 x − 1) 2

 2 x − 1) + 1 (  2

=

=

(u − 1)

2

and u = x + 2 , then y =

[ 2 ⋅ ( x + 1) ]− [ 2 ( x + 1) 2

2 −1

]

⋅ ( 2 x + 4)

2

=

2

( x + 1) 4 2 x 2 + 4 x + 2 − 4 x 2 − 12 x − 8

( x + 1)

4

Hamilton Education Guides

=

( x + 1)

4

=

(x

-4x 2

+ 2)

2

(1 − x ) 2 5 ( 1 − x ) 2 + 2 ( 1 − x ) ( 5 x − 4)

=

( 1 − x) 4

-5x 2 + 8x - 3

(1 - x )

4

and

[ ] [ (2x − 1) + 1]

2

2

1 4

=

−2

[ ( 2 x − 1)

[ ( 2 x − 1)

2

2

−1

+1

]

2

]

and

3

( x + 2 − 1)

−2 x 2 − 8 x − 6

2

5x − 4

=

2 1 − (2 x − 1)2

2( x + 2)

=

+ 2)

2

2 2 ⋅ ( 2 x − 1) + 1 −  4 ( 2 x − 1) ⋅ ( 2 x − 1)    2 ( 2 x − 1)2 + 1  

and u = ( 2 x − 1) 4 , then y = ( 2 x − 1) 4 +

2u

4

( 2 x − 1) 2 + 1

( 2 x − 1) + 1   2

=

(x

2

2x − 1

=

2 − 2 ( 2 x − 1)

=

2

(1 − x )

2

2 x3 + 4 x − 2 x3 − 8 x

=

5 (1 − x ) − −  2 (1 − x )( 5 x − 4 )  4 (1 − x )

(1 − x )

= 4 ( 2 x − 1)4 −1 ⋅ 2 + 0 = 4 ( 2 x − 1)3 ⋅ 2 = 8 ( 2x - 1)

g. Given y =

y′

1 4

2

1 − 5 + 5x

=

2

5 x 2 − 10 x + 5 − 10 x 2 + 18 x − 8

=

2 2 −1 2 ⋅ ( 2 x − 1) + 1 −  2 ( 2 x − 1) ⋅ 2 ⋅ ( 2 x − 1)      2 2 ( 2 x − 1) + 1  

f. Given y = u + y′

4

and u = 2 x − 1 , then y =

2

2

2

[ 5 ⋅ (1 − x) ]− [ 2 ( 1 − x) y′ = =

2

2

and u = 1 − x , then y =

2

2

2

2 −1

2

u

2

2

2

=

2x + 4

( x + 1) 2

2 ( x + 1) − 2 ( x + 1)( 2 x + 4 ) 2

( x + 1)

=

(

4

− 2 x 2 + 4x + 3

( x + 1)

4

=

2 ( x 2 + 2 x + 1) − 2 ( 2 x 2 + 6 x + 4 )

( x + 1)

4

)

42

Basic Differentiation

1.4 The Chain Rule

h. Given y = u 4 − 1 and u = y′

1+ x    1− x 

 = 4

4 −1

 (1 + x )3

⋅

[1⋅ (1 − x )] − [− 1⋅ (1 + x )] (1 − x )2

  3 2  (1 − x ) (1 − x ) 

= 4

4

1+ x 1+ x  , then y =   − 1 and  1− x  1− x

2

⋅

=

8 (1 + x )3

(1 − x )3+ 2

−0

=

=

 1 + x  (1 − x ) + (1 + x ) 4  ⋅ 2 1− x  (1 − x ) 3

=

3

2 1+ x  4  ⋅ 2  1 − x  (1 − x )

8 (1 + x )3

(1 − x )5

Section 1.4 Practice Problems - The Chain Rule 1. Find the derivative of the following functions. Do not simplify the answer to its lowest term.

(

)



1   x2 

a. y = x 2 + 2 d. y = 1 − 

(

3 2

x +1   3 

g. y = x 2 

(

3

)

4

(x + 1)−2 x

5

e. y = 2 x 3 +

1

1+ x f. y =  3   r 

2





3x 2

k.

 t2 y= 1+ t 2 

   

]

3

3

x 3

x3

o. y =

−1



(

2

1

4

i. y =  − 2 x 3 

)

−1

l. y = 1 + x −2

 1  + 3 x 1− x  

n. y = 

3

)

c. y = x 3 − 1

h. y = x ( x + 1)2 + 2 x

j. y = x + 3x + 1 m. y =

(

−2

[

3

2

)

b. y = x 2 + 1

− x2

3

x +2

2. Find the derivative of the following functions at x = 0 , x = 1 , and x = −1 .

(

)

a. y = x 3 + 1

(

b. y = x 3 + 3x 2 − 1

)

2

d. y = x x + 1   2  x +1 

g. y = 

x

)

(

5

2

5

3

(

4

)

2

e. y = x + 2 x + 1

(

)

2

 x2 y= 1+ x 2 

   

f.

 x3 

1

3

h. y = x 2 + 1 ⋅

3

x   x  +1

c. y = 

3

2

 + 5x i. y =  x −1 

x2





3. Use the chain rule to differentiate the following functions. a.

d dt

 (t + 1)3     t 2 

(

=

)

2 d  3  d. x − 1 ( 2 x + 1) 3  =  dx  

Hamilton Education Guides

(

)

b.

 d  u 2 +1 du  3u 4 

e.

d ds

3

   

 3 1  s − 2  s + 6 

= 2

=

 ( 2 x + 1) 3     ( 1 − x ) 2 

c.

d dx

f.

3  d  t 2 −1  dt  t 2 + 1   

(

)

=

=

43

Basic Differentiation

g.

2 3 1   d  2  u +1    du   u + 1   

(

)

1.4 The Chain Rule

=

h.

d dθ

 θ 2 +3    3  ( θ − 1) 

2

=

i.

 r7 d  dr  2  r + 2r

(

)

  3 

=

4. Given the following y functions in terms of u find y ′ . a. y = 2u 2 − 1 and u = x − 1 d. y = u 2 −

1 2

and y = x 4

Hamilton Education Guides

b. y =

u u −1

and u = x3

e. y = u 4 and u =

1 1− x 2

c. y = f. y =

u

1 + u2 u2

(u + 1)3

and u = x 2 + 1 and u = x − 1

44

Basic Differentiation

1.5

1.5 Implicit Differentiation

Implicit Differentiation

In many instances an equation is explicitly represented in the form where y is the only term in the left-hand side of the equation. In these instances y ′ is obtained by applying the differentiation rules to the right hand side of the equation. However, for cases where y is not explicitly given, we must either first solve for y (if y can be factored) and then differentiate or use the implicit differentiation method. For example, to differentiate the equation x y = x 2 + y we can either solve for y by bringing the y terms to one side of the equation and then differentiate as follows: x y = x2 + y

y′

[2x =

2 −1

; x y − y = x 2 ; y( x − 1) = x 2 ; y =

][ ]

⋅ (x − 1) − 1 ⋅ x 2

(x − 1)2

x2 x −1

therefore, y ′ is equal to:

[2 x(x − 1)] − x 2 = = (x − 1)2

2x 2 − 2x − x 2

( x − 1) 2

=

x 2 − 2x

( x − 1) 2

or, we can use the implicit differentiation method as shown below. x y = x2 + y

; (1⋅ y + y ′ ⋅ x ) = 2 x 2 −1 + y ′ ; y + y ′x = 2 x + y ′ ; y ′x − y ′ = 2 x − y ; y ′( x − 1) = 2 x − y ; y ′ =

Substituting y =

y′

=

2x − y x −1

=

x2 into the y ′ equation we obtain: x −1 x2 x −1 x −1

2x −

=

2 x( x − 1) − x 2

2 x( x − 1) − x 2

x −1 x −1

x −1 x −1 1

=

[2 x( x − 1) − x ]⋅1 = = 2

( x − 1) ⋅ ( x − 1)

2x − y x −1

x 2 − 2x

( x − 1)2

Note that the key in using the implicit differentiation method is that the chain rule must be applied each time we come across a term with y in it. Following are additional examples showing the two methods of differentiation when y is not explicitly given: Example 1.5-1: Given xy + x = y + 3 , find y ′ . Solution: First Method: Let’s solve for y by bringing the y terms to the left-hand side of the equation, i.e., xy + x = y + 3 ; y (x − 1) = − x + 3 ; y =

−x + 3 x −1

We can now solve for y ′ using the differentiation rule for division. y′

=

[− 1⋅ (x − 1)] − [1⋅ (− x + 3)] = (x − 1)2

− x/ + 1 + x/ − 3

(x − 1)

2

=

−2

( x − 1)2

Second Method: Use the implicit differentiation method to solve for y ′ , i.e., given xy + x = y + 3

then, (1 ⋅ y + y ′ ⋅ x ) + 1 = y ′ + 0 ; y +1 = y ′ − y ′x ; y ′(1 − x ) = y + 1 ; y ′ =

Hamilton Education Guides

y +1 1− x 45

Basic Differentiation

1.5 Implicit Differentiation

−x + 3 x −1

Substituting y =

y′

=

=

y +1 y′ = 1− x

−x + 3 +1 x −1 1− x

=

2 ⋅1

(x − 1)⋅ (1 − x )

into the y ′ equation we obtain: −x + 3 1 + x −1 1 1− x

=

2

=

(x − 1)⋅ −(x − 1)

= −

[(− x + 3)⋅1] + [1⋅ (x − 1)] (x − 1)⋅1 = =

− x/ + 3 + x/ − 1 x −1 1− x

1− x

2

(x − 1)(x − 1)

=

=

2 x −1 1− x

=

2 x −1 1− x 1

−2

( x − 1)2

Example 1.5-2: Given x 2 y + 5 = y + 2 x , find y ′ . Solution: First Method: Let’s solve for y by bringing the y terms to the left-hand side of the equation,

(

)

i.e., x 2 y − y = 2 x − 5 ; y x 2 − 1 = 2 x − 5 ; y =

2x − 5 x2 −1

We can now solve for y ′ using the differentiation rule for division.

[2 ⋅ (x − 1)]− [2x = (x − 1) 2

y′

2 −1

⋅ (2 x − 5)

]

2

2

2 x 2 − 2 − 4 x 2 + 10 x

=

( x − 1) 2

2

=

−2 x 2 + 10 x − 2

( x − 1) 2

2

Second Method: Use the implicit differentiation method to solve for y ′ , i.e., given x 2 y + 5 = y + 2x

(

(

)

then, 2 x ⋅ y + y ′ ⋅ x 2 + 0 = y ′ + 2 x 1−1 ; 2 xy + y ′x 2 = y ′ + 2 ; y ′x 2 − y ′ = 2 − 2 xy

)

; y ′ x 2 − 1 = 2 x − 2 xy ; y ′ = Substituting y =

y′

=

=

2 − 2 xy x2 −1

=

2x − 5 x2 −1

2 − 2x ⋅

2 − 2 xy x2 −1

into the y ′ equation we obtain: 2x − 5

x2 −1 x2 −1

=

2−

−2 x 2 + 10 x − 2

−2 x 2 + 10 x − 2

x2 −1 x2 −1

x2 −1 x2 −1 1

=

(

) (

2 x 2 − 1 − 4 x 2 − 10 x

4 x 2 − 10 x x2 −1 x2 −1

x2 −1 x2 −1

=

)

2 x 2 − 2 − 4 x 2 + 10 x

=

x2 −1 x2 −1

(−2 x + 10x − 2) ⋅1 = −2x + 10 x − 2 = ( x − 1) ⋅ ( x − 1) ( x − 1) 2

2

2

2

2

2

In the previous examples, to find y ′ we could either first solve for y and then differentiate or use the implicit differentiation rule. However, sometimes we can not simply solve for y by bringing the y terms to the left-hand side of the equation. In these instances, as is shown in the following examples, we can only use implicit differentiation in order to differentiate y .

Hamilton Education Guides

46

Basic Differentiation

1.5 Implicit Differentiation

Example 1.5-3: Given x 2 y 2 + y = 3 y 3 − 1 , find y ′ = Solution: d 2 2 d y) ( x y += ( 3 y 3 − 1) dx dx

;

−2 xy 2 y′ ( 2 x 2 y − 9 y 2 + 1) =

;

2 x 2 y y′ − 9 y 2 y′ + y′ = −2 xy 2

;

y′ = -

dy dx

Solution: d d (10 x ) ( xy + x 2 y 2 + y 3 ) = dx dx

x y′ + +2 x 2 y y′ + 3 y 2 y′ = −2 x y 2 − y + 10

Solution:

;

;

y′ ( 9 x 3 y 2 + 4 y − 1) = −9 x 2 y 3

dy dx

Solution:

;

;

y′ ( x + 3 x 3 y 2 ) = −3 x 2 y 3 − y

(

)

Solution: d d 2 ( 3x y + y=) ( x + y2 ) dx dx 3 x y′ + y′ − 2 y y′ = 2 x − 3 y

y′ = -

;

y′ = -

dy dx

; 3 ( 1⋅ y + y ′ ⋅ x ) + y ′ = 2 x + 2 y y ′ ; ;

Example 1.5-8: Given

Hamilton Education Guides

y′ ( 3 x + 1 − 2 y ) = 2 x − 3 y

1 1 + = 10 x , x y2

find

;

dy dx

y′ =

-2x y 2 - y + 10 x + +2x 2 y + 3y 2

in terms of x and y .

0 ; ; (1 ⋅ y + y′ ⋅ x ) + ( 3x 2 ⋅ y 3 + 3 y 2 y′ ⋅ x3 ) =

x y′ + 3 x 3 y 2 y′ = −3 x 2 y 3 − y

;

9 x 2 y 3 + 9 x 3 y 2 y′ + 4 y y′ = y′

9x 2 y 3 9x y 2 + 4 y - 1 3

in terms of x and y .

Example 1.5-7: Given 3x y + y = x 2 + y 2 , find

;

dy dx

)

Example 1.5-6: Given x y + x 3 y 3 = 5 , find

;

in terms of x and y .

; 3 3x 2 y 3 + 3 y 2 y ′x 3 + ( 4 y y ′) = y ′ + 0 ;

−9 x 2 y 3 9 x 3 y 2 y′ + 4 y y′ − y′ =

d d ( 5) ( x y + x3 y 3 ) = dx dx

2xy 2 2x y - 9y 2 + 1 2

y′ ( x + +2 x 2 y + 3 y 2 ) = −2 x y 2 − y + 10

(

d ( 3x3 y 3 + 2 y 2 ) = dxd ( y + 1) dx

2 xy 2 + 2 yy′x 2 + y′ = 9 y 2 y′

10 ; ( y + x y′ ) + ( 2 x y 2 + 2 x 2 y y′ ) + 3 y 2 y′ =

Example 1.5-5: Given 3x 3 y 3 + 2 y 2 = y + 1 , find

;

.

; ( 2 x ⋅ y 2 + 2 y 2 −1 y′ ⋅ x 2 ) + y′ = ( 3 ⋅ 3) y 3−1 ⋅ y′ − 0 ;

Example 1.5-4: Given xy + x 2 y 2 + y 3 = 10 x , find

;

dy dx

y′ =

y + x y′ + 3 x 2 y 3 + 3 x 3 y 2 y′ = 0 3x 2 y 3 + y x + 3x 3 y 2

in terms of x and y . 3 y + 3 x y′ + y′ = 2 x + 2 y y′

2x - 3y 3x - 2y + 1

in terms of x and y .

47

Basic Differentiation

1.5 Implicit Differentiation

Solution: d  1 1  d (10 x ) = + dx  x y 2  dx

; y′ =

x −2 + 10 − 2y

−3

d −1 d (10 x ) ( x + y −2 ) = dx dx

; 1

2 ; y′ = x −2

x2 −2

; y′ =

3

y

; y′ =

Solution:

;

(

y 3 1 + 10 x 2 − 2x

3

Example 1.5-9: Given xy 2 + yx 2 = x 2 , find d d 2 xy 2 + yx 2 ) = ( (x ) dx dx

− x −2 − 2 y −3 y′ = 10

1 + 10 x 2

+ 10

y

;

dy dx

y′ ( x 2 + 2 x y ) = − y2 − 2x y + 2x

; 2

Example 1.5-10: Given y 3 + x 3 y = y , find Solution:  d d  23  y + x3 y  =   dx ( y ) dx  

;

)

2

−2 y −3 y′ = x −2 + 10

; y′ = −

y3 2

 1   + 10  2 x 

in terms of x and y .

2x ; ; (1 ⋅ y 2 + 2 y y′ ⋅ x ) + ( y′ ⋅ x 2 + 2 x ⋅ y ) =

2 x y y′ + x 2 y′ = − y2 − 2x y + 2x

;

dy dx

;

y 2 + 2 x y y′ + x 2 y′ + 2 x y = 2x

y′ =

-y 2 - 2x y + 2x x 2 + 2x y

in terms of x and y .

(

)

2 23 −1 2 −1 y ⋅ y ′ + 3x 2 ⋅ y + y ′ ⋅ x 3 = y ′ ; y 3 y ′ + 3x 2 y + x 3 y ′ = y ′ 3 3

  2 −1 − 3x 2 y 2 − 13 y y ′ + x 3 y ′ − y ′ = −3 x 2 y ; y ′ y 3 + x 3 − 1 = −3 x 2 y ; y ′ = 3 2 − 13  3 y + x3 −1 3 1 dy Example 1.5-11: Given xy + y 2 = y 8 , find in terms of x and y . dx

;

Solution:

( )

d d 1 xy + y 2 ) = y 8 ( dx dx

;

yy′ ; (1 ⋅ y ) + ( y′ ⋅ x ) + 2=

1 18 −1 y ⋅ y′ 8

1 −7 1 −7   x y′ + 2 y y′ − y 8 y′ = − y ; y′  x + 2 y − y 8  = −y 8 8  

; y′ = −

Example 1.5-12: Given xy 2 + y = x 2 + 3 , find Solution: d d 2 xy 2 + y= ( ) ( x + 3) dx dx

;

y′ ( 2 xy + 1) = 2 x − y 2

;

; (1 ⋅ y 2 ) + ( 2 y y′ ⋅ x ) + y′ = y′ =

Hamilton Education Guides

1 − 78 y + x y′ + 2 y y′ = y y′ 8

;

dy dx

2x + 0

y x + 2 y − 18 y

−7

8

in terms of x and y . ;

y 2 + 2 xy y′ + y′ = 2x

;

2 xy y′ + y′ = 2 x − y 2

2x - y 2 2xy + 1

48

Basic Differentiation

1.5 Implicit Differentiation

dy dx

Example 1.5-13: Given x 4 y 3 + y 2 = x + 4 , find Solution: d 4 3 ( x y + y 2 )= dxd ( x + 4 ) dx

;

3 x 4 y 2 y′ + 2 y y′ = 1 − 4 x3 y 3

1 0 ; ; ( 4 x3 ⋅ y 3 ) + ( 3 y 2 y′ ⋅ x 4 ) + 2 y y′ =+

y′ ( 3 x 4 y 2 + 2 y ) = 1 − 4 x3 y 3

;

;

y′ =

Example 1.5-14: Given y 6 + x 3 y 5 + x 2 = 5 , find Solution: d 6 d ( 5) ( y + x3 y 5 + x 2 ) = dx dx

;

in terms of x and y .

;

4 x 3 y 3 + 3 x 4 y 2 y′ + 2 y y′ = 1

1 - 4x 3 y 3 3x 4 y 2 + 2y

dy dx

in terms of x and y .

0 6 y 5 y′ + ( 3 x 2 ⋅ y 5 ) + ( 5 y 4 y′ ⋅ x 3 )  + 2 x = 0 ; 6 y 5 y′ + 3 x 2 y 5 + 5 x 3 y 4 y′ + 2 x =

6 y 5 y′ + 5 x 3 y 4 y′ = −3 x 2 y 5 − 2 x

;

y′ ( 6 y 5 + 5 x 3 y 4 ) = −3 x 2 y 5 − 2 x

;

y′ = -

3x 2 y 5 + 2x 6y 5 + 5x 3 y 4

Section 1.5 Practice Problems - Implicit Differentiation Use implicit differentiation method to solve the following functions. a. x 2 y + x = y

b. x y − 3x 2 + y = 0

c. x 2 y 2 + y = 3y 3

d. x y + y 3 = 5 x

e. 4 x 4 y 4 + 2 y 2 = y − 1

f. x y + x 2 y 2 − 10 = 0

g. x y 2 + y = x 2

h. x y 3 + x 3 y = x

i. y 2 + x 2 y = x

j. x 2 y + y 2 = y 4

k. x + y 2 = x 2 − 3

l. x 4 y 2 + y = −3

m. y 7 − x 2 y 4 − x = 8

n. (x + 3) 2 = y 2 − x

o. 3x 2 y 5 + y 2 = − x

1

Hamilton Education Guides

1

49

Basic Differentiation

1.6

1.6 The Derivative of Functions with Fractional Exponents

The Derivative of Functions with Fractional Exponents

The derivative of a function f ( x ) with fractional exponent is obtained by applying the chain rule in the following way: d f ( x) dx

[

a

]b

=

a

a f ( x) b

a

] b −1 ⋅ dxd [ f ( x )]

[

For example, the derivative of f ( x ) = x b is equal to a

d b x dx

=

a

a b −1 d x ⋅ x b dx

=

a

a b −1 x ⋅1 b

=

a

a b −1 x b

=

a −b

a b x b

Note that the steps in finding the derivative of a function with fractional exponent is similar to finding the derivative of a function that is raised to a power as discussed in Section 1.4. The following examples illustrate how to obtain the derivative of exponential functions: Example 1.6-1: Find the derivative for the following exponential expressions.

( )

2

b. y = 3x 2

a. y = x 3

(

d. y = 3x 2 + 6 x

m.

(

3 5

2

1

)

1 4

2 7

(

i. y = ( x + 1) 2 x 2 + 3

1

l. y =

x3

)

1 3

x3 2

(x + 1) 3

1

y = (x + 1) ⋅

a. Given

)

( x + 3) 5 k. y = 2

x

)

f. y = 3x 2 + 8

h. y = x x 2 + 1 3

1 2

Solutions:

(

3

1 3

2

(

c. y = 3x 3 + 2 x

e. y = (2 x + 1) 4

( ) + (2x + 1) ( x + 1) y=

g. y = x 2 j.

)

2 5

1 3

1

x7

then y ′ =

2

y = x3

2 2 2 −1 2 −1 2 23 −1 2 2 −3 2 1 2 1 x = x 3 1 ⋅1 = x 3 = x 3 = ⋅ 1 = ⋅ 3 = 3 3 3 3 3 3 3 x 3 x x3

Note that the answer does not necessarily need to be in radical form. We can simply stop when y ′ =

2 − 13 x 3

. However, for review purposes only, the answer to some of the problems are

shown in radical form (see Sections 1.1 and 1.2 from Mastering Algebra – Advanced Level to review the subjects of exponents and radicals). b. Given = 2x

y = ( 3x 2 ) 3 1

1

3

( 3x )

2 2

then

y′

=

1

2x

Hamilton Education Guides

3

=

9 x4

( )

1 3x 2 3

=

2 x/ 3

x/ 9 x

1 −1 3

=

⋅ 6x

=

( )

6 x 3x 2 3

1 −1 3 1

( )

= 2 x 3x 2

1−3 3

( )

= 2 x 3x 2

−2 3

= 2x

1

( 3x ) 2

2 3

9x

50

2 3

Basic Differentiation

1.6 The Derivative of Functions with Fractional Exponents

(

c. Given y = 3x 3 + 2 x =

(

1 3x 3 + 2 x 4

) ( 9x 1− 4 4

(

2

d. Given y = 3x + 6 x =

(

12 3x 2 + 6 x 5

)

−3

1 4

)

then

y ( 2 x + 1) then e. Given = 3 4

=

3 1 ⋅ 2 (2 x + 1) 14

3 1 ⋅ 2 4 2x + 1

=

=

(

)

−5

( )

g. Given y = x 2 =

=

7

1 3

12 x ⋅ 7

( 3x

( 3x

) )

3 5

+ 6x

)

=

(

9x 2 + 2

4 3x 3 + 2 x

12 ⋅ 5 5

x +1 2

3 1 6 (2 x + 1) 4 − 1 4

=

)

2

2 3x 2 + 8 7

(

)

1

12 x ⋅ 7 7

+8

)

=

5 7

2 −1 7

3

+ ( 2 x + 1) 5 = x 3 + ( 2 x + 1) 5

(

)

(

)

= x 2 +1

2 3

+

2 3

(

then

)

2−3 3

4x 2 2 x +1 3

1

(

)

i. Given y = ( x + 1) 2 x 2 + 3

Hamilton Education Guides

1 3

y′

1 −1 4 1

+ 6x

)

)

( 3x

4

2 −1 5 1

+2

)

3

+ 2x

5

5

2

3

(

( x + 1) =

( 3x

)

12 3x 2 + 6 x 5

12 ( x + 1)

=

3− 4 3 (2 x + 1) 4 2

=

2

9x 2 + 2

4

3

=

⋅ 6x

( 3x

then

(

12 x 3x 2 + 8 7

1 2

+8

y′

=

)

5

)

2 −1 7 1

+ 6x

)

)

2−5 5

( x + 1)

3

1 3 (2 x + 1) − 4 2

=



(

)

2

7

7

( 3x



(

)

= x 2 +1

then

y′

2 3

+

1

+8

(

2 5

(

)

1

2−7 7

5

−1

3

(

(

)

= x 2 +1

)

1

2 3

 

3

3 x

+

6 5 5 (2 x + 1)2

(

)

= x 2 +1

+

(x 3

 1

2

=

)

4x 2 2 x +1 3

)

)

3 3 1 2 2 −1 6 2 23 −1 3 x + (2 x + 1) 5 −1 ⋅ 2 = x 3 1 + (2 x + 1) 5 − 1 3 5 3 5

=  ( x + 1) 2 −1 ⋅1 ⋅ x 2 + 3 3  +   2

2

2 −1     2 2 x + 1 3 ⋅ 2 x ⋅ x       3

 = 1 ⋅ x 2 + 1 3  +  

(

12 x 3x 2 + 8 7

=

12 x

=

3− 5 2 6 1 2 1 2 2 −3 3 6 2 −1 6 x + (2 x + 1) 5 = x 3 + (2 x + 1) − 5 = ⋅ 1 + ⋅ 3 3 5 (2 x + 1) 3 5 3 5 x

h. Given y = x x 2 + 1

3 4

(

( 3x

) ( 9x

=

12 3x 2 + 6 x 5

( 6 x + 6) =

=

(

1 3x 3 + 2 x 4

+2 =

3

2

3

+2

2 −1 5

1 2

2

)

2

2 2x + 1

2 7

12 x 3x 2 + 8 7

4

(

4

y ( 3 x 2 + 8 ) then y ′ = f. Given=

=

1 −1 4

−3

3 3 (2 x + 1) 4 −1 ⋅ 2 4

=

y′

) ( 9x

) ( 9x

2 3x 2 + 6 x 5

=

y′

12 ( x + 1)⋅ 5

( x + 1) =

5

(

1 3x 3 + 2 x 4

+2 =

2 5

(

1 3x 3 + 2 x 4

=

y′

)

2

)

then

(

4x 2

)

3 x 2 +1 2

+3

)

1 −1 3

1 3

2 3

+

=

(

)

4x 2 2 x +1 3 3

(x

2

+1

)

2

2 −1 3 1

+

4x 2 3

1 ⋅ 2 x ⋅ ( x + 1) 2  

51

3

x2 +1

Basic Differentiation

=

=

(

1 −1 2  ( x + 1) 2 ⋅ x + 3 2 

(x

)

2

2 ( x + 1)

x2

1 3

+3

j. Given

=

1.6 The Derivative of Functions with Fractional Exponents

1 2

2 x ( x + 1) 2

+

2

(

)

+ 1)

( x + 1) 2

x

then

(

)

then

2

x3

x

2

2

=

3 x 2 (x + 1) 3 −

( x + 1)

2

( x + 1) 3

5(x + 3) 5

m. Given y = (x + 1) ⋅

=

7 2

x

− 4

then y ′ =

2 x3

1 3 ( x +1) 3 4 3

7

1 −1 2

2

+3

)

3x 2

1 1 x7

x

3

=

7

=

Hamilton Education Guides

x +1

7

⋅

1 1

( x + 1) 2

  

(

)

x +1 x

3

1

=

2x 3 33 x + 1

=

)

)

−1

2

(

)

− x 2 +1

1 2

x2

=

−1 x2 x2 + 1

1 1 2 1 −4  2 −  x 3 ( x + 3 ) 5  −  x 3 ( x + 3) 5    5   3

=

4

x3

2(x + 3) 5 3x 3

(

x2 + 1 x2

=

2

x2

5 5 ( x + 3) 4 3

−

25 x + 3 33 x

x4

4

y′

(

−1

2

(x + 1) 3 −

  1 1  +  2 x ⋅ ( x + 1) 2 ⋅   3   x2 + 3

2x 2 2 x +1 2

=

x 2 − x 2 +1

=

2

(x + 1)4

then

x 17 x−

1 2

(

   ⋅ 2 x  ⋅ x  − 1 ⋅ x 2 + 1   

3 x 3−1 ⋅ ( x + 1) 23  −  2 ( x + 1) 23 −1 ⋅ x 3     3  

(x + 1)2

=

1 7 6

3

)

1 3

2

x3

x3

7 x − x +1 ⋅

(x

1

4

4

3

4 x3

x3

=

)

(

=

2  1 1 −1 1 2 2 −1  ( x + 3 ) 5 ⋅ x 3  −  x 3 ⋅ ( x + 3) 5   5   3 

x3

l. Given y =

3

− x2 + 1

x2 + 1

=

1

x3 1 2 ( x + 3) 5 ⋅ − ⋅ 1 5 (x + 3) 54 3 x3

 1 ⋅ x2 + 3 2 

2x x + 1

+

)

x2

y′

  

1 2

x2

=

1 5

⋅ ( x + 1)

3

  1 2   x +1   2

1 2

2

2

=

x2 + 3

2 x +1

=

y′

− x 2 +1

( x + 3)

y=

)

− 2

(

1 2

x

1 2

3

=

2 3

3 x2 + 3

1

k. Given

(

  2x 2 x +3 +   3 1

(x

y=

)

1 3

3x 2

=

3

( x + 1)2

−

=

3 3 x 2 ( x + 1) 32  −  2 x ( x + 1)− 13       3  4

( x + 1) 3

2x 3 33 x + 1

( x + 1) 3 x + 1

1    1 1 −1 1 ⋅ x 7  −  x 7 ⋅ (x + 1) 7     2

x7

=

1 x7

x + 1 − 76 − x 7 2

x7

1

x7 −

=

x +1 1 ⋅ 6 7 x7 2

x7

x +1 7

7 x6 x2

52

2 3

   

Basic Differentiation

1.6 The Derivative of Functions with Fractional Exponents

d of the following functions. dx

Example 1.6-2: Find a.

d  − 23  x  dx  

d.

d 3x 2 + 4 x dx

b.

1 d (x + 1)− 4 dx

=

e.

−2 d x x 2 +1 3 dx

  =  

h.

d 2 x + 3x + 5 dx

)

)

=

k.

d dx

   

= − x

=

(

)

1 d  ( x + 1)− 2 g. dx  x 

j.

(

7 8

−1 d 2 x +5 6 dx

)

(

(

(

)

=

f.

d 3x 3 + 4 dx

=

i.

5 d  (x + 1)(x − 1) 4  =  dx  

l.

d  x 2 dx  ( x + 1)− 13 

1 4

  

=

(

Solutions: d  − 23 x dx 

b.

1 d (x + 1)− 4 dx

1 4

c.

3 2 x ⋅ 8

(x + 1) 3

(

)

7 8

7 3x 2 + 4 x 8

(

(

)

−2 3

7 8

−1

)

)

)

=

=

−2 d x x 2 +1 3 dx

= x 2 +1

)

−1

3

   

=

=

(

−

8

(

3x 2

7

)

⋅

)

8 x3 +1 8

(

7 3x 2 + 4 x 8

⋅ (6 x + 4 )

(

)

(

)

= x 2 +1

7 ⋅ 8

= −2 3

⋅

− 2 −1 2x 2 x +1 3 1 3

Hamilton Education Guides

3

8

1 −1 8 1

⋅ 3x 2

(

)

1−8 3 2 3 x x +1 8 8

=

=

(

)

−7 3 2 3 x x +1 8 8

(x + 1)

7

3

(

d 3x 2 + 4 x dx 6x + 4

(3x

2

+ 4x

)

1 8

=

(

) = 78 (3x

)

= x 2 +1

−2 3

+ 4x

)

7 −1 8 1

⋅ (6 x + 4 )

=

(

7 3x 2 + 4 x 8

)

7 −8 8

⋅ (6 x + 4 )

8

8 3 x2 + 4 x

)

−

2

7(6 x + 4 )

−2 d d 2 x + x⋅ x +1 3 dx dx

(

5

3 x2

8

7 −1 8

1 4

= − (x + 1)− 4

4 ( x + 1) 4 x + 1

) = 18 (x + 1)

=

−1− 4 4

1

= −

4 4 ( x + 1)5

(

1

(

1

1 4

1 1

= − (x + 1)− 4 − 1 ⋅1 = − (x + 1)

1 −1 1 3 d 3 x +1 8 ⋅ x +1 8 dx

=

d 3x 2 + 4 x dx

=

e.

)

1 4

d (x + 1) dx

1

= −

5 4

(x + 1)

1 d 3 x +1 8 dx

=

d.

1 4

= − (x + 1)− 4 −1 ⋅

1

(

=

−2 2 1 2 − 23 −1 d 2 − 2 −1 2 −2 −3 2 −5 −2 ⋅ ( x ) = − x 3 1 ⋅1 = − x 3 = − x 3 = − ⋅ 5 = = 3 3 3 3 3 dx 3 3 3 3x x 2 3 x5 x

a.

= − ⋅

1

d 3 x +1 8 dx

1 4

 3 2 x 5 x +1 

)

(

c.

=

(

(

)

)

= x2 + 1

= x 2 +1

2x 2 x +1 3

−2 −3 3

−2 3

(

⋅1 − x ⋅

)

(

)

− 2 −1 2 2 x +1 3 3

−2 3

−

(

)

−5 2x 2 x +1 3 3

53

Basic Differentiation

=

f.

1

(x + 1)

2 3

(

)

2

d 3x 3 + 4 dx

2x ⋅ 3

−

−1

(

h.

i.

2

)

−4 3

1

=

5 3

(

)

= − 3x 2 ⋅

3

(3x

2

2

− 1 −1

⋅

3

(

d 3x 3 + 4 dx

1 3

2x

−

+4

)

4 3

3

(x + 1)

5

2

) = − 13 (3x 3x 2

= −

3

(3x

3

+4

1

=

3

(x + 1)

3

+4

)

− 1 −1

= −

)

4

2

2

3 1

⋅ 9x 2

−

2x

(

) (x + 1) 3

3 x2 +1

= −

2

2

(

9x 2 3x 3 + 4 3

)

−1−3 3

3 x2

(3 x + 4) 3

3

3 x3 + 4

1  d     −1   −1 d − 1 −1 d −1    x ⋅ dx ( x + 1) 2  − ( x + 1) 2 ⋅ dx ( x )  x ⋅ − 2 ( x + 1) 2 ⋅ dx ( x + 1) − ( x + 1) 2 ⋅1      =  =  2 2  x x 

x2

(

(x + 1)

3

 x − 1 −1  −1 − 2 ( x + 1) 2 1 ⋅1 − ( x + 1) 2  

d 2 x + 3x + 5 dx

=

(x + 1) 1 3

= − 3x 2 3x 3 + 4

=

1

= − 3x 3 + 4

3

1 d  ( x + 1)− 2 g. dx  x 

1.6 The Derivative of Functions with Fractional Exponents

)

1 4

=

)

(

−1− 2 1 x (x + 1) 2 − (x + 1)− 2 2 x2

(

1 −1 1 2 d 2 x + 3x + 5 x + 3x + 5 4 ⋅ 4 dx

=

)

(

−

3

− 1 2 x + 3 x + 5 4 (2 x + 3) 4

2x + 3 ⋅ 4

=

1

(x2 + 3x + 5)

3 4

=

−

) = 14 (x2 + 3x + 5) −

−3 1 x ( x + 1) 2 − ( x + 1)− 2 2 x2

1 1 4 1

⋅ (2 x + 3)

2x + 3 1 ⋅ 4 3 4 2 x + 3x + 5

=

)

(

=

=

(

1 2 x + 3x + 5 4

)

1− 4 4

⋅ (2 x + 3)

2x + 3 4 4  x 2 + 3 x + 5   

3

5 5 5 5 5 d  (x + 1)(x − 1) 4  = (x − 1) 4 d (x + 1) + (x + 1) d (x − 1) 4  = (x − 1) 4 ⋅1 + (x + 1)⋅ 5 (x − 1) 4 −1 ⋅ d (x − 1)  4 dx dx dx dx           5 1 5− 4  5 5 5 1  5 5 5   = (x − 1) 4 + (x + 1) ⋅ (x − 1) 4 − 1 ⋅1 = (x − 1) 4 + (x + 1) ⋅ (x − 1) 4  = ( x − 1)4 + ( x + 1)( x − 1)4

4



j.

(

)

−1 d 2 x +5 6 dx

x 3

= − ⋅

k.

d dx

(x

= −

1 2

+5

(

)

)

7 6



(

)

(

− 1 −1 d 1 2 x +5 6 ⋅ x2 + 5 6 dx

x 3 6

= − ⋅

(

(x

1 2

+5

4



= −

)

7

)

) = − 16 (x

(

2

+5

4



)

− 1 −1

6 1

⋅ 2x

= −

(

2x 2 x +5 6

)

−1− 6 6

= −

(

)

−7 x 2 x +5 6 3

x 2

3 x +5

)

6

x2 + 5

(

)

(

)

(

)

(

)

1 1 1 1 −1 1  2  3 2   d 53   53 d 2 d 2 3 3 −1   3 1 2 x  + x x +1 4 ⋅ x +1 4  =  x 2 +1 4 ⋅ x 5  + x 5 x +1  x 5 x +1 4  =  x +1 4 dx dx dx 5 4          

3  1 1 −1 1− 4  1 3 −1   3 3− 5  5  3 2  3 2 1 x x ⋅ 2 2  ⋅ x 2 +1 4  =  x + 1 4 ⋅ x 5 1  +  x 5 x + 1 4 1 ⋅ 2 x =  x + 1 4 ⋅ x 5  +   4  5     5   4  

(

)

Hamilton Education Guides

(

)

(

)

(

)

54

Basic Differentiation

=

l.

1.6 The Derivative of Functions with Fractional Exponents

8  −2 3 1   3 x 5 ⋅ x2 + 1 4  +  x 5 ⋅ x2 + 1 − 4    5   2    

(

d  x 2 dx  ( x + 1)− 13 

)

(

  = d  dx 

)

1 1 d   x 2 ( x + 1) 13  2  2 d   = ( x + 1) 3 dx x  +  x dx ( x + 1) 3 

1 2 1 1 1 1  1 x2 1 d 1    = (x + 1) 3 ⋅ 2 x  +  x 2 ⋅ (x + 1) 3 −1 ⋅ (x + 1) = 2 x (x + 1) 3 +  x 2 ⋅ (x + 1) 3 − 1 ⋅1 = 2 x ( x + 1)3 + ( x + 1)− 3



3

 

dx



3



3



Example 1.6-3: Find y ′(1) in example 1.6-1 a through i . Solutions: In Example 1.6-1 we obtained the derivative of the exponential functions a through e to be equal to the following: a. Given y ′ =

2

then y ′(1) =

3

3 x

2x b. Given y ′ = 3

c. Given y ′ =

9x 4

9x 2 + 2

4

(3x

3

+ 2x

5

5

(3x

2

)

+ 6x

3

3 1

)

3

2 2x + 1

= 0.67 6

2 ⋅ −3 3

= −3

9 ⋅ 14

(3 ⋅1

4

3

5

2 2 ⋅1 + 1

2

=

9

3

)

+ 6 ⋅1

3

3 4

= −

0.33

=

)

+ 2 ⋅1

(3 ⋅1

3 4

9

12(1 + 1)

5

6

= −

9 ⋅ 12 + 2

then y ′(1) =

then y ′(1) =

4

2 3

=

then y ′(1) =

3

12(x + 1)

d. Given y ′ =

e. Given y ′ =

then y ′(1) =

2 3

2 2 +1

11

=

4 3

5

=

=

6 2.08

11 125

24 5

5 729

3 4

2 3

= −2.88

=

=

0.25

=

11 3.343

= 3.29

=

24 5 ⋅ 3.74

24 5 ⋅ 729 3

2⋅3

0.25

0.2

=

3 2 ⋅1.32

= 1.28

= 1.14

Section 1.6 Practice Problems - The Derivative of Functions with Fractional Exponents 1. Find the derivative of the following exponential expressions. 1

( ) y = (x ) − (3 x − 1) 1

d. y = 2 x 2 + 1 8 g.

j. y =

( ) y = (2 x + 3 x )

b. y = 4x 3

a. y = x 5

1

3 2

e. 1 3

x +1 2

x3

Hamilton Education Guides

1 2

3

1

h. y = x 2 (x + 1) 8 k.

(x + 1) y= 2

x2

1 2

1

c. y = (2 x + 1) 3 3 5

( ) y = (x + 1)

f. y = x 3 + 8 i.

l. y =

2 3

2 5

3

1

+x2

(x + 1)2 1

x3

55

Basic Differentiation

2. Use the

d dx

notation to find the derivative of the following exponential expressions.

2

a.

d  15 x dx 

d.

−1 d 3 x +1 4 dx

g.

d dx

j.

(

   

)

(

1.6 The Derivative of Functions with Fractional Exponents

=

b. =

=

1 d  ( x − 1) 2  = dx  x 2     d  3 1 h. x ⋅ 1 dx   x 2 +1 2

e.

)( )  =

 3 2  x +1 x 

1 d (x − 1) 2 dx

1 3



1 1 d  (x − 1) 2 (x + 1) 3  =  dx  

Hamilton Education Guides

(

k.

d dx

(

)

)

1  3 2 x x +1 2   

  =  

(

)

c.

1 d 2 x +1 3 dx

f.

d 3 x + 2x dx

i.

 d  x5 dx  3  x + 1

l.

d dx

(

(

)

)

(

= 1 8

2 3

=   =  

)

 3 2 −1  x x +1 3  =  

56

Basic Differentiation

1.7

1.7 The Derivative of Radical Functions

The Derivative of Radical Functions

In this section finding the derivative of radical expressions and the steps as to how it is calculated is discussed. The derivative of radical functions is found by using the following steps: 3

x 3 as x 2 .

First - Write the radical expression in its equivalent fraction form, i.e., write

Second - Apply the differentiation rules to find the derivative of the exponential expression. Third – Change the answer from an expression with fractional exponent to an expression with radical expression (optional). The following examples show the steps in solving functions containing radical terms. Students who have difficulty with simplifying radical expressions may want to review radicals addressed in Chapter 4 of “Mastering Algebra - An Introduction”. Example 1.7-1: Find the derivative for the following radical expressions. a. f ( x ) = x 3 + 1

b. f ( x ) = x 2 + 3x + 1

c. f ( x ) = 2 x 5 + 1

d. f (u) = 5 u3 + 3u

e. f (t ) = t 2 + t + 1

f. g( x ) = x 2 x 3 + x − 5

g. h( w) = 3 w 2 + 1

h. f ( z ) = 4 z 3 − z 2 + z

i. f ( x ) =

j. f ( x ) = m. g(u) =

x

k. r (θ ) =

x2 −1 u −1

n. h(t ) =

u +1

Solutions: a. Given f ( x ) = f ′( x ) =

(

(

)

θ +1

(

x +1

r3

l. p(r ) =

θ 2 +1 3 2

t +1

o. s(r ) =

t3

r3 −1 r2 −1 r −1

1

= x 3 + 1 2 , then

(

)

)

(

)

1 −1 −1 −1 1 3 3 1 3 3 x + 1 2 ⋅ 3x 2 = x 2 x 3 + 1 2 = x + 1 2 ⋅ 3 x 3−1 = 2 2 2 2

b. Given f ( x ) = f ′( x ) =

x3 +1

3

1 2

x 2 + 3x + 1

(

)

x2

(x + 1) 3

1 2

3 2

=

x2 x3 + 1

1

= x 2 + 3x + 1 2 , then

) (

)

(

)

1 −1 −1 1 2 1 2 x + 3 x + 1 2 ⋅ 2 x 2−1 + 3 x1−1 = x + 3 x + 1 2 ⋅ (2 x + 3) = 2 2

(

)

(

2x + 3

)

2 x 2 + 3x + 1

1 2

=

2x + 3 2 x 2 + 3x + 1

1

c. Given f ( x ) = 2 x 5 + 1 = 2 x 5 + 1 2 , then

(

) (

)

(

)

1 −1 −1 1 1 2 x 5 + 1 2 ⋅ 2 × 5 x 5−1 + 0 = 2 x 5 + 1 2 ⋅10 x 4 = f ′( x ) = 2 2

5

(

1/ 0/ x 4

)

2/ 2 x 5 + 1

1 2

=

5 x4 2 x5 + 1

3

d. Given f (u) = 5 u3 + 3u = u 5 + 3u , then f ′(u) =

3 −2 3 −2 3 53 −1 u + 3u 1−1 = u 5 + 3u 0 = u 5 + 3 = 5 5 5

Hamilton Education Guides

3 5u

2 5

+3 =

3 5

5 u2

+3

57

Basic Differentiation

1.7 The Derivative of Radical Functions 1

e. Given f (t ) = t 2 + t + 1 = t 2 + (t + 1) 2 , then 1 2

1 1 (t + 1) 2 −1 ⋅ t 1−1 2

f ′(t ) = 2t 2−1 +

1

1

= 2t + (t + 1)− 2 ⋅ t 0 = 2t +

(

)

2(t + 1)

1

= 2t +

1 2

2 t +1

1

f. Given g( x ) = x 2 x 3 + x − 5 = x 2 x 3 + x − 5 2 , then

(

)

(

1   1 g ′( x ) = 2 x 2−1 x 3 + x − 5 2  +  x 3 + x − 5    2

(

)

)

(

1

= 3x 2 + 1 x 2 = 2 x x 3 + x − 5 2 +

(

)

) ⋅ (3x 1 −1 2

3−1

(3x + 1) x 2 (x + x − 5) 2

2

3

)

(

 + x1−1 + 0  x 2 = 2 x x 3 + x − 5 

= 2 x x3 + x − 5 +

1 2

)

1 2

(

+

(

)

−1 1 3 x + x−5 2 2

)

x2 3 x2 + 1

2 x3 + x − 5

1

g. Given h( w) = 3 w 2 + 1 = w 2 + 1 3 , then h ′( w) =

(

) (

1 −1 1 2 w + 1 3 2 w 2−1 + 0 3

4 3

h. Given f ( z ) =

(

= z3 − z2

) (3z

2 −1

(

1 −1 4

1

3−1

− 2z

)

1 4

3

+z,

(z

)+ 1 =

− z2

)

= x 2 +1

x +1

− 1 −1

2

−1

2

x

x −1

2

1 2

1

=

−

(x − 1) 2

1 2

x 2 −1 −

1

(x2 − 1) (x2 − 1) 1 2

Hamilton Education Guides

(x − 1) ⋅ (x − 1) (x − 1) 1 2

2

x2

= −

=

2

4

3

3

(w

2

+1

2

− 2z

)

2

) + 1 = 3(3z − 2 z ) + 1 = 3z 4 (z 4(z − z ) 2

3 2 4

3

⋅ 2/ x =

−x

(x + 1) 2

(x − 1) = 1 2

4

−x

=

3 2

2

− 2z

3

−z

)

2 3

+1

2

( x + 1) 2

−x

=

3

(x + 1) 2

(x − 1)

1 +1 2

=

−

1

(x2 − 1)

3 2

(

(x − 1)

)

2

 1 −1 −  x 2 − 1 2 ⋅ 2/ x 2    2/

(x − 1) − x (x − 1) 1+1 2 2

2

=

x 2 −1

1 2

=

x 2 −1

− x2

1 2

2

1

2

−3

2

x 2 −1 2

) (3z

)

 − 0 ⋅ x 

2 −1

 



(x − 1)

=

1 −1 2

2

)

x2 +1

, then

1 2

2

1 2

2

f ′( x ) =

x

=

(x − 1)  1  1 ⋅ (x − 1)  −  (x − 1) (2 x  2  2

(

3 w2 +1

2w

=

2 3

, then

−3

2

2 −1

2

j. Given f ( x ) =

1 2

2

2w

4

(x + 1) 1 (2 x + 0) = − 12/ (x + 1) f ′( x ) = − (x + 1) 2 2

=

⋅ 2w

then

3

(

1

=

−2

2

z − z2 + z

1 3 z − z2 f ′( z ) = 4

i. Given f ( x ) =

) = 13 (w + 1)

= −

1 2

2

x 2 −1

1

(x − 1) 2

3

= −

2

1 2

−

x2

(x − 1) 2

x 2 −1

x/ 2/ − 1 − x/ 2/

=

(x − 1) 2

1 2

x 2 −1

1

(x − 1) 2

x2 −1

58

1 2

Basic Differentiation

θ3 +1

k. Given r (θ ) =

r ′(θ ) =

=

=

(

)  1 + 0 )(θ + 1)  −  (θ  2

(

+1

1 2

(

2θ θ + 1

=

θ 2 +1

, then

) (2θ

2

+1

)

2 θ 2 +1

 3−1 3r 

1 2

1 −1 2

2 −1

)

)(

(

 + 0 θ 3 +1  

r3

(

)

2 θ 2 +1

=

r3

1 2

=

3−1

(

3r 5

)

3r 5 − 6r 2

)(

)

2 r 3 −1 r 3 −1

1 2

=

(

)

3r 2 r 3 − 1

1 2

2 r 3 −1

r 3 −1

6θ 4 + 6θ 2 − 2θ 4 − 2θ

=

(

) (θ

2 θ 2 +1

1 2

2

)

+1

=

2

2

3

(

1 2

(

−

)

2 r 3 −1

)

1 +1 2

(

3r 5

(

)

2 r 3 −1

(

)

r 3 −1

3r 2 r 3 − 2

=

(

)

1   2 3  −1 2 − 1 r 3 − 1 2 3r 2 ⋅ r 3 r r − 3 1        2 

=

r 3 −1

3r 2 r 3 − 2

(

3

)

2 r 3 −1

3 2

)=

( ) 2(r − 1)

6r 2 r 3 − 1 − 3r 5 1 2

3

=

(

)

2 r 3 −1

3

)=

(

(

(

3r 2 r 3 − 2

2 r 3 −1

)

)

2 r 3 −1

=

r 3 −1

3r 2 r 3 − 2

(

6r 5 − 6r 2 − 3r 5

1 2

1 2

r 3 −1

)

r 3 −1

1

u +1

1 −1 1  1 1 −1 1 1  2 (u − 1) 2 (u + 1) 2  −  2 (u + 1) 2 (u − 1) 2      u +1

  (u − 1) 12 −   2(u + 1) 12   u +1

   

4/

=

1 2

4/ (u + 1) (u − 1) (u + 1)

Hamilton Education Guides

1 2

(u − 1) 2 , then = 1 (u + 1) 2

u −1

 (u + 1) 12   2(u − 1) 12 

)

θ 2 +1

)

 −0  

r 3 −1

(

(

2 θ 2 +1

2

1 −1 2

3

−

)

θ 2 +1

3

 

1 2

(

 1 2 −1  − θ +1 2  2

, then

1 2

3

1 2

)

1 2

6θ 2 θ 2 + 1 − 2θ 4 − 2θ

3 2

2

(r − 1) (r − 1)  −  12 (r − 1) (3r 3

)

)

 2 2 3θ θ + 1 

) = θ (2θ + 3θ − 1) = θ (2θ + 3θ − 1) (θ + 1) θ + 1 (θ + 1) 2/ (θ + 1)

(

=

r3 −1

(

)

2/ θ 2θ 3 + 3θ − 1

=

1 +1 2

(

2θ θ 3 + 1

−

θ 2 +1

4θ 4 + 6θ 2 − 2θ

(

)

3θ 2 θ 2 + 1

)

3

m. Given g(u) =

=

1 2

θ 2 +1

=

=

2

 

(

g ′( u) =

(θ

2

3r 2 r 3 − 1

=

θ 3 +1

=

θ 2 +1

 3−1  3θ 

l. Given p(r ) =

p ′( r ) =

1.7 The Derivative of Radical Functions

1 2

=

1 1  1 1 −1 −1  2 (u − 1) 2 (u + 1) 2  −  2 (u + 1) 2 (u − 1) 2      u +1

2(u + 1) − 2(u − 1)

=

1 2

4(u − 1) (u + 1) u +1

=

1 1 2

(u + 1) (u + 1) (u − 1)

4

2/ u/ + 2 − 2/ u/ + 2

1 2

1 2

1 2

4(u − 1) (u + 1) u +1

=

(u + 1)

1 2

=

1 2

1

4(u − 1) (u + 1) 2 u +1

1 u+1

u −1

59

Basic Differentiation

n. Given h(t ) =

h ′( t ) =

1.7 The Derivative of Radical Functions

3 2

t +1 t3

(

1 −1 3

t 2 −1

, then

3 2

)

(

)

(

)

1  2 1+ 3 2 −2  3 1 2  t 2 t +1 3  −  t 2 t +1 3    3   2

=

t3 5

2t 2 −

(

r2 −1

r 2 −1

=

r −1

1

(r − 1) 2

(

=

3r 2 − 4r + 1 1

2(r − 1) (r − 1) 2

=

)

(

)

3t 3 t 2 + 1

(

)   

)

=

2 3

=

2t 2 t − 3t 3

3

2t

(

(

5 2

)

3 t 2 +1

(

2 3

)

1 3 12 2 t t +1 3 −2 1

t3

)

9 t t 2 +1 2

(t

2

)

+1

2

, then

)

(

(

1+ 2 9 12 2 t t +1 3 3 2

5

)

 2r 2−1 − 0 (r − 1) 12  −  1 (r − 1) 12 −1 r 2 − 1      2  r −1

 2 2r (r − 1) 12  −  r − 1    1  2(r − 1) 2 = r −1

2 3

 1 1  −  3 t 2 t 2 +1 3     2    t3

2t 2 −

=

t3

o. Given s(r ) =

)

t3

2 3

2 3

2

=

)

( ) ( 3(t + 1)

1 3 12 2 t t +1 3 ⋅3 t 2 +1 2

=

(

)

1 1 2  3 3 1 2 −2  t + 1 3 ⋅ 2t  t 2 −  t 2 t + 1 3   3  2  

=

5   2t 2   3 t 2 + 1

(

)

(

)

1  3  3 3 −1 + 0  t 2 −  t 2 ⋅ t 2 +1 3    2 

t3

(

s ′( r ) =

2

) ⋅ (2t

1 2  t +1  3

1 3

(t + 1) =

=

 2r (r − 1) 12   1 

=

(

)

1  1   −1 2 2r (r − 1) 2  −  2 (r − 1) 2 r − 1  r −1

(

) 

  2  −  r −1 1     2(r − 1) 2 r −1

 

(

)

4r (r − 1) − r 2 − 1

=

1

2(r − 1) 2 r −1

4r 2 − 4r − r 2 + 1 1

=

2(r − 1) 2 r −1

3r 2 − 4r + 1

2(r − 1) r − 1

Example 1.7-2: Use the chain rule to differentiate the following radical expressions. a. d

  dx 

x−

  dx 

1   x 

b. d

 x3    dt  x + 1   

d. d

e. d

x + 2  3x + 1 

 3 x2

dx 

  dx 

c. d

+ x −2  

x 2  x 2 + 1 

f. d 5 x3 + 1 dx

Solutions: 



d  12 1  d  12 − 12  1 1 −1 1 − 1 −1 d 12 d − 12 d  1  x − 1 = x − x = x2 + x 2 x −x a. =  x −  =   2 2 dx dx dx  dx  dx  x  x 2  

Hamilton Education Guides

60

Basic Differentiation

=

b.

1.7 The Derivative of Radical Functions

1 − 12 1 − 32 x + x 2 2

1

=

1

+

1 2x 2

d  x+2 2   dx  3 x + 1 

=

1

1

+

2 x

1  x+2 2   2  3x + 1 

=

 3/ x/ + 1 − 3/ x/ − 6  1   = = 1 1  2    x+2 2  x + 2  2  ( 3 x + 1) 2 2    3x + 1   3x + 1  1

= −

= −

c.

5 2 (x + 2)

2

(

1 2

x+2

(

)

5

)

2x x +1

1 2

=

)

( x + 1) 2

2

( x + 1)

=

1 2

2

(

)

1 −  x 2 +1  2/

−1

2

) ( x + 1) ( x + 1)

(

1 2

2

 ⋅ 2/ x  ⋅ x 2 

x3

1 2

2

=

1 2−

x 3 + 2x

( x + 1) 2

3

=

(x

(

x x2 + 2 2

+1

)

2x x

1  x+2    2  3x + 1 

=

2

     1

 5 (3 x + 1) 2 5  = − = − 1 1  2 (x + 2 ) 2 ( 3 x + 1) 2   x+2 2 2 2  ( 3 x + 1)  3x + 1 

 −5   ( 3 x + 1) 2 

1 2

 [ 1 ⋅ ( 3 x + 1) ] − [ 3 ⋅ ( x + 2 ) ]  ( 3x + 1) 2  

−1

( 3x + 1)

= −

2− 1

2

5 2 (x + 2)

1 2

( 3x + 1)

5

= −

4 −1 2

2 (x + 2 )

1 2

3

( 3x + 1) 2

5

1 2

(

)

d 2 d x − x2 x 2 +1 dx dx x 2 +1

=

(

)

2x x 2 +1

( ) ( x + 1)

1 2−

(

1 2

(

)

(

1  1 2−

=

x3 x 2 +1

x 2 +1

)

2x x 2 + 1

) ( 2x 1 −1 2

2  x +1 2

2 −1

x 2 +1

(

)

2x x 2 +1

−1

2

=

1 2−

x3

( x + 1) 2

2

1 2

x 2 +1

=

2x 3 + 2x − x 3

=

1 2

x 2 +1

x 3 + 2x

( x + 1) ( x + 1) ( x + 1) 2

1 2

2

2

1 +1 2

=

x 3 + 2x

( x + 1) 2

3 2

)

2

x +1

   d 32   32 d 3 32 − 1   32  3  ( ) ( ) ( ) x x x 1 + ⋅ x x + ⋅ − x + 1 1      −  x ⋅1   dx dx 2 d  x 3  d  x2          d. =  =  =   2 2 dt  x +1  dt  x +1  (x + 1) (x + 1)    

=

3 3 12 x (x + 1) − x 2 2 (x + 1) 2

=

Hamilton Education Guides

3 3 12 + 1 3 12 x + x −x2 2 2 (x + 1) 2

=

3 3 32 3 12 x + x −x2 2 2 (x + 1) 2

=

3  3 3 1  − 1 x 2 + x 2 2 2 

(x + 1) 2

=

)

 + 0  x2 

2x x 2 +1 − x 3

=

x 2 +1

1

x + 2 (3 x + 1) 3 x + 1

x 2 +1

2x x 2 +1

=

x

(

d  x+2  ⋅   dx  3 x + 1 

2 (x + 2)

2

2

2 x

+

5

= −

−1

= −

( 3x + 1) 3

d = dx

 d  x   dx  x 2 + 1  2

2

( 3x + 1) ( 3x + 1) 2

1

=

2 x3

1 −1

1

d  x+2    dx  3 x + 1 

=

3 2x 2

1 32 3 12 x + x 2 2 (x + 1) 2 61

Basic Differentiation

3

=

e.

1

x 2 + 3x 2 2 (x + 1) 2

3

1

x 2 + 3x 2

=

2 (x + 1)

x3 + 3 x

=

2

2 (x + 1)

x x +3 x

=

2

2 (x + 1)

=

2

x ( x + 3)

2 ( x + 1) 2

2−3  d 3 2 d  23 2 −1 d 23 d − 2 2 23 − 1 2 3 −2  − 2 −1  x + x− 2  = x + x 2 − x x x − 2 x − 3 = x 3 − 2 x −3 = =  x +x  =   dx  dx  3 dx dx 3 3  

=

f.

1.7 The Derivative of Radical Functions

2

−

1 3x 3

d 5 3 x +1 dx

=

2 x

2

=

3

3

3 x

(

)

d x3 +1 dx

=

)

(

3x 2 x 3 + 1 5

2

−

x3 1 5

−4 5

=

(

3x 2

(

)

1 3 x +1 5

=

)

5 x3 +1

4 5

1 −1 5

⋅

(

3x 2

=

(x

5

5

) = 15 ( x + 1) ( 3x

d x3 +1 dx

3

+1

)

−4

3

5

3 −1

+0

) = 15 ( x + 1) 3

−4 5

⋅ 3x 2

4

Example 1.7-3: Use the chain rule to differentiate the following radical expressions. a. y = x 3 x 2 − 1

b. y = 3x 2 + x + 1

c. y = x 3 + x 2 + 1

d. y = 5 x 5 + x 2 − 1

e. y = x 2 − 1 ⋅ x + 1

f. y = x ( x + 1)3

h.

i. y = (x − 1) 3 3 x 5

g. y =

(x

2

)

+ 5x − 1

5

Solutions:

(

)

y = x2 3 x + 1

1

a. Given y = x 3 x 2 − 1 = x 3 x 2 − 1 2 , then y′



(

)

(

)

1

(



)

 1 2 x −1   2

= 3x 2 ⋅ x 2 − 1 2  +  1− 2   4 2 + x x −1 2   

1 −1 2

(

(

)

= 3x 2 x 2 − 1

)

  ⋅ 2 x ⋅ x 3  = 3 x 2 x 2 − 1   1 2

(

)

+ x 4 x 2 −1

−1

2

1 2

(

)

  2x 4 2 x −1 +   2

(

)

( x − 1) 2

)

 2 2  = 3 x x − 1  

x4

1

= 3x 2 x 2 − 1 2 +

(

1 −1 2 1

1 2

= 3x2 x2 − 1 +

1 2

   x4 x2 − 1

1

b. Given y = 3x 2 + x + 1 = 3x 2 + (x + 1) 2 , then y′

1 −1

1 2

= (3 ⋅ 2) x 2 − 1 + (x + 1) 2

1 2

= 6 x + (x + 1)

(

)

1− 2 2

1 2

1

= 6 x + (x + 1) − 2 = 6 x +

1 2 (x + 1)

1 2

= 6x +

1 2 x +1

1

c. Given y = x 3 + x 2 + 1 = x 3 + x 2 + 1 2 , then y′

=

(

)

1 3 x + x 2 +1 2

1 −1 2

Hamilton Education Guides

(

)

(

)

 3x 2 + 2 x  3  x + x 2 +1 ⋅ 3x 2 + 2 x =    2  

1 −1 2 1

(

)

 3x 2 + 2 x  3  x + x2 + 1 =   

2



1− 2 2

62

Basic Differentiation

1.7 The Derivative of Radical Functions

(

)

 3x 2 + 2 x  3  x + x 2 +1 =   2





−1

2

 3x 2 + 2 x  ⋅ =   2



)

(



(x

1 3

)

+ x 2 +1

3x2 + 2x

=

1 2

2 x3 + x2 + 1

1

d. Given y = 5 x 5 + x 2 − 1 = x 5 + x 2 − 1 5 , then y′

=

(

)

1 5 x + x 2 −1 5

1 −1 5

(

(

)

 5x 4 + 2x  5  x + x 2 −1 =   5





(

)

)

 5x 4 + 2x  5  x + x 2 −1 ⋅ 5x 4 + 2 x =    5   −4 5

 5x 4 + 2 x  ⋅ =   5



(

)



1

(x

1 5

)

+ x 2 −1

5





5x 4 + 2x

=

4 5

(

)

 5x 4 + 2 x  5  x + x 2 −1 =  

1 −1 5 1

5

5

(x

5

+ x2 −1

)

1− 5 5

4

1

e. Given y = x 2 − 1 ⋅ x + 1 = x 2 − 1 2 ⋅ (x + 1) 2 , then y′

1

(

)

1 −1 2 ⋅ 2x ⋅

(

)

−1

=  x 2 −1 2 

= x x 2 −1 

2

1

1 −1

 

(x + 1) 2  + (x + 1) 2  

(

(

)

1  1 1 ⋅ (x + 1) 2  +  (x + 1) − 2 ⋅ x 2 − 1   2

  1  ⋅ x 2 − 1 ⋅ x + 1 +    2 x +1  x 2 −1 



1

= x ⋅  

)

1 2

1 2

  

⋅ x 2 −1

( x − 1)  2/  2/ x

  

1 −1 2 1





=  x ⋅ 

x x +1

=

2

= 

+

x2 −1

1

( x − 1) 2

1 2

(

)

1  1 1 1 ⋅ (x + 1) 2  +  (x + 1) 2 − 1 ⋅ x 2 − 1   2

  1 1 ⋅ (x + 1) 2  +    2 ( x + 1)

1 2

(

)

⋅ x 2 −1

1 2

1 2

   

x2 −1 2 x +1

1

f. Given y = x ( x + 1)3 = x 2 (x + 1) 3 , then y′

1

1 −1

=  x2 2

1   ⋅ (x + 1) 3  + 3 (x + 1) 2 ⋅ x 2    

 (x + 1) 3

= 

2



 (x + 1) 3

1 −1  1+ 3

1 2  (x + 1) x 2    

⋅x2

=  

2

x

−1

2

1   2  + 3 ( x + 1) x 2    

 (x + 1) 3 ⋅ 1  + 3 ( x + 1) 2 x 12  ( x + 1) 3 + 3 x ( x + 1) 2 =   =  1  

g. Given y = y′

=

 x2  

2

(

)

2

x + 5x − 1

(

)

5 2 x + 5x − 1 2

= h. Given

5 −1 2

5

(

y = x2 3 x + 1

)

5 =  x 2 + 5 x − 1   

⋅ ( 2 x + 5) =

5 ( 2 x + 5) ⋅ x 2 + 5x − 1 2

(

2 x



)

3 2

=

1 2

(

)

5 ( 2 x + 5) ⋅ x 2 + 5x − 1 2

5 ( 2 x + 5) ⋅ 2

5

= x 2 + 5 x − 1 2 , then

(

(x

)

2

)

+ 5x − 1

5 −1 2 1

3

=

=

5 ( 2 x + 5) ⋅ x 2 + 5x − 1 2

(

(

5 ( 2 x + 5) x 2 + 5 x − 1 2

)

)

5−2 2

x 2 + 5x − 1

1

= x 2 (x + 1) 3 , then

Hamilton Education Guides

63

  

Basic Differentiation

1.7 The Derivative of Radical Functions

1  2 1 1  x 1  = 2 x ⋅ (x + 1) 3  +  (x + 1) 3 − 1 ⋅ x 2  = 2 x (x + 1) 3  +  (x + 1)

y′

 3



1

= 2 x (x + 1) 3 +



2 x2 (x + 1) − 3 3

1

= 2 x (x + 1) 3 +

( )

i. Given y = (x − 1) 3 3 x 5 = (x − 1) 3 x 5 5



 5  



3 = 3 x x 2 ( x − 1) 2 +

Example 1.7-4: Find a.

x+

d.

2x + 1 = y 2

g.

5 −1

= 3 (x − 1) 2 ⋅ x 3  +  x 3 3

y′

  3



1 3

x2 1 ⋅ 3 (x + 1) 23

1 −1 3 1



1

= 2 x (x + 1) 3 +

= 2x 3 x + 1 +

1− 3 x2 (x + 1) 3 3

x2 3 3 ( x + 1) 2

5

= (x − 1) 3 x 3 , then  5

5



 ⋅ (x − 1) 3  

= 3x 3 (x − 1) 2  +  x 3

5−3 3

2  53 2  5 3  3 x (x − 1)  +  x 3 (x − 1)     3



 





(x − 1) 3  =

53 2 x ( x − 1) 3 3

dy by implicit differentiation. dx 3

x2 + 5 y3 = 2

x +1 = y 3

f.

4

x2 y2 = x

x3 + y 3 = 2

i.

2

5 2

e. h.

7 x y 2 = 3x 2

c.

x2 + y2 = x

b.

y = 10

x 2 −1 = x y

Solutions: d dx

a.

(

)

x+ y =

d (10) dx

;

1 d  12 x + y2  = 0   dx  

1 −1 1 −1 1 12 − 1 d 1 1 −1 d x ⋅ (x ) + y 2 ⋅ ( y ) = 0 ; x 2 ⋅1 + y 2 ⋅ y ′ = 0 2 2 dx dx 2 2

;

1

1 1 − y y −y2 1 1 1 −1 1 − 1 1 y′ 1 y′ 1 1 ; y 2 y ′ = − x 2 ; ⋅ 1 = − ⋅ 1 ; 2/ y 2 ⋅ ⋅ 1 = 2/ y 2 ⋅ − ⋅ 1 ; y ′ = 1 ; y ′ = ; y′ = − x 2 2 / 2 2 2 2 2/ 2 x y x x2 y2 x2

(

d  2 d 2 (x ) ; d x 2 + y 2  x +y = dx dx   dx

b.

;

c.

2x + 2 y y′

(

2 x2 + y2

)

1 2

=1

(

) 2

1 2=1

; 2x + 2 y y′ = 2 x +

(

1 2 x + y2 2

;

)

1

y2 2

( ) +(y )

 d 3 2 5 3  d (2) ; d  x 2  x + y = dx  dx   dx

1 3

1

3 5

;

(

)

(

)

−1 1 2 d 2 d 2 x + y 2 2 (2 x + 2 y y ′) = 1 y  =1 ; x + 2 dx   dx

1 −1  2 

2 x2 + y2 y′ = 2y

)

1 2

− 2x

; y′ =

x2 + y2 − x y

3  2 2 −1 d 3 3 −1 d d  23 =0 ;  x + y 5  = 0 ; x 3 ⋅ (x ) + y 5 ⋅ ( y ) = 0 dx dx 3 5 dx    2

1 2 10 5 y 2 3y′ − 10 y 5 2 −1 3 −2 3 −2 − 2 − 13 −2 3 y ′ = −10 y 5 ; y ′ = y′ = − x 9 x = ; ; ; ; x 3 ⋅ 1 + y 5 ⋅ y′ = 0 ; y 5 y ′ = 1 2 1 5 3 3 5 93 x 9x 3 5 y 5 3x 3

Hamilton Education Guides

64

Basic Differentiation

d dx

d.

(

1.7 The Derivative of Radical Functions

( );

)

1 d (2 x + 1) 2 = 2 y ⋅ d ( y ) dx dx

d 2 y dx

2x +1 =

; 2 y y ′ = x (2 x + 1)

−1

2

;

(

)

1 2 x +1 5

−4 5

⋅ 2x =

; y′ =

f.

(

2 − 13 ⋅ y′ y 3

2 xy 2 + 2 x 2 y y ′

5

(

)

4 x 2 y 2 −1

(

5

(x

)

2 x 2 y 2 −1

3 4

−

+1 4

)

1 4=1

)

( )

; y′ =

( )

d 7 2  d d 3x 2 ; xy 2  xy  = dx  dx  dx

g.

;

y 2 + 2 y y ′x 7

( )

6

xy 2 7

= 6x

3 4

(

)

−1

2

+

;

)

5

(

1 2 2 x y 4

;

) = 1 ; xy

(

2 x2 y2 −1 2

1 7

(

=0 ;

(

)

Hamilton Education Guides

)

−1

)

d  2 d  d (x y ) x 2 −1 ;  x −1  = dx dx   dx

(

)

1 2 x +1 5

−4

2 y′

;

1 3y 3

=

( )

d  2 23 − 1 d d 2 ⋅ (y)  dx x + dx (1) = 3 y dx  

1 −1  5 ⋅

2x

(

)

5 x 2 +1

4 5

(

)

4

1

; 10 x 2 + 1 5 y ′ = 6 x y 3

2

1 2

1 2 = 1⋅

(

;

)

+ x 2 y y′ = 2 x 2 y 2 −1

−

)

(

3 4

)

3

; x 2 y y ′ = 2 x 2 y 2 − 1 4 − xy 2

y x 1 −1 7

⋅

( )

d xy 2 = 6 x dx

; 2 y y ′x = 42 x xy 2

(

1 3 x + y3 2

)

;

6 7−

( ) (1 ⋅ y

1 xy 2 7

y 2 ; y′ =

−6 7

42 x

13 7

2

)

+ 2 y y′ ⋅ x = 6x 12

y 7 − y2 2x y

(

) (

)

−1 1 3 d 3 d 3 x + y 3 2 ⋅ 3x 2 + 3 y 2 y ′ = 0 x + y =0 ; 2 dx  dx 

(

d (y) dx

;

1 −1  2 ⋅

3y 2 y′ 3 x + y3 2

y + x⋅

) (

(

)

( )

6 xy 2 7

=0

(

−3 1 2 2 d 2 2  4 2 xy 2 + 2 x 2 y y ′ = 1 y x − 1 1 = x y ;   dx 4  

1 −1  4

( )

( )

(

3 4

)

1 xy 2 7

;

= 6x

; y 2 + 2 y y ′x = 6 x ⋅ 7

3y 2 y′ 3 x + y3 2

)

2

x y

d d  3 3 (2) ; d x 3 + y 3  x +y = dx dx   dx 3x 2 3 x + y3 2

1 1 (2 x + 1) − 2 ⋅ 2 x = 2 y ⋅ y ′ 2

)

2

(

x2 y

2 32 −1 d ⋅ (y) y 3 dx

(

4 x 2 y 2 −1

xy 2

=

2 − 13 2x 2 y y′ = x +1 3 5

2 xy 2 + x 2 y y ′

=1 ;

3 4

x2 y

(

1 5

3x3 y

; y′ =

4 5

;

(

; y′ =

i.

)

)

d x 2 +1 dx

;

d 4 2 2  d (x ) ; d x 2 y 2  x y = dx  dx  dx

;

h.

(

10 x 2 + 1

;

1

1

6x y 3

1− 2 2

x x (2 x + 1) − 2 x ; y′ = ; y′ = ; y′ = 1 2y 2 y 2x + 1 2 y (2 x + 1) 2

2 d 5 2  d  y 3   x +1 = dx   dx  

e.

1 2

; 6 x + (x + 1)

(

)

−1

2

=−

)

1 2 x −1 2

(

3x 2 3 x + y3 2

1 −1 2

)

−1

2

; y 2 y′ = − x 2 ; y ′ = −

(

)

⋅ 2x = y + x y ′ ; x x 2 −1

−1

2

= y + x y′

65

x2 y2

Basic Differentiation

(

2

1.7 The Derivative of Radical Functions

)

(

−1

; y + x y′ = x x −1

)

2

; x y′ = x x −1

2

−1

2

−y

(

x x2 −1 y′ = x

;

)

−1

2

−y

Example 1.7-5: Compute the derivative for the following Radical expressions.

(

)=

a.

d 2x + x dx

d.

d  3 x+ x  dx  

g.

d  x +1   dx  x 

=

=

(

)=

b.

d dx

e.

d  5   x − 3 dx  

h.

d  1  = dx  x +1 

x +1

=

c.

d 3 2   x +1 = dx  

f.

d x3 x +1 dx

i.

d  x 2 dx  3 x

(

)=

  =  

Solutions:

(

)=

1 1 1 −1 1 −1 1 d d 12 2x + x = 2+ x2 = 2+ x 2 = 2+ 1 = 2+ 2 2 dx dx 2 x 2x 2

a.

d 2x + x dx

b.

d dx

c.

d 3 2 d  x 2 +1  x +1 = dx  dx 

)

1 3

d.

3 d  d  3 x+ x2 x+ x  = dx  dx  

   

e.

 5x x 5 d  5 5 32 5 52 − 1 d  52 d 52 d   x − 3 = x3 = x = 0 + x ( ) x + − 3 = =  x − 3 =   2 2 2 2 dx  dx  dx dx  

f.

d x3 x +1 dx

(

x +1

)=

1 d (x + 1) 2 dx

(

(

= 3x (x + 1) 2

=

g.

d d 2x + dx dx

)=

1 2

d  x +1   dx  x 

=

=

=

1 1 (x + 1) 2 − 1 ⋅1 2

=

=

(

)

1 2 x +1 3

1 1 (x + 1) − 2 2

=

1 −1 3

⋅ 2x

=

=

1 2 (x + 1)

(

)

2x 2 x +1 3

=

1 2

−2

=

3

1 2 x +1

(

2x

)

3 x 2 +1

3 3 1 3 3 −1 d d 32 x+ x = 1+ x 2 = 1+ x 2 = 1 + 2 2 2 dx dx

2 3

= 3

3

(x

2x 2

+1

)

2

x

1 1 −1 1 d 1  x 3 (x + 1) 12  2 3 1 3 3 d  = (x + 1) 2 dx x + x dx (x + 1) 2 = (x + 1) 2 ⋅ 3 x + x ⋅ 2 (x + 1) 2 

1 x3 (x + 1) − 2 + 2

6x 3 + 6x 2 + x 3 2 x +1

d dx

=

x

7x 3 + 6x 2 2 x +1

  d  x +1 dx  12   x 

Hamilton Education Guides

=

1 2

= 3x (x + 1) + 2

=

1

1

x3 1

2 (x + 1) 2

=

3 x 2 (x + 1) 2 ⋅ 2 (x + 1) 2 + x 3 1

2 (x + 1) 2

=

6 x 2 (x + 1) + x 3 2 x +1

x 2 (7 x + 6 ) 2 x +1

 12 d   d 12  x   x ⋅ (x + 1) − (x + 1) ⋅ dx dx     x

=

 12   1 1 −1   x ⋅1 − (x + 1) ⋅ x 2  2     x

66

Basic Differentiation

1 x2

=

1.7 The Derivative of Radical Functions

1 x2

 1 −1 − (x + 1) ⋅ x 2  2   x

−

=

1

x 2 ⋅ 2 x 2 − (x + 1)

1 2x 2

1 2x 2

=

x

=

−

1 1 (x + 1) − 2 2 x +1

d  x 2 i. dx  3 x

=

 2   = d x  dx  13  x

x −1

2x − x −1

x −1 = 2 x =

2 x x

=

x

x

2x x

1 1 d   d  1 1 0 − (x + 1) 2 − 1  (x + 1) 2 ⋅ dx (1) − 1 ⋅ dx (x + 1) 2       2 =  = x +1 x +1 

d  1 d  1  d  1    = h. = dx  (x + 1) 12 dx  x +1  dx  x +1  

−

1

x +1

1 1

2 (x + 1) 2 (x + 1)

    

=

= −

1 2 (x + 1)

d  2 − 13  x ⋅x   dx  

1 2

= −

(x + 1)

1 2 (x + 1)

d 2 − 13 d x x = = dx dx

6−1 3

= −

3 2

1 2 5

d 3 x dx

=

(x + 1) 3 =

5

= −

5 3 −1 x 3

1

2 ( x + 1) x + 1 2

5 3 x 3

=

=

3

5 x2 3

Example 1.7-6: Evaluate the derivative of the following equations at the given values. a. y =

( x + 1) 2

3

at x = 2

b. y =

d. y = 3 2 x + 3 x 2 at x = 2

e. y =

5x + 1 2x + 1

3

6x + 1 x

2

c. y = x x 2 + 1 at x = 5

at x = 10

x + x +2

at x = 3

f. y =

at x = 3

h. y = x x 2 − 10 + 2 x at x = 5

2

at x = 0

k. y =

a. Given y =

( x + 1)

g. y = j. y =

x2 +1 1− x

x5 +1 2x

1 2

x +1

at x = −5

i. y = x 2 ( x + 1)3 at x = 1

at x = 2

l. y =

x +1 x3 +1

at x = 1

Solutions:

y′

=

2

3

) ( 2x

(

3 2 x +1 2

3 −1 2

(

)

3

= x 2 + 1 2 , then

2 −1

)

+0 =

(

)

3 2 x +1 2/

3−2 2

⋅ 2/ x

)

(

= 3x x 2 + 1

1 2

= 3x x 2 + 1

Substituting x = 2 in place of x in the y ′ equation we obtain the following value: y′

= (3 ⋅ 2) 2 2 + 1 =

b. Given y = y′

=

3

6 5

x + x +2

=

13.42 1

1

= x 3 + x 2 + 2 , then

1 −2 1 −1 1 13 − 1 1 12 − 1 x + x +0 = x 3 + x 2 = 3 2 3 2

1 2 3x 3

+

1 1 2x 2

=

1 3

3 x2

+

1 2 x

Substituting x = 10 in place of x in the y ′ equation we obtain the following value:

Hamilton Education Guides

67

Basic Differentiation

y′

=

1 2 3 ⋅10 3

+

1.7 The Derivative of Radical Functions

1

=

1 2 ⋅10 2

1 1 + 3 ⋅ 4.634 2 ⋅ 3.162

(

)

= 0.072 + 0.158 = 0.23

1

c. Given y = x x 2 + 1 = x x 2 + 1 2 , then

(

)

(

)

1   1 y ′ = 1 ⋅ x 2 + 1 2  +  x 2 + 1   2 

(

)

2

x2

1 2

= x +1 +

(x + 1) 2

(

(

)

)

 ⋅ 2 x 2 − 1 + 0 ⋅ x = x 2 + 1 

(x + 1) (x + 1) (x + 1) 1 2

2

=

1 2

1 −1 2

2

1 2

+ x2

1 2

2

=

1 2

(

)

1 +  x 2 +1  2/

(x + 1) + x (x + 1) 1+1 2 2

2

1 2

2

2

−1

=

2

)

(

 ⋅ 2/ x 2  = x 2 + 1  x 2 +1+ x 2

(x + 1) 2

1 2

=

1 2

)

(

+ x 2 x 2 +1

2x 2 + 1 x 2 +1

Substituting x = 5 in place of x in the y ′ equation we obtain the following value: y′

=

2 ⋅ 52 +1

(5 + 1) 2

1 2

=

2 ⋅ 25 + 1

(25 + 1)

=

1 2

50 + 1 1 26 2

=

51 5.1

= 10

2

1

d. Given y = 3 2 x + 3 x 2 = (2 x ) 3 + x 3 , then y′

=

2 −1 1 2 −1 1 2 1 + (2 x ) 3 − 1 + 2 x 3 = 1 (2 x )− 3 + 2 x 3 = 1 2 + 21 = 3 3 3 3 3 3(2 x ) 3 3 x 3 33 (2 x )2 3 x

Substituting x = 2 in place of x in the y ′ equation we obtain the following value: y′

=

1 3 ⋅ (2 ⋅ 2)

e. Given y =

y′

=

2 3

+

2

=

1 3⋅ 2 3

6x + 1 x2

2 3⋅ 4 3

+

2 1 3⋅ 2 3

=

2 1 + 3 ⋅ 2.53 3 ⋅1.26

= 0.132 + 0.53 = 0.662

1

(6 x ) 2 + 1 , then = 2 x

1 −1 1  1   2   2 −1  ⋅ (6 x ) 2 + 1    (6 x ) 2 ⋅ 6 + 0 ⋅ x  − 2 x     2  

x4 1

=

1

=

1  1  2    −1   (6 x ) 2 ⋅ 6 ⋅ x  − 2 x ⋅ (6 x ) 2 + 1      2  

x4

1

3 x 2 (6 x ) − 2 − 2 x(6 x ) 2 − 2 x x4

Substituting x = 3 in place of x in the y ′ equation we obtain the following value: 1

y′

= =

1

3 ⋅ 3 2 ⋅ (6 ⋅ 3) − 2 − 2 ⋅ 3 ⋅ (6 ⋅ 3) 2 − 2 ⋅ 3 3

4

=

1

1

27 ⋅ (18) − 2 − 6 ⋅ (18) 2 − 6 81

=

(27 ⋅ 0.236) − (6 ⋅ 4.243) − 6 81

25.086 6.372 − 25.458 − 6 = − = −0.31 81 81

Hamilton Education Guides

68

−1

2

Basic Differentiation

1.7 The Derivative of Radical Functions

1

f. Given y =

= =

1

, then

(x + 1)   1 ⋅ (2 x 0 ⋅ (x + 1)  −  (x + 1)  2  x2 +1 1 2

2

y′

=

1 2

2

1 −1 2

2

2−1

 



x 2 +1 −x

(x + 1)(x + 1) 2

2

=

1 2

−x

(

)

(x + 1) 2

−1

2

 ⋅ 2/ x  

=

x 2 +1

=

3 2

)

1 0 −  x 2 +1  2/

=

−x

=

1+ 1 x 2 +1 2

(

)

 + 0 ⋅1 

−x

(x + 1) 2

3

(

)

− x x 2 +1

−1

2

x 2 +1

−x

=

(x + 1) 2

x 2 +1

Substituting x = −5 in place of x in the y ′ equation we obtain the following value: y′

= −

−5

[(− 5) + 1] 2

g. Given y =

y′

=

5

=

3

(25 + 1)

3

5

=

=

17576

5 = 0.038 132.57

1

5x + 1 (5 x + 1) 2 , then = 2x + 1 2x + 1

(

)

(

)

1 1 −1 1  1−1 + 0 ⋅ (2 x + 1) −  2 x1 − 1 + 0 (5 x + 1) 2   2 (5 x + 1) 2 ⋅ 5 x    

(2 x + 1)2

=

2.5(5 x + 1) − 12 ⋅ (2 x + 1) − 2(5 x + 1) 12     

(2 x + 1)2

Substituting x = 3 in place of x in the y ′ equation we obtain the following value: y′

=

2.5(15 + 1) − 12 ⋅ (6 + 1) − 2(15 + 1) 12     

=

(6 + 1)2

(

)

17.5 ⋅ (16) − 12  − 2⋅ (16) 12      72

[17.5 ⋅ 0.25] − [2 ⋅ 4] =

=

49

−0.074

1

h. Given y = x x 2 − 10 + 2 x = x x 2 − 10 2 + 2 x , then y′



(

)

1

 

(

= 1 ⋅ x 2 − 10 2  + 0.5 x 2 − 10 

 

)

1 −1 2

(

)

(

 ⋅ 2 x 2 − 1 − 0 ⋅ x  + 2 x1 − 1 = x 2 − 10 

)

1 2

(

 +  x 2 x 2 − 10 

)

−1

2

 +2 

Substituting x = 5 in place of x in the y ′ equation we obtain the following value: y′

(

= 5 2 − 10

)

1 2

(

 + 5 2 5 2 − 10 

)

−1

2

 +2 

1



= 15 2 + 25 ⋅15 

−1

2

 +2 

. + [ 25 ⋅ 0.258] + 2 = 12.323 = 3873

3

i. Given y = x 2 ( x + 1)3 = x 2 (x + 1) 2 , then y′

(

)

3 3 3 1 3 3  = 2 x 2−1 ⋅ (x + 1) 2  +  (x + 1) 2 − 1 ⋅ x1−1 + 0 ⋅ x 2  = 2 x(x + 1) 2  +  x 3 (x + 1) 2    2   2  

Substituting x = 1 in place of x in the y ′ equation we obtain the following value: y′

3   3 1  3 1 3 . = 2(1 + 1) 2  +  (1 + 1) 2  =  2 ⋅ 2 2  +  ⋅ 2 2  = 5.657 + 2121 = 7.778 2   2 

Hamilton Education Guides



 



69

Basic Differentiation

1.7 The Derivative of Radical Functions 1

x2 +1

 x 2 +1 2   , then =  1− x 2  1− x2  

j. Given y =

) ( ( )

(

1 −1 

)

1

1

− −  2 2  2  2  2 x − 2/ x 3/ + 2/ x 3/ + 2 x   2  2 4x  2x 1 − x + 2x x +1  = 1  x + 1    = 1  x +1 2     2  1 − x 2  2  1 − x 2  2 2 2 2 1− x 1− x 1− x 2    

1  x 2 + 1  2 y′ =  2  1 − x 2 

(

)

(

)

Substituting x = 0 in place of x in the y ′ equation we obtain the following value: 1  0 +1 y′ =   2  1− 0 

)

(

=

2

⋅

1 5  x +1  2/

4⋅0

1 − 12 ⋅1 ⋅ 0 2

=

(1 − 0)2

x5 +1 2x

k. Given y =

y′

−1

(x + 1) = 5

1 2

2x

1 −1 2 5x 4

= 0

, then

(

)

  ⋅ 2/ x  − 2 x1 − 1 ⋅ x 5 + 1  

1 2

(2 x )2

(

  

=

)

 5 5 5 x x + 1 

−1

2

(

)

  5  − 2 ⋅ x + 1  

1 2

  

4x 2

Substituting x = 2 in place of x in the y ′ equation we obtain the following value: y′

=

160(32 + 1) − 12  − 2 ⋅ (32 + 1) 12      4⋅4 x +1

l. Given y =

x +1 1 2



1 −1 2

3

2

 

(x + 1)

1 2

=

3 2 x (x + 1) 2 − 1

(x + 1) 3

(x + 1) 3

2

2 x 3 + 2 − 3x 3 − 3x 2

(

) (x + 1)

2 x3 +1

1 2

3

1

=

x3 +1

=

. 27.8 − 115 = 1.02 16

=

16

)

  + 0  (x + 1)   

x3 +1

3

=

1 2

3

[160 ⋅0.174] − [2 ⋅ 5.744]

, then

(x + 1)     1 (3x 1 ⋅ (x + 1)  −   (x + 1)    2 3

3

y′ =

x +1

=

=

1 2

−

(

)

x3 +1 − x 3 − 3x 2 + 2

(

) (x + 1) 1 2

3

3

=

1 2

1 2

=

=

)

(x + 1) 2(x + 1) − 3x 2(x + 1) 1 2

1 2

3

3

3

)

2 x3 +1

1 +1 2

=

(

1 2

)

2 x 3 + 1 − 3x 3 − 3x 2

− 3x 2

x3 +1

− x 3 − 3x 2 + 2

(

(

   1  −1 −   x 3 + 1 2 3 x 2  (x + 1)     2  x3 +1

3

3x 3 + 3x 2 2 x3 +1

2 x3 +1

(x + 1)

=

(

)

2 x3 +1

1 2

x3 +1

− x 3 − 3x 2 + 2

(

)

2 x3 +1

3 2

Substituting x = 1 in place of x in the y ′ equation we obtain the following value: y′

=

( ) 2(1 + 1)

− 13 − 3 ⋅12 + 2 3

3 2

Hamilton Education Guides

=

−1 − 3 + 2 3 2⋅2 2

=

−2/ 2/ ⋅ 2.828

=

−1 2.828

= −0.354

70

Basic Differentiation

1.7 The Derivative of Radical Functions

Section 1.7 Practice Problems - The Derivative of Radical Functions 1. Find the derivative of the following radical expressions. Do not simplify the answer to its lowest term. b. y = x 3 + 3x − 5

a. y = x 2 + 1 d. y = g. y = 2. Use the

x +1 x x2 + 3

f. y = x 3 + 3x 2

2

x −1 4

h. y =

x +1 d dx

x2

e. y =

c. y = x 2 + x − 1

x 3 −1

x3

i. y =

x

x2 x

notation to find the derivative of the following radical expressions.

a.

d  2 1  x +  dx  x

d.

d  x + 5  = dx  x 

=

b.

d  x  = dx  x −1 

c.

d  x 3  = dx  x +1 

e.

d  3 x  = x +  dx  x 

f.

d  2 x 1+ dx  x3

c.

d (x y ) = d dx dx

  =  

3. Find the derivative of the following radical expressions. a.

d  3  d (x )  x + y= dx   dx

d.

d dx

g.

d d (2) x y2 + x = dx dx

(

)

y + x3 = 0

(

)

(

)

d (2) dx

( x)

b.

d dx

e.

d  4 d 2 (x )  x +y = dx   dx

f.

d dx

(

x +1 =

h.

d  3 d (x y ) = 0  x + dx   dx

i.

d dx

(

x + 3y =

x + y3 =

)

)

( )

d 3 y dx

d (y) dx

4. Evaluate the derivative of the following radical expressions for the specified values of x . a. y = 3x 3 + x 2 at x = 1 d. y =

x x 2 +1

at x = 1

Hamilton Education Guides

b. y = x 2 + 1 x at x = 0

(

)

c. y =

e. y = x 3 + 1 + 4 x 3 at x = 0

f. y =

x 2 −1 4x 2 x 2 +1 x3

at x = 2 at x = 3

71

Basic Differentiation

1.8

1.8 Higher Order Derivatives

Higher Order Derivatives

If the function y = f ( x ) is differentiable, then we can form a new function y ′ = f ′( x ) which is referred to as the first derivative of y = f ( x ) . Consequently, if y ′ = f ′( x ) is differentiable, then we can form another new function y ′′ = f ′′( x ) called the second derivative of y = f ( x ) . This process of obtaining a new function can continue as long as we have differentiability. First, second, third, and higher order derivatives are denoted in various notations. In general, however, first, second, third, and nth derivatives are shown in the following forms: d dy f ( x) = D y = D f ( x) = dx dx

y′

= f ′( x ) =

y ′′

= f ′′( x ) =

y ′′′

= f ′′′( x ) =

d2y dx 2

=

d  dy  d d d2  f ( x ) = D 2 y = D 2 f ( x ) f ( x) =   = 2 dx  dx  dx  dx  dx

=

 d  d2 d  d2 y d3  =  f x f x = = D3 y = D3 f ( x ) ( ) ( )   2 2 3 dx  dx  dx  dx dx 

=

d  d n −1 y    dx  dx n −1 

d3y dx 3

 yn

dny

= f n ( x) =

dx n

=

dn dx n

f ( x)

=

 d  d n −1 n n  n −1 f ( x )  = D y = D f ( x ) dx  dx 

Students are encouraged to become familiar with these notations for finding the derivative of different functions. Note that the prime notation is not used beyond the third derivative. In general, the fourth or higher derivatives are shown as y 4 = f 4 ( x ) ; y 5 = f 5 ( x ) ; y 6 = f 6 ( x ) ;  ; yn = f

n

( x ) instead of

f ( x) = x 6 + x 3 + 1 , f 5 ( x ) = 720 x , f

6

then

y iv = f

iv

(x ) ;

(x ) ; y v i = f v i (x ) , etc. For example, given f ′′( x ) = 30 x 4 + 6 x , f ′′′( x ) = 120 x 3 + 6 , f 4 ( x ) = 360 x 2 ,

yv = f

f ′ ( x ) = 6 x 5 + 3x 2 ,

v

( x ) = 720 , and all derivatives higher than

are equal to zero. The following examples show in detail how higher order derivatives are obtained: 7

Example 1.8-1: Find the second derivative of the following functions. a. f ( x ) = 5x 8 − 3x 3 + 1 d. f (u) =

u3 − 1 u +1

(

)

)

−1

t 3 + t 2 +1 10

h. r (θ ) = θ 2 + k. p( r ) = r 2 −

x

1 x

f. h( x ) = a 2 + x 3

3

1

(θ + 1) 1 r

c. f ( x ) = x 2 +

(

1

e. g( x ) = x 2 +

g. f ( x ) = x 2 + 1 j. f ( t ) =

(

b. f ( x ) = x 3 x 2 + x + 5

3

(

2

)

i. s(r ) = r 2 r 2 + 1 l. f ( x ) =

)

3

x3 x +1

Solutions: a. Given f ( x ) = 5x 8 − 3x 3 + 1 , then f ′( x ) = ( 5 ⋅ 8) x 8−1 − ( 3 ⋅ 3) x 3−1 + 0 = 40 x 7 − 9 x 2 and

Hamilton Education Guides

72

Basic Differentiation

f ′′( x )

1.8 Higher Order Derivatives

= ( 40 ⋅ 7) x 7 −1 − ( 9 ⋅ 2) x 2 −1 = 280 x 6 − 18 x

(

)

b. Given f ( x ) = x 3 x 2 + x + 5 , then f ′( x )

[

)] [(

(

) ] [ (

)] [

= 3x 3−1 ⋅ x 2 + x + 5 + 2 x 2−1 + 1x 1−1 + 0 ⋅ x 3 = 3x 2 x 2 + x + 5 + ( 2 x + 1) ⋅ x 3

]

= 3x 4 + 3x 3 + 15x 2 + 2 x 4 + x 3 = 5 x 4 + 4 x 3 + 15 x 2 and f ′′( x ) = ( 5 ⋅ 4) x 4 −1 + ( 4 ⋅ 3) x 3−1 + (15 ⋅ 2) x 2 −1 = 20 x 3 + 12 x 2 + 30 x 1 x

c. Given f ( x ) = x 2 + , then f ′( x ) = 2 x 2 −1 +

[0 ⋅ x] − (1⋅ 1) = 2

2x +

x

0−1 x

2

1

= 2x −

x2

A second way is to rewrite f ( x ) as f (x ) = x 2 + x −1 and find its derivative, i.e., f ′( x ) = 2 x 2 −1 − 1 ⋅ x −1−1 = 2 x − x −2 = 2 x − f ′′( x )

1 x2

and

= 2 x 1−1 + ( −1⋅ −2) x −2 −1 = 2 x 0 + 2 x −3 = 2 + 2 x −3 = 2 + u3 − 1 , then u +1

d. Given f (u) =

[ (3u f ′( x ) =

3−1

)]

)(

=

(u + 1)2

[ (2 ⋅ 3u f ′′( x ) = [ (6u =

] [(

)

− 0 ⋅ (u + 1) − u 1−1 + 0 ⋅ u 3 − 1

2

3−1

][

)

x3

(

) = 3u

3u 2 ⋅ ( u + 1) − u 3 − 1

( u + 1) 2

3

+ 3u 2 − u 3 + 1

( u + 1) 2

=

2u 3 + 3 u 2 + 1

( u + 1)2

)]

(

+ 3 ⋅ 2u 2−1 + 0 ⋅ (u + 1)2 − 2(u + 1)2−1 ⋅ 2u 3 + 3u 2 + 1

(u + 1)4

][

)

)]

(

+ 6u (u + 1)2 − 2(u + 1) 2u 3 + 3u 2 + 1

(u + 1)

e. Given g( x ) = x 2 + g ′( x ) = 2 x 2−1 +

2

1

x3

4

=

[ (6u

2

][

)

(

)]

+ 6u (u + 1)2 + (− 2u − 2 ) 2u 3 + 3u 2 + 1

(u + 1)

4

, then

(0 ⋅ x )− (3x 3

x

6

3−1

)

⋅1

= 2x +

−3x 2 x

6

= 2x −

3x 2/ x

6/ = 4

= 2x −

3 x4

A second way is to rewrite g( x ) as g (x ) = x 2 + x −3 and find its derivative, i.e., g ′( x ) = 2 x 2 −1 − 3 ⋅ x −1− 3 = 2 x − 3x −4 = 2 x −

Hamilton Education Guides

3 x4

and

73

Basic Differentiation

1.8 Higher Order Derivatives

12

g ′′( x ) = 2 x1−1 + (− 3 ⋅ −4 )x −4−1 = 2 x 0 + 12 x −5 = 2 + 12 x −5 = 2 +

(

) ) ⋅ (0 + 3x ) = 2(a

x5

2

f. Given h( x ) = a 2 + x 3 , then

(

h ′( x ) = 2 a 2 + x 3

2 −1

3−1

2

)

(

)

+ x 3 ⋅ 3x 2 = 6 x 2 a 2 + x 3 = 6a 2 x 2 + 6 x 5 = 6 x 5 + 6a 2 x 2

)

(

h ′′( x ) = ( 6 ⋅ 5) x 5−1 + 6a 2 ⋅ 2 x 2 −1 = 30 x 4 + 12a 2 x

(

)

g. Given f ( x ) = x 2 + 1

(

)

 f ′( x ) = −1 ⋅ x 2 + 1 

f ′′( x )

)

( 

)



= −2 x 2 + 1

h. Given r (θ ) = θ 2 +

) = −( x + 1)

(

2

2 −1 +0  ⋅ 2x 

= −21⋅ x 2 + 1 

, then

−1−1 

(



−1

−2

−2

(

)

 + −2 x 2 + 1 

(

)

− 4x x 2 + 1 1

(θ + 1) 3

−2 −1

−3 

 

(

)

(

3−1

(θ + 1)

= 2θ +

6

[ 0 ⋅ (θ + 1) ]− [2(θ + 1) 2

2 −1

⋅3

]

= 2−

(θ + 1)4

(

)

)

= −2 x 2 + 1

−2

(

)

= −2x x 2 + 1

( 

)



= −2 x 2 + 1

(

)

+ 8x x 2 + 1

−2

−2

(

)

 + −2 x 2 + 1 

−3

 ⋅ 2 x  

−3

, then

3

r ′′(θ ) = 2θ 1−1 −

 ⋅ 2x 

 ⋅ 2 x 2 −1 + 0   

[ 0 ⋅ (θ + 1) ]− [3(θ + 1) ⋅1]

r ′(θ ) = 2θ 2−1 +

−2 

0 − 3 (θ + 1)

(θ + 1)

2

6

0 − 6 (θ + 1)

(θ + 1)

= 2θ −

= 2+

4

3 (θ + 1)

(θ + 1)

= 2θ −

6/ = 4

6 (θ/ + 1/ )

(θ + 1)

2/

4/ = 3

= 2+

3

(θ + 1) 2

and

6

(θ + 1) 3

3

i. Given s( r ) = r 2 r 2 + 1 , then s ′( r )

)   (

(



) ⋅ (2r

3

= 2r 2 −1 ⋅ r 2 + 1  + 3 r 2 + 1 

(

)

)

(

3

= 2r r 2 + 1 + 6 r 3 r 2 + 1

(

)

2

(

) ⋅ ( 2r

( 

)

3

(  

)

= 2  r 2 + 1 + 3 r 2 + 1

Hamilton Education Guides

3−1

2 −1

)

 + 0 ⋅r2 



)   (

(

)

3

2



= 2 r r 2 + 1  + 3 r 2 + 1 ⋅ 2 r  ⋅ r 2 



and

3   s ′′( r ) = 2 1 ⋅ r 2 + 1  + 3 r 2 + 1   



3−1

3−1

2 −1

)

(

)

(

) ⋅ ( 2r

2     + 0  ⋅ r  + 63r 3−1 ⋅ r 2 + 1  + 2 r 2 + 1     

(

) [(

2 −1

2 −1

)

  + 0 ⋅r3  

) ]

2     ⋅ 2r  ⋅ r  + 63r 2 r 2 + 1  + 2 r 2 + 1 ⋅ 2r ⋅ r 3      

74

Basic Differentiation

(

1.8 Higher Order Derivatives

)

(

)

) [ (

(

)]

3  2  2    = 2  r 2 + 1 + 6 r 2 r 2 + 1   + 6   3 r 2 r 2 + 1  + 4 r 4 r 2 + 1       

j. Given f (t ) = f ′( t )

[ (3t =

f ′′( t ) =

3−1

t 3 + t 2 +1 , then 10

) ][ (

)]

+ 2t 2−1 + 0 ⋅10 − 0 ⋅ t 3 + t 2 + 1

=

10 2

[ (3 ⋅ 2t

2 −1

) ][ (

+ 2t 1−1 ⋅10 − 0 ⋅ 3t 2 + 2t 10

k. Given p(r ) = r 2 −

)]

2

(

)

10 3t 2 + 2t − 0 100

10 ( 6t + 2) − 0

=

100

=

=

(

/ / 3t 2 + 2t 10 / / / = 10 100

/ / ( 6t + 2) 10 / / / = 10 100

=

)=

3 t 2 + 2t 10

2/ (3t + 1) 1/ 0/ = 5

=

and

3t + 1 5

1 which is equal to p(r ) = r 2 − r −1 , then r

(

)

p ′(r ) = 2r 2 −1 − −1 ⋅ r −1−1 = 2r + r −2 and p ′′( r ) = 2r 1−1 − 2r −2 −1 = 2r 0 − 2r −3 = 2 − 2r −3

l. Given f ( x ) =

[ 3x f ′( x ) =

3−1

x3 , then x +1

] [(

) ] = 3x

⋅ ( x + 1) − 1 ⋅ x 1−1 + 0 ⋅ x 3

{[(2 ⋅ 3) x f ′′( x ) =

( x + 1) 2 2

]

+ ( 3 ⋅ 2) x ⋅ ( x + 1)

2

d3y dx 3

3

+ 3x 2

4

3x 3 + 3x 2 − x 3

( x + 1) 2

)} = [ (6 x

2

)

=

2x 3 + 3 x 2

and

( x + 1) 2

][

(

+ 6 x ( x + 1) 2 − 2( x + 1 ) 2 x 3 + 3 x 2

)]

( x + 1)

4

for the following functions.

(

a. y = (1 − 5x )3 1 5

( x + 1) − x 3 = ( x + 1) 2

} − {2( x + 1) ⋅ (2 x

( x + 1)

Example 1.8-2: Find

2

b. y = a − bx 2

1 4

d. y = x 5 + x 4 + x

e. y =

Solutions:

)

−2

ax 2 + b c

c. y = f. y =

x 2 + 3x + 1 x +1 x2 +1 x3

a. Given y = (1 − 5x )3 , then

(

)

y′

2 = 3(1 − 5x ) 3−1 ⋅ 0 − 5x 1−1 = 3 (1 − 5x ) 2 ⋅ ( −5) = −15 (1 − 5 x)

y ′′

= ( −15 ⋅ 2) (1 − 5x ) 2 −1 ⋅ 0 − 5x 1−1 = −30 (1 − 5x ) ⋅ ( −5) = 150 (1 − 5 x)

y ′′′

= 150 0 − 5x 1−1 = 150 ⋅ ( −5) = −750

(

(

(

b. Given y = a − bx 2

) )

−2

Hamilton Education Guides

)

, then

75

Basic Differentiation

(

1.8 Higher Order Derivatives

)

= −2 a − bx 2

y ′′

= ( 4b ⋅ 1) a − bx 2

(

 

(



= 4b a − bx 2 

y ′′′

)

)

  ⋅ ( 0 − 2bx )  ⋅ ( 4bx )   

)

−3−1

(



= 24b 2 x a − bx 2 

c. Given y =

[ (2x y′ = (2 x =

2

[ (2x y ′′ =

)

(

)

)

−3−1

−3

−4 

)(

(

)

−4

−4

(

+ 192b 3 x 3 a − bx 2

( x + 1) 2 2 −1

=

2 x 2 + 5x + 3 − x 2 − 3x − 1

( x + 1) 2

][

)

(

+ 2 x1−1 + 0 ⋅ (x + 1)2 − 2(x + 1)2−1 ⋅ x 2 + 2 x + 2

)]

(x + 1)4 2

2

+ 2x + 2

2 x 3 + 6x 2 + 6x + 2 − 2 x 3 − 6x 2 − 8x − 4

( x + 1)

[ (− 2x y ′′′ =

1−1

1 5

)

(

−3 

 2  + 6bx a − bx  

)

−4

 ⋅ ( 4bx )  

2 −4 −1

)

 ⋅ ( −2bx )  

−5 

 

=

x 2 + 2x + 2

( x + 1) 2

[ (2x + 2)(x + 1) ]− [2(x + 1) (x = 2

2

+ 2x + 2

)]

(x + 1)4

)]

(x + 1)4

4

)

[ (2 x + 3)⋅ (x + 1) ] − [1⋅ (x 2 + 3x + 1) ] = (x + 1)2

)]

)(

)



2 2

(x + 1)2

) (

(



= 4b a − bx 2

) + (−96b x )(a − bx )

+ 3 x1−1 + 0 ⋅ (x + 1) − x1−1 + 0 ⋅ x 2 + 3 x + 1

+ 2 x + 3x + 3 − x 2 + 3x + 1

−3

 

(

] [(

)

)

−4 

 2 2  + 48 b x a − bx  

[ (2x + 2) (x + 1) ]− [2(x + 1) (x = =

⋅ ( −2bx ) = 4 bx a − bx 2

  ⋅ ( −2bx )  +  48b 2 x ⋅ a − bx 2  

x 2 + 3x + 1 , then x +1

2 −1

(

 2  + −3 a − bx  

−3 

 2 2 2  +  24 b x a − bx  

= −12b a − bx 2 

)

−3 

(



(

(

⋅ ( 0 − 2bx ) = −2 a − bx 2

−2 −1

y′

=

][

[2x =

+ 6 x 2 + 6 x + 2 − 2 x 3 + 6 x 2 + 8x + 4

]

(x + 1)4

−2 x − 2

( x + 1)4

+ 0 ⋅ (x + 1)4 − 4(x + 1)4−1 ⋅ (− 2 x − 2 )

(x + 1)8

][

3

]

=

[− 2 (x + 1) ]+ [ (x + 1) (8 x + 8) ] 4

3

( x + 1)8

1 4

d. Given y = x 5 + x 4 + x , then y′

=

1 1 ⋅ 5x 5−1 + ⋅ 4 x 4 −1 + x 1−1 = x 4 + x 3 + 1 4 5

Hamilton Education Guides

y ′′

= 4 x 4 −1 + 3x 3−1 + 0 = 4 x 3 + 3 x 2 and

76

Basic Differentiation

y ′′′

= ( 4 ⋅ 3) x 3−1 + ( 3 ⋅ 2) x 2 −1 = 12x 2 + 6 x ax 2 + b , then c

e. Given y = y′

=

[ (a ⋅ 2x

=

=

) ][ (

+ 0 ⋅ c − 0 ⋅ ax 2 + b

x2 +1

[ (2x

x3

2 −1

)]

) ][

)

− x 2/ x 2 + 3

= −

x 6/ = 4

= −

2 −1

)]

(

=

) ][ =

x8

)]

(

y ′′

(

x 3/ 2 x 2 + 12 x 8/ =5

) ][

−6 x 6 − 60 x 4

)=

(

3

2

2

=

2a c

and y ′′′ = 0

x6

2 x 4 − 3x 4 − 3x 2

=

x6

[2x ⋅ x ]− [4x ⋅ (x + 3) ] = − 4

3

2

=

− x 4 − 3x 2 x6

)]

(

/ / =6 x 10

2 x 5 − 4 x 5 − 12 x 3 x8

x5

[ 4x ⋅ x ]− [5x ⋅ (2x =

) = − 6x

x 4/ −6 x 2 − 60

= −

2 x 2 + 12

5

4

2

+ 12

x10

x10

=

[2x ⋅ x ]− [3x ⋅ (x + 1) ]

x8

+ 0 ⋅ x 5 − 5 x 5−1 ⋅ 2 x 2 + 12

x 10

2ax c

x4

x8

2 −1

c

=

2/ =1

x2 + 3

+ 0 ⋅ x 4 − 4 x 4−1 ⋅ x 2 + 3

−2 x 5 − 12 x 3

[ (4x y ′′′ =

c

2acx /

=

2

, then

+ 0 ⋅ x 3 − 3 x 3−1 ⋅ x 2 + 1

(

( 2ax ⋅ c) − 0

=

x6

[ (2x y ′′ = −

=

2 −1

c2

f. Given y = y′

1.8 Higher Order Derivatives

2

+ 60

x

6

)]

=

4 x 6 − 10 x 6 − 60 x 4 x 10

Example 1.8-3: Find f ′′(0) and f ′′(1) for the following functions. a. f ( x ) = 6x 7 + 7 x 2 − 2 d. f ( x ) =

x3 +1 x

b. f (x ) = x 5 (x − 1)2 e. f ( x ) = x 3 −

g. f ( x ) = ( x − 1) −2

h. f ( x ) =

j. f ( x ) = (1 + 5x )3

k. f (x ) =

Hamilton Education Guides

1 x +1

( x + 1) 2 x

1+ x x

3

c. f ( x ) = x −

1 x

f. f ( x ) = (ax + b) 2

(

)

i. f ( x ) = ( x + 1) x 2 + 1 + 5 1 3

1 2

l. f ( x ) = x 3 + x 2 + x + 10

77

Basic Differentiation

1.8 Higher Order Derivatives

Solutions: a. Given f ( x ) = 6 x 7 + 7 x 2 − 2 , then f ′( x ) = 6 ⋅ 7 x 7−1 + 7 ⋅ 2 x 2−1 − 0 = 42 x 6 + 14 x and f ′′( x ) = 42 ⋅ 6 x 6−1 + 14 ⋅ 1x 1−1 = 252 x 5 + 14

Therefore, f ′′(0) = 252 ⋅ 05 + 14 = 0 + 14 = 14 and f ′′(1) = 252 ⋅15 + 14 = 252 + 14 = 266 b. Given f ( x ) = x 5 ( x − 1) 2 , then

(

)

[

2

]

2

2

f ′( x ) = 5x 5−1 ⋅ 1 ⋅ ( x − 1) + 2( x − 1) ⋅ 1 ⋅ x 5 = 5x 4 ( x − 1) + 2 x 5 ( x − 1) = 5x 4 ( x − 1) + 2 x 6 − 2 x 5 and

[

]

2

2

f ′′( x ) = 5 ⋅ 4 x 4−1 ( x − 1) + 2( x − 1) ⋅ 5x 4 + 2 ⋅ 6 x 6−1 − 2 ⋅ 5x 5−1 = 20 x 3 ( x − 1) + 10 x 4 ( x − 1) + 12 x 5 − 10 x 4

(

)

= 20 x 3 x 2 − 2 x + 1 + 10 x 5 − 10 x 4 + 12 x 5 − 10 x 4 = 20 x 5 − 40 x 4 + 20 x 3 + 22 x 5 − 20 x 4 = 42 x 5 − 60 x 4 + 20 x 3 Therefore, f ′′(0)

= 42 ⋅ 0 5 − 60 ⋅ 0 4 + 20 ⋅ 0 3 = 0 and f ′′(1) = 42 ⋅15 − 60 ⋅14 + 20 ⋅13 = 42 − 60 + 20 = 2 1 x

c. Given f ( x ) = x − , then f ′( x ) = 1 −

0 ⋅ x − 1⋅1 x

= 1+

2

2

Therefore, f ′′(0) = −

0

1 x

[3x ⋅ x]− [1⋅ (x + 1) ] f ′( x ) = 2

3

=

x2

f ′′( x )

2

2 0

= −

3

0 ⋅ x − 2x ⋅1 x

4

= −

2x x

4

2

= −

which is not defined and f ′′(1) = −

x3

2

2 = −2 1

= −

3

1

x3 +1 , then x

d. Given f ( x ) =

[6x =

and f ′′( x ) = 0 +

2

][ (

3x 3 − x 3 − 1 x2

)]

⋅ x 2 − 2x ⋅ 2x 3 −1 x4

(

e. Given f ( x ) = x 3 −

)=

2 03 + 1

Therefore, f ′′(0) =

f ′( x ) = 3x 3−1 −

=

0

3

2 0

=

2x 3 − 1 x2

6x 4 − 4x 4 + 2x x4

=

and 2x 4 + 2x x4

=

(

) = 2(x + 1)

2 x/ x 3 + 1

3

x 4/ =3

x3

which is not defined and f ′′(1) =

(

)=

2 13 + 1 3

1

4 1

= 4

1 , then x +1

0 ⋅ ( x + 1) − 1 ⋅ 1

( x + 1)

Hamilton Education Guides

2

= 3x 2 +

1

( x + 1) 2

and

78

Basic Differentiation

1.8 Higher Order Derivatives 2

0 ⋅ ( x + 1) − 2( x + 1)

f ′′( x ) = 3 ⋅ 2 x 2−1 +

f ′′(0) = 6 ⋅ 0 −

( x + 1)

2

(0 + 1)

= −

3

2−1

⋅1

= 6x −

4

2( x + 1)

( x + 1)

4

= 6x −

2

( x + 1)3

Therefore,

2 1 2 2 2 = −2 and f ′′(1) = 6 ⋅ 1 − = 6 − 3 = 6 − = 6 − = 5.75 3 1 4 8 2 (1 + 1)

f. Given f ( x ) = (ax + b) 2 , then f ′( x ) = 2(ax + b)

2−1

⋅ a = 2a(ax + b) = 2a 2 x + 2ab and f ′′( x ) = 2a 2 + 0 = 2a 2

Therefore, f ′′(0) = 2a 2 and f ′′(1) = 2a 2 Note that since f ′′( x ) is independent of x , f ′′( x ) is equal to 2a 2 for all values of x . g. Given f ( x ) = ( x − 1) −2 , then f ′( x ) = −2( x − 1) f ′′(0) = 6(0 − 1)

h. Given f ( x ) = f ′( x ) =

f ′′( x ) =

−2 −1

−4

⋅ 1 = −2( x − 1)

=

6

( −1)

=

4

−3

[

]

and f ′′( x ) = −2 ⋅ −3( x − 1) −3−1 ⋅ 1 = 6( x − 1) −4 Therefore,

6 6 6 = 6 and f ′′(1) = 6(1 − 1) −4 = 4 = which is undefined 0 1 0

( x + 1) 2 , then

[2 (x + 1)

x

2 −1

][

⋅1 ⋅ x − 1 ⋅ (x + 1)2 x2

[ 2x ⋅ x ]− [2x ⋅ (x − 1) ] 2

2

x

4

2

Therefore, f ′′(0) =

0

(

3

2 0

=

=

]

=

2 x ( x + 1) − ( x + 1) x

2x 3 − 2x 3 + 2x x

4

2

=

2

=

2x x

4

=

2x 2 + 2x − x 2 − 2x − 1 x

2

=

x2 −1 x2

and

2

x3

which is undefined and f ′′(1) =

2

13

=

2 = 2 1

)

i. Given f ( x ) = ( x + 1) x 2 + 1 + 5 = x 3 + x + x 2 + 1 + 5 = x 3 + x 2 + x + 6 , then f ′( x ) = 3x 2 + 2 x + 1 and f ′′( x ) = 6 x + 2

Therefore, f ′′(0) = 6 ⋅ 0 + 2 = 2 and f ′′(1) = 6 ⋅ 1 + 2 = 8 j. Given f ( x ) = (1 + 5x )3 , then f ′( x )

= 3(1 + 5x )3−1 ⋅ 5 = 15(1 + 5x ) 2 and f ′′( x ) = 0 ⋅ (1 + 5x ) 2 + 2(1 + 5x ) ⋅ 5 ⋅ 15 = 150(1 + 5x )

Therefore, f ′′(0) = 150(1 + 5 ⋅ 0) = 150 and f ′′(1) = 150(1 + 5 ⋅ 1) = 150(1 + 5) = 150 ⋅ 6 = 900

Hamilton Education Guides

79

Basic Differentiation

k. Given f ( x ) =

1+ x x3

, then

1 ⋅ x 3 − 3x 2 (1 + x )

f ′( x ) =

f ′′(x )

1.8 Higher Order Derivatives

x

=

6

x 3 − 3x 2 − 3x 3 x

[ − 2 ⋅ x ]− [4x (− 2x − 3) ] = 4

3

x

l. Given f ( x ) =

6 ⋅ 0 + 12 0

5

=

12 0

−2 x 3 − 3x 2 x

6

− 2 x 4 + 8 x 4 + 12 x 3 x

8

=

=

− x 2/ (2 x + 3) x

6/ = 4

6 x 4 + 12 x 3 x

8

=

which is not defined and f ′′(1) =

=

−2 x − 3 x4

6 x 3/ (x + 2 ) x

6 ⋅1 + 12 5

1

=

6 x + 12

18 1

= 18

8/ = 5

=

x5

1 3 1 2 x + x + x + 10 , then 3 2

3 3−1 2 2−1 + x + x 1−1 + 0 x 2 3

f ′( x ) =

=

8

Therefore, f ′′(0) =

6

=

= x 2 + x + 1 and f ′′( x ) = 2 x 2 −1 + x1−1 + 0 = 2 x + 1

Therefore, f ′′(0) = 2 ⋅ 0 + 1 = 1 and f ′′(1) = 2 ⋅ 1 + 1 = 3 dy d 2 y d3y , 2 , and 3 for the following functions. dx dx dx 1 b. y = x + a. y = x 4 + 5x 3 + 6 x 2 + 1 x

Example 1.8-4: Find

(

)

d. y = x 2 + 1 g. y = x −

−2

1 x

c. y = x( x + 1)3

e. y = x 3 + 3x 2 + 10

f. y =

h. y = ax 3 + bx

i. y =

1 1+ x x3 +1 x2

Solutions: a. Given y = x 4 + 5x 3 + 6 x 2 + 1 , then d 4 d d d dy x + 5 x 3 + 6 x 2 + 1 = 4 x 4−1 + (5 ⋅ 3) x 3−1 + (6 ⋅ 2) x 2−1 + 0 = 4 x 3 + 15 x 2 + 12 x = dx dx dx dx dx d2y dx

2

d3y dx

3

= 4

d d 3 d x + 15 x 2 + 12 x = (4 ⋅ 3) x 3 −1 + (15 ⋅ 2) x 2 −1 + (12 ⋅ 1) = 12 x 2 + 30 x + 12 dx dx dx

= 12

d 2 d d x + 30 x + 12 = (12 ⋅ 2) x 2 −1 + (30 ⋅ 1) x1−1 + 0 = 24 x 1 + 30 x 0 = 24 x + 30 dx dx dx

b. Given y = x +

1 x

which is the same as y = x + x −1 , then

(

)

dy d d −1 x+ x = 1 + −1 ⋅ x −1−1 = 1 − x −2 = dx dx dx d2y dx

2

=

(

)

d d −2 1− x = 0 − −2 ⋅ x −2−1 = 2 x −3 dx dx

Hamilton Education Guides

80

Basic Differentiation

d3y dx

3

= 2

1.8 Higher Order Derivatives

d −3 x = (2 ⋅ −3) x −3 −1 = −6 x −4 dx

c. Given y = x( x + 1)3 , then dy d   d   = ( x + 1)3 x  +  x ( x + 1)3  = dx dx   dx   d2y dx

2

=

[( x + 1) ⋅1] + [ x ⋅ 3( x + 1) ⋅1] = ( x + 1) 3

3−1

[

3

+ 3 x( x + 1)

][

2

][

]

d d d ( x + 1)3 + 3( x + 1) 2 x  + 3x ( x + 1) 2  = 3( x + 1)3−1 ⋅ 1 + 3( x + 1) 2 ⋅ 1 + 3x ⋅ 2 ( x + 1) 2−1 ⋅ 1 dx dx dx    

2 = 3( x + 1) 2 + 3( x + 1) 2 + 6 x ( x + 1) = 6( x + 1) + 6 x ( x + 1)

d3y dx

3

= 6

d (x + 1)2 +  6(x + 1) d dx dx 

  d  x  + 6 x (x + 1)  dx   

=

[12(x + 1)

2 −1

]{

[

⋅1 + [6(x + 1) ⋅1] + 6 x ⋅(x + 1)1−1 ⋅1

]}

= 12(x + 1) + 6(x + 1) + 6 x(x + 1)0 = 18(x + 1) + 6 x = 18 x + 18 + 6 x = 24 x + 18 = 6(4 x + 3) d. Given y = ( x + 1) −2 , then

[

]

dy d d = ( x + 1) −2 = −2( x + 1) −2 −1 ⋅ ( x + 1) = −2( x + 1) −3 ⋅1 = −2 ( x + 1) −3 dx dx dx d2y dx

2

d3y dx

3

= −2 = 6

d ( x + 1) −3 dx

[

] dxd ( x + 1) = 6 ( x + 1)

−4

⋅ 1 = 6 ( x + 1)

[(6⋅ −4) ( x + 1) ] ⋅ dxd ( x + 1) = −24 ( x + 1)

−5

⋅ 1 = −24 ( x + 1)

= ( −2 ⋅ −3) ( x + 1) −3−1 ⋅

d ( x + 1) −4 = dx

−4 −1

−4

−5

e. Given y = x 3 + 3x 2 + 10 , then d dy d 3 d 3 d d x + x + 3 x 2 + 0 = 3x 3−1 + (3 ⋅ 2) x 2−1 = 3 x 2 + 6 x 3x 2 + 10 = = dx dx dx dx dx dx d2y dx

2

d3y dx

3

=

d d d d 3x 2 + 6 x = 3 x 2 + 6 x = (3 ⋅ 2) x 2 −1 + (6 ⋅ 1) = 6 x + 6 dx dx dx dx

=

d d d 6x + 6 = 6 x + 0 = 6 ⋅1 = 6 dx dx dx

f. Given y =

1 1+ x

, then

d   d   (1 + x ) dx 1 − 1 ⋅ dx (1 + x ) 0 − (1 ⋅ 1) 1 dy     = = = − 2 2 dx (1 + x ) (1 + x ) (1 + x ) 2

Hamilton Education Guides

81

Basic Differentiation

2

d y dx

=

2

d3y dx

=

3

1.8 Higher Order Derivatives

 2 2 d   d (1 + x ) dx 1 − 1 ⋅ dx (1 + x )  −

(1 + x )

= −

4

d  3 d   3 (1 + x ) dx 2 − 2 ⋅ dx (1 + x )     

(1 + x )

(1 + x )

[

(1 + x )

]

4

0 − (2 ⋅ 3) (1 + x )3−1

=

6

[

0 − (1 ⋅ 2 ) (1 + x )2−1

]

=

= −

6

2(1/ + x/ )

(1 + x)

4/ = 3

6 (1 + x )

(1 + x)

2/

6/ = 4

=

2

(1 + x ) 3

= −

6

(1 + x ) 4

1 which is the same as y = x − x −1 , then x

g. Given y = x −

dy d d −1 x− x = 1 + x −1−1 = 1 + x −2 = dx dy dx d2y dx

=

2

d3y dx

d d −2 x 1+ dy dx

= −2

3

d −3 x dy

= 0 − 2 x −2−1 = −2 x −3

= ( −2 ⋅ −3) x −3−1 = 6 x −4

h. Given y = ax 3 + bx , then dy d d d 3 d bx = a ax 3 + x + b x = (a ⋅ 3) x 3−1 + (b ⋅ 1) x 1−1 = 3ax 2 + bx 0 = 3ax 2 + b = dx dx dx dx dx d2y dx

2

d3y dx

3

=

d d 3ax 2 + b dx dx

=

d d 6ax = 6a x = 6a ⋅ 1 = 6a dx dx

i. Given y =

2

d y dx

2

x2

(

d 2 x +0 dx

= (3a ⋅ 2) x 2−1 = 6ax

, then

)

(

)

d 2  2 d 3   3  x dx x + 1  −  x + 1 dx x     

dy = dx

=

x3 +1

= 3a

x 3x 4 − 2 x 4 − 2 x x4

(

= =

=

4

x 4 − 2x

=

=

x4

)

(

(

x/ x 3 − 2 x 4/ =3

)

d 3  3 d 3   3  x dx x − 2  −  x − 2 dx x      x 3x 5 − 3x 5 + 6 x 2 x

6

Hamilton Education Guides

6

=

6 x 2/ x

(

6/ = 4

=

)

d  d x 2  x 3 + 1 − x 3 + 1 ⋅ 2 x dx dx   x4

=

)

=

=

[

] (

)

x 2 3x 3−1 + 0 − 2 x x 3 + 1 x

4

x3 − 2 x3

(

)

d  d x3  x3 − 2  − x 3 − 2 ⋅ 3x 2 dx dx   x6

=

[

]

(

x 3 3 x 3−1 − 0 − 3 x 2 x 3 − 2 x

6

6

x4

82

)

Basic Differentiation

d3y dx

3

x4

=

1.8 Higher Order Derivatives

( x ⋅ 0) − (6 ⋅ 4)x =

d 4 d 6−6 x dx dx x8

4

x

4−1

=

8

0 − 24 x 3 x

= −

8

24 x 3/ x

= −

8/ =5

24

x5

Example 1.8-5: Find y ′ and y ′′ for the following functions. Do not simplify the answer to its lowest term. b. xy + y 2 = 1

a. x 2 + y 2 = 2

d. x 3 y + y = 1

c. 1 + x 2 y 2 = x

Solutions: −2 x 2y

a. Given x 2 + y 2 = 2 , then y ′ is equal to 2 x 2−1 + 2 y ⋅ y ′ = 0 ; 2 x + 2 y y ′ = 0 ; 2 y y ′ = −2 x ; y ′ = ; y′ = −

x (1⋅ y ) − ( y ′ ⋅ x ) y ′′ = − y − y ′ x and y ′′ = − ; y y2 y2

b. Given xy + y 2 = 1 , then y ′ is equal to (1 ⋅ y + y ′ ⋅ x ) + 2 y ⋅ y ′ = 0 ; y + y ′ x + 2 y y ′ = 0 ; y ′ ( x + 2 y ) = − y

; y′ = −

[y ′ ⋅ (x + 2 y ) ] − [ (1 + 2 y ′)⋅ y ] y and y ′′ = − x + 2y ( x + 2 y )2

(

c. Given 1 + x 2 y 2 = x , then y ′ is equal to 0 + 2 x ⋅ y 2 + 2 y y ′ ⋅ x 2

2

2

; 2 y y ′ x = 1 − 2x y ; y′ =

; y ′′ =

(

1 − 2 x y2

and

2x2 y

) (

)(

)

( x + 2 y) 2

; 2x y 2 + 2 y y ′ x 2 = 1

=1

+ 2 yy ′ ⋅ x

xy ′ − y

) ] (2 x y ) }− {2(2 xy + x y ′)(1 − 2 xy )} (2 x y ) 2

2

2

2

2

)

4x y

(

= −

=

2

( x + 2 y)

; y ′′ = −

2

4 2

)

d. Given x 3 y + y = 1 ; y x 3 + 1 = 1 ; y =

y ′′

{[0 − 2(1⋅ y y ′′ =

−4 x 2 y y 2 + 2 xyy ′ − 2 2 xy + x 2 y ′ 1 − 2 xy 2

xy ′ + 2 y y ′ − y − 2 y y ′

; y ′′ = −

1 x3 +1

) [( ) (x + 1)

(

2  2 3 3 2 − 6 x ⋅ x + 1  − 2 x + 1 ⋅ 3 x ⋅ −3 x  4

3

(

6 x − x 3 − 1 + 3x 3

(x + 1) 3

3

Hamilton Education Guides

)

=

(

, then y ′ =

]

=

(

[0 ⋅ (x + 1) ]− [3x ⋅1 ] (x + 1) 3

2

2

3

)

2

(

3

4

− 3x 2

and

(x + 1)

2

3

) = 6 x(x + 1)[ − (x + 1)+ 3x ] (x + 1)

− 6 x x 3 + 1 + 18 x 4 x 3 + 1

(x + 1)

=

3

3

3

3

4

)

6x 2x 3 −1

(x + 1) 3

3

83

Basic Differentiation

1.8 Higher Order Derivatives

Example 1.8-6: Find the first, second, and third derivative of the following functions. a. f (x ) = 6 x 7 + 7 x 2 − 2

b. f ( x ) = 3x 4 − 2 x 2 + 5x + 9

c. f ( x ) = x −4 + 3x −3 + x −2 + x

d. f ( x ) = x 7 + 6 x 5 + 8 x + 3x −3

e. f ( x ) =

3 x

3

+

g. f ( x ) = x 3 +

2 x

2

+

1 x

1

x3

f. f ( x ) = x (2 x + 1)3 h. f ( x ) = ( x + 1) 2 − x 3

Solutions: a. Given f ( x ) = 6 x 7 + 7 x 2 − 2 , then f ′( x ) = (6 ⋅ 7) x 7−1 + (7 ⋅ 2) x 2−1 − 0 = 42 x 6 + 14 x f ′′( x ) = (42 ⋅ 6) x 6−1 + 14 x 1−1 = 252 x 5 + 14 f ′′′( x ) = (252 ⋅ 5) x 5−1 + 0 = 1260 x 4

b. Given f ( x ) = 3x 4 − 2 x 2 + 5x + 9 , then f ′( x )

= (3 ⋅ 4) x 4−1 − (2 ⋅ 2) x 2−1 + 5x1−1 + 0 = 12 x 3 − 4 x + 5

f ′′( x ) = (12 ⋅ 3) x 3−1 − 4 x 1−1 = 36 x 2 − 4 f ′′′( x ) = (36 ⋅ 2) x 2−1 − 0 = 72x

c. Given f ( x ) = x −4 + 3x −3 + x −2 + x , then f ′( x )

= −4 x −4−1 + (3 ⋅ −3) x −3−1 − 2 x −2−1 + x1−1 = −4 x −5 − 9 x −4 − 2 x −3 + 1

f ′′( x ) = ( −4 ⋅ −5) x −5−1 + ( −9 ⋅ −4) x −4−1 + ( −2 ⋅ −3) x −3−1 + 0 = 20 x −6 + 36 x −5 + 6 x −4

f ′′′( x ) = (20 ⋅ −6) x −6−1 + (36 ⋅ −5) x −5−1 + (6 ⋅ −4) x −4−1 = −120 x −7 − 180 x −6 − 24 x −5

d. Given f ( x ) = x 7 + 6 x 5 + 8 x + 3x −3 , then f ′( x ) = 7 x 7−1 + (6 ⋅ 5) x 5−1 + 8 x 1−1 + (3 ⋅ −3) x −3−1 = 7 x 6 + 30 x 4 + 8 − 9 x −4 f ′′( x ) = (7 ⋅ 6) x 6−1 + (30 ⋅ 4) x 4−1 + 0 + ( −9 ⋅ −4) x −4−1 = 42 x 5 + 120 x 3 + 36 x −5 f ′′′( x )

= (42 ⋅ 5) x 5−1 + (120 ⋅ 3) x 3−1 + (36 ⋅ −5) x −5−1 = 210 x 4 + 360 x 2 − 180 x −6

Hamilton Education Guides

84

Basic Differentiation

e. Given f ( x ) =

1.8 Higher Order Derivatives

3

2 1 + 2 + x x x 3

which is equal to f ( x ) = 3x −3 + 2 x −2 + x −1 , then

f ′( x ) = (3 ⋅ −3) x −3−1 + (2 ⋅ −2) x −2−1 − x −1−1 = −9 x −4 − 4 x −3 − x −2 f ′′( x ) = ( −9 ⋅ −4) x −4−1 + ( −4 ⋅ −3) x −3−1 + ( −1 ⋅ −2) x −2−1 = 36 x −5 + 12 x −4 + 2 x −3 f ′′′( x ) = (36 ⋅ −5) x −5−1 + (12 ⋅ −4) x −4−1 + (2 ⋅ −3) x −3−1 = −180 x −6 − 48 x −5 − 6 x −4

f. Given f ( x ) = x (2 x + 1)3 , then f ′( x )

]

][

[

3 2 = 1 ⋅ (2 x + 1)3 + 3(2 x + 1)3−1 ⋅ 2 ⋅ x = ( 2 x + 1) + 6 x ( 2 x + 1)

[

f ′′( x ) = 3(2 x + 1)

3−1

][

2

⋅ 2 + 6 ⋅ (2 x + 1) + 2 (2 x + 1)

2−1

⋅ 2 ⋅ 6x

] = 6(2x + 1)

2

2

+ 6(2 x + 1) + 24 x (2 x + 1)

2 = 12( 2 x + 1) + 24 x ( 2 x + 1)

f ′′′( x )

]

][

[

= (12 ⋅ 2)(2 x + 1) 2−1 ⋅ 2 + 24 ⋅ (2 x + 1) + (2 x + 1)1−1 ⋅ 2 ⋅ 24 x = 48 (2 x + 1) + [24 (2 x + 1) + 48 x ] = 72 ( 2 x + 1) + 48 x

g. Given f ( x ) = x 3 +

1

x3

which is equal to f ( x ) = x 3 + x −3 , then

f ′( x ) = 3x 3 −1 − 3x −3 −1 = 3 x 2 − 3 x −4 f ′′( x ) = (3 ⋅ 2) x 2−1 + ( −3 ⋅ −4) x −4−1 = 6 x + 12 x −5 f ′′′( x ) = 6 x 1−1 + (12 ⋅ −5) x −5−1 = 6 x 0 − 60 x −6 = 6 − 60 x −6

h. Given f ( x ) = ( x + 1) 2 − x 3 , then

[

f ′( x ) = 2( x + 1)

2−1

]

⋅ 1 − 3x 3−1

= 2( x + 1) − 3x 2 = 2 x + 2 − 3x 2 = −3 x 2 + 2 x + 2

f ′′( x ) = ( −3 ⋅ 2) x 2−1 + 2 x 1−1 + 0 = −6 x + 2 x 0 = −6 x + 2 f ′′′( x )

= −6 x1−1 + 0 = −6 x 0 = −6 Section 1.8 Practice Problems - Higher Order Derivatives

1. Find the second derivative of the following functions. a. y = x 3 + 3x 2 + 5 x − 1 Hamilton Education Guides

b. y = x 2 (x + 1)2

c. y = 3x 3 + 50 x 85

Basic Differentiation

d. y = x 5 + g. y = x 4 +

1 x

2

x 8 − 7 x 5 + 5x 10

1.8 Higher Order Derivatives

e. y =

(

x3 − 5x 2 x +1

)

f. y = x 3 x 2 − 1 1

h. y = x 2 −

1 x +1

i. y =

b. y = x 2 +

1 x

c. y = 4 x 3 (x − 1)2

− 3x

x2

2. Find y ′′′ for the following functions. a. y = x 5 + 6 x 3 + 10 d. y =

x x +1

e. y = x 8 − 10 x 5 + 5 x − 10

f. y =

x −1 x2

+ 5x 3

3. Find f ′′(0) and f ′′(1) for the following functions. a. f (x ) = 6 x 5 + 3x 3 + 5

b. f (x ) = x 3 (x + 1)2

d. f (x ) = (x − 1)−3

e. f (x ) = (x − 1) x 2 + 1

Hamilton Education Guides

(

c. f (x ) = x + (x − 1)2

)

(

)

2

f. f (x ) = x 3 − 1 + 2 x

86

Appendix – Exercise Solutions Section 1.1 Practice Problems – The Difference Quotient Method 1. Find the derivative of the following functions by using the difference quotient method.

[(x + h) − 1]− (x − 1) = x/ 2

f (x + h ) − f (x ) = h

a. Given f (x ) = x 2 − 1 , then

= h + 2 x . Therefore, f ′(x ) = lim h→0

h/ (h + 2 x ) + h 2 + 2hx − 1/ − x/ 2 + 1/ h 2 + 2hx = = h/ h h

2

h

f (x + h ) − f (x ) h

f (x + h ) − f (x ) = h

b. Given f (x ) = x 3 + 2 x − 1 , then

2

= lim h→0 (h + 2 x ) = 0 + 2 x = 2 x

[(x + h) + 2(x + h) − 1]− [x 3

3

]

+ 2x − 1

h

(

)

h3 + 3 x 2 h + 3 xh 2 + 2h h/ h 2 + 3 x 2 + 3 xh + 2 x/ 3 + h3 + 3 x 2 h + 3 xh 2 + 2/ x/ + 2h − 1/ − x/ 3 − 2/ x/ + 1/ = = = h 2 + 3 x 2 + 3 xh + 2 . h h h/ f (x + h ) − f (x ) Therefore, f ′(x ) = lim h→0 = lim h →0 h 2 + 3 x 2 + 3 xh + 2 = 02 + 3 ⋅ x 2 + 3 ⋅ 0 ⋅ h + 2 = 3 x 2 + 2 h

=

(

)

(x + h )(x − 1) − x(x + h − 1) (x + h − 1)(x − 1)

x+h x − f (x + h ) − f (x ) x x + h − 1 x −1 = , then = c. Given f (x ) = h x −1 h x/ 2 − x/ + h/ x/ − h − x/ 2 − h/ x/ + x/

=

(

h x − x + hx − h − x + 1

d. Given f (x ) = −

=

1

(

2

x

2

2

hx x + h + 2hx

f ′(x ) = lim h →0

)

=

2

(

h/ (h + 2 x ) 2

2

h/ x x + h + 2hx

)

1

x 2 − 2 x + hx − h + 1

(x + h )(x − 1) − x(x + h − 1) h [ (x + h − 1)(x − 1)]

. Therefore,

−

1

(x + h )

2

+

− x 2 + (x + h )2

1 x

2

h =

)

4

x 2 (x + h )2 h

=

h + 2x

x + h 2 x 2 + 2hx3

− x 2 + (x + h )2

=

=

hx (x + h ) 2

2

(

− x 2 + x 2 + h 2 + 2hx 2

(

2

2

hx x + h + 2hx

)

)

. Therefore,

f (x + h ) − f (x ) 2 2 x/ 0 + 2x h + 2x   = lim h →0  4 = 4/ =3 =  = 4 2 2 3 2 2 3 h x + 0 ⋅ x + 2⋅0⋅ x x x3  x + h x + 2hx 

e. Given f (x ) = 20 x 2 − 3 , then

[

][

] = [20(x

20(x + h )2 − 3 − 20 x 2 − 3 f (x + h ) − f (x ) = h h

2

) ][

]

+ 2hx + h 2 − 3 − 20 x 2 − 3 h

40hx + 20h 2 2/ 0/ x + 40hx + 20h 2 − 3/ − 2/ 0/ x + 3/ h/ (40 x + 20h ) = = = 40 x + 20h h h h/ 2/

2/

Therefore, f ′(x ) = lim h→0 f. Given f (x ) =

=

(

= −

h/ x − 2 x + hx − h + 1

f (x + h ) − f (x ) = h

, then

− x/ 2 + x/ 2 + h 2 + 2hx 2

h/ 2

=

f (x + h ) − f (x ) −1 −1 1 −1   = lim h →0  2 = 2 = −  = 2 h x − 2x + 1 x − 2x + 0 ⋅ x − 0 + 1 ( x − 1)2  x − 2 x + hx − h + 1 

f ′(x ) = lim h→0

=

= −

)

2

h

x3 , then

f (x + h ) − f (x ) = lim h →0 (40 x + 20h ) = 40 x + (20 ⋅ 0 ) = 40 x + 0 = 40 x h

f (x + h ) − f (x ) = h

x3 + h3 + 3 x 2 h + 3 xh 2 − x3 h x3 + h3 + 3 x 2 h + 3 xh 2 + x3

Hamilton Education Guides

=

(x + h )3 − h

(

x3

=

h/ h 2 + 3 x 2 + 3 xh

(x + h )3 −

)

h/ x3 + h3 + 3 x 2 h + 3 xh 2 + x3

h

=

x3

⋅

(x + h )3 + (x + h )3 +

x3 x3

=

h

h 2 + 3 x 2 + 3 xh x3 + h3 + 3 x 2 h + 3 xh 2 + x3

(x + h )3 − x3 (x + h )3 + x3 . Therefore,

87

Basic Differentiation

 f (x + h ) − f (x ) = lim h →0  h  

f ′(x ) = lim h→0

3x 2

=

x3 + x3

= 10 ⋅

3x 2

=

2 x3

10

g. Given f (x ) =

= 10 ⋅

(

3x 2

=

3

x−5 − x+h−5

(h/ ⋅

3

f (x + h ) − f (x ) h

10 x+h−5 h

=

x−5 + x+h−5

⋅

h⋅ x + h −5 ⋅ x −5

  =  x3 + h3 + 3 x 2 h + 3 xh 2 + x3  h 2 + 3 x 2 + 3 xh

3 2 − 32 3 2− 2 3 x ⋅x = = x x 2 2 2

=

2x 2

, then

x−5

= 10 ⋅

x−5 + x+h−5

− h/

)(

x+h−5 ⋅ x−5 ⋅

x−5 + x+h−5

  f (x + h ) − f (x ) = lim h →0  h  

f ′(x ) = lim h→0

=

Solutions

−10

)(

x−5 ⋅ x−5 ⋅

x−5 + x−5

)

=

(

=

)

(h ⋅

4 −3 2

x 3 + 03 + 3 x 2 ⋅ 0 + 3 x ⋅ 0 2 + x 3

1

3 3 2 x x = 2 2

=

10

−

02 + 3x 2 + 3x ⋅ 0

10 x − 5 − 10 x + h − 5

x−5 =

x+h−5 ⋅ x−5 h

x/ − 5/ − x/ − h + 5/

)(

x+h−5 ⋅ x−5 ⋅

x−5 + x+h−5

−10

(

)(

x+h−5 ⋅ x−5 ⋅

x−5 + x+h−5

   = x−5 + x+h−5  

− 10

)(

x+h−5 ⋅ x−5 ⋅

5

−10

(x − 5) ⋅ (2

)

x−5

)

=

− 1/ 0/ 2/ (x − 5)

(

x−5

)

=

h⋅ x + h −5 ⋅ x −5

)

. Therefore,

)

−10

)(

(

x+0−5 ⋅ x−5 ⋅

−5

=

1+ 1

(x − 5)

10 x − 5 − 10 x + h − 5

=

2

−5

( x − 5)

3 2

x−5 + x+0−5 5

= −

( x − 5)3

 a (x + h ) + b   ax + b  cx(ax + ah + b ) − (cx + ch ) (ax + b )  −  ( ) + c x h cx f (x + h ) − f (x ) ax + b cx(cx + ch )    , then =  = h. Given f (x ) = cx h h h

=

a/ c/ x/ 2 + a/ c/ h/ x/ + b/ c/ x/ − a/ c/ x/ 2 − b/ c/ x/ − a/ c/ h/ x/ − bch −bc/ h/ −b acx 2 + achx + bcx − acx 2 − bcx − achx − bch = = = 2 . c/ h/ x(cx + ch ) chx(cx + ch ) chx(cx + ch ) cx + chx

Therefore, f ′(x ) = lim h→0

f (x + h ) − f (x ) b −b b   = lim h →0  2 = − 2  = − 2 h cx + c ⋅ 0 ⋅ x cx  cx + chx 

2. Compute f ′(x ) for the specified values by using the difference quotient equation as the lim h→0 . a. Given f (x ) = x3 , then using f ′(x ) = lim h →0 f ′(1) = lim h →0

(1 + h )3 − 13 = h

lim h →0

(x + h )3 − x3 at x = 1 f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h h

(

b. Given f (x ) = 1+ 2 x , then using f ′(x ) = lim h →0 f ′(0 ) = lim h →0

f ′(−1) = lim h →0

(

f (x + h ) − f (x ) 1 + 2(x + h ) − (1 + 2 x ) we obtain f ′(x ) = lim h →0 at x = 0 h h

1 + 2(0 + h ) − (1 + 2 ⋅ 0 ) 1/ + 2h − 1/ 2h/ = lim h →0 = lim h →0 = lim h →0 2 = 2 h h/ h

c. Given f (x ) = x 3 + 1 , then using f ′(x ) = lim h →0

= lim h →0

)

1/ + h3 + 3h + 3h 2 − 1/ h/ h 2 + 3 + 3h = lim h →0 = lim h →0 h 2 + 3 + 3h = 02 + 3 + 3 ⋅ 0 = 3 h h/

[(− 1 + h) + 1]− [(− 1) + 1] 3

)

3

h

(

[

](

)

(x + h )3 + 1 − x3 + 1 f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h h

= lim h →0

(− 1/ + h

3

)

(

at x = −1

)

+ 3h + 3h 2 + 1/ − (− 1/ + 1/ ) h3 + 3h + 3h 2 − 0 = lim h →0 h h

)

h/ h 2 + 3 + 3h = lim h →0 h 2 + 3 + 3h = 02 + 3 + 3 ⋅ 0 = 3 h/

Hamilton Education Guides

88

)

Basic Differentiation

Solutions

d. Given f (x ) = x 2 (x + 2 ) = x 3 + 2 x 2 , then using f ′(x ) = lim h →0

[(x + h) + 2(x + h) ]− (x 3

f ′(x ) = lim h →0

3

+ 2x2

)

at x = 2 f ′(2) = lim h →0

h

[8 + h

= limh →0

2

3

]

+ 12h + 6h 2 + 8 + 2h 2 + 8h − 16

(

= lim h →0

h

)

f (x + h ) − f (x ) we obtain h

[(2 + h) + 2(2 + h) ]− (2 3

2

3

+ 2 ⋅ 22

)

h

(

1/ 6/ + h3 + 20h + 8h 2 − 1/ 6/ h/ h 2 + 8h + 20 = lim h →0 h h/

)

= lim h →0 h 2 + 8h + 20 = 02 + 8 ⋅ 0 + 20 = 20 1

e. Given f (x ) = x −2 + x −1 + 1 =

x

2

1 + 1 + h + (1 + h )2

(1 + h )

2

f ′(1) = lim h →0

=

2

3/ + h + 3h − 3/ − 3h − 6h

(

2

h 1 + h + 2h

(−2 ⋅ 0) − 3 =

( 10 + h )−

1 10 + 10

1 + 1 + 12 2

1

10

h

=

1 + x + h + (x + h )2

= lim h →0

2

− 2h − 3h

(

2

h 1 + h + 2h

)

1 + x + x2 x2

at x = 1

h

1 + h 2 + 2h h

= lim h →0

−

(x + h )2

2 + h + 1 + h 2 + 2h

x + 2 , then using f ′(x ) = lim h →0

at x = 10 f ′(10) = lim h →0

=

)

)

= lim h →0

−3

= lim h →0

h/ (−2h − 3)

(

2

h/ 1 + h + 2h

)

1 2 10

=

h

( 10 + h )+ 10 ( 10 + h )+ 10

⋅

= lim h →0

(

(

2

h 1 + h + 2h

= lim h →0

f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h

( 10 + h + 2)− ( 10 + 2) = lim

(

3 + h 2 + 3h − 3 1 + h 2 + 2h

−3 = −3 1

1 + 02 + 2 ⋅ 0

= lim h →0

−

h

2

f. Given f (x ) =

(

1 x + x2 x + x 2 + x3 1 + x + x2 x/ 1 + x + x 2 = = = , then using +1 = + 1 x x3 x3 x3/ = 2 x2

f (x + h ) − f (x ) we obtain f ′(x ) = lim h →0 h

f ′(x ) = lim h →0

= lim h →0

+

(

)

)

−2h − 3

1 + h 2 + 2h

)(

x+h +2 − h

x +2

)

10 + h + 2/ − 10 + 2/ 10 + h − 10 = lim h →0 h h

h →0

1/ 0/ + h − 1/ 0/

h ⋅ 10 + h + 10

= lim h →0

)

(

h/

h/ ⋅ 10 + h + 10

)

=

1 10 + 0 + 10

1 1 = = 0.158 6.32 2 ⋅ 3.16

Section 1.2 Solutions - Differentiation Rules Using the Prime Notation 1. Find the derivative of the following functions. Compare your answers with the practice problem number one in Section 1.1. a. Given f (x ) = x 2 − 1 , then f ′(x ) = 2 x 2 −1 − 0 = 2 x b. Given f (x ) = x 3 + 2 x − 1 , then f ′(x ) = 3x 3−1 + 2 ⋅1x1−1 − 0 = 3x 2 + 2 x 0 = 3 x 2 + 2 c. Given f (x ) =

[1 ⋅ (x − 1)] − [1 ⋅ x] = x/ − 1 − x/ = − 1 x , then f ′(x ) = x −1 (x − 1)2 (x − 1)2 ( x − 1)2

d. Given f (x ) = −

1 x

2

, then f ′(x ) = −

(0 ⋅ x )− (2 x ⋅1) = − 0 − 2 x = 2

x

4

x

4

2 x/ x

4/ =3

=

2

x3

e. Given f (x ) = 20 x 2 − 3 , then f ′(x ) = (20 ⋅ 2)x 2 −1 − 0 = 40 x 3

f. Given f (x ) = x 3 = x 2 , then f ′(x ) =

Hamilton Education Guides

3 3− 2 3 32 −1 3 3 1 = x 2 = x2 = x x 2 2 2 2

89

Basic Differentiation

10

g. Given f (x ) = =

−5

(x − 5)

x−5

=

1 +1 2

Solutions

=

1

(x − 5) 2

−5

(x − 5)

0 ⋅ (x − 5) 12  − 10 ⋅ 1 (x − 5)− 12  −1     −5 2  = 0 − 5(x − 5) 2 = , then f ′(x ) =  1 x−5 x−5 (x − 5) 2 ⋅ (x − 5)

10

−5

=

3 2

(x − 5)

3

= −

5

( x − 5)

x−5

ax + b a ⋅ cx − c ⋅ (ax + b ) acx − acx − bc b bc , then f ′(x ) = = = − 2/ =1/ 2 = − 2 2 2 cx (cx ) (cx ) cx c x

h. Given f (x ) =

2. Differentiate the following functions: a. Given f (x ) = x 2 + 10 x + 1 , then f ′(x ) = 2 x 2−1 + 10 x 1−1 + 0 = 2 x + 10 x 0 = 2 x + 10 b. Given f (x ) = x 8 + 3 x 2 − 1 , then f ′(x ) = 8 x 8−1 + (3 ⋅ 2 )x 2−1 − 0 = 8 x 7 + 6 x c. Given f (x ) = 3 x 4 − 2 x 2 + 5 , then f ′(x ) = (3 ⋅ 4 )x 4−1 − (2 ⋅ 2 )x 2−1 + 0 = 12 x 3 − 4 x

(

)

d. Given f (x ) = 2 x 5 + 10 x 4 + 5 x = 2 x 5 + 20 x 4 + 10 x , then f ′(x ) = (2 ⋅ 5)x 5−1 + (20 ⋅ 4 )x 4−1 + (10 ⋅1)x 1−1 = 10 x 4 + 80 x 3 + 10 x 0 4

3

= 10 x + 80 x + 10

( )

( )

e. Given f (x ) = a 2 x 3 + b 2 x + c 2 , then f ′(x ) = 3 ⋅ a 2 x3−1 + 1 ⋅ b 2 x1−1 + 0 = 3a 2 x 2 + b 2 x 0 = 3a 2 x 2 + b 2 f. Given f (x ) = x 2 (x − 1) + 3 x = x3 − x 2 + 3 x , then f ′(x ) = 3 x 3−1 − 2 x 2−1 + 3 x1−1 = 3 x 2 − 2 x + 3 x 0 = 3 x 2 − 2 x + 3

(

)(

[ (

)

)]

)] [ (

g. Given f (x ) = x3 + 1 x 2 − 5 , then f ′(x ) = 3 x 2 x 2 − 5 + 2 x x3 + 1

(

)

h. Given f (x ) = 3 x 2 + x − 1 (x − 1) , then f ′(x ) =

= 3 x 4 − 15 x 2 + 2 x 4 + 2 x = 5 x 4 − 15 x 2 + 2 x

[ (6 x + 1) ⋅ (x − 1) ] + [1 ⋅ (3x 2 + x − 1) ]

= 6 x 2 − 6 x + x − 1 + 3x 2 + x − 1

= (6 + 3)x 2 + (− 6 + 1 + 1)x + (− 1 − 1) = 9 x 2 − 4 x − 2

(

)

i. Given f (x ) = x x 3 + 5 x 2 − 4 x = x 4 + 5 x 3 − 4 x , then f ′(x ) = 4 x 4−1 + (5 ⋅ 3)x 3−1 − (4 ⋅1)x 1−1 = 4 x 3 + 15 x 2 − 4 x 0 = 4 x 3 + 15 x 2 − 4 j. Given f (x ) = k. Given f (x ) = =

9x 6 + 6x 9x

4

=

l. Given f (x ) =

[(3x + 0)⋅ x]− [1 ⋅ (x + 1)] 2

x3 +1 , then f ′(x ) = x

x 5 + 2x 2 −1 3x 2

(

3/ x/ 3 x5 + 2 3 = 9/ x

4/ =3

3

, then f ′(x ) =

) = 3x

5

[(5x

x 4

=

2

3x 3 − x 3 − 1 x

) ][ ( (3x )

=

2

)]

+ 4 x ⋅ 3x 2 − 6 x ⋅ x5 + 2 x 2 − 1

2x3 − 1 x2

=

2 2

15 x 6 + 1/ 2/ x/ 3/ − 6 x 6 − 1/ 2/ x/ 3/ + 6 x 9x4

+2

3x3

[ ]

[ 2 x ⋅ (4 x − 1) ] − 4 ⋅ x 2 x2 x2 , then f ′(x ) = = (x − 1) + 3x 4 x − 1 (4 x − 1)2

=

8x 2 − 2x − 4x 2

(4 x − 1)

2

=

4x 2 − 2x

(4 x − 1)

2

=

2 x (2 x − 1)

(4 x − 1)2

1 x2  m. Given f (x ) = x 2  2 +  = 2 x 2 + = 2 x 2 + x , then f ′(x ) = (2 ⋅ 2 )x 2−1 + x 1−1 = 4 x + x 0 = 4 x + 1 x x  n. Given f (x ) = (x + 1) ⋅ =

(4 − 2)x 2 − 4 x − 2 (x − 1)2

o. Given f (x ) =

2x 2x 2 + 2x , then f ′(x ) = = x −1 x −1

=

[ (4 x + 2) ⋅ (x − 1) ] − [1 ⋅ (2 x 2 + 2 x ) ] (x − 1)2

=

4 x 2 − 4 x + 2/ x/ − 2 − 2 x 2 − 2/ x/

(x − 1)2

2x2 − 4x − 2

x 3 + 3x − 1 x4

Hamilton Education Guides

( x − 1)2

[ (3x + 3)⋅ x ]−  4 x ⋅ (x 2

, then f ′(x ) =

4

3

x8

3

)

 + 3x − 1  

=

3 x 6 + 3 x 4 − 4 x 6 − 12 x 4 + 4 x 3 x8

90

Basic Differentiation

=

Solutions

− x 6 − 9x 4 + 4x 3 x

=

8

(

(

x 3/ − x 3 − 9 x + 4 x

8/ =5

)= − x

3

+ 9x − 4 x5

(

)

)(

2 3  3    2x 3 + 5   , then f ′(x ) = 2 x ⋅  2 x + 5  +  6 x ⋅ x − 2 x + 5 p. Given f (x ) = x 2 − 1  2   x   x  x  

(

)

(

) ⋅ (x − 1) = 4 x  2



3

+ 10

)

5 3 2 5 3 2  3   6x 3 − 2x 3 − 5  2  ⋅ x − 1 = 4 x 3 + 10 +  4 x − 5  x 2 − 1 = 4 x 3 + 10 + 4 x − 4 x − 5 x + 5 = 8 x − 4 x + 5 x + 5 + 2 2 2 2  x    x x x    

q. Given f (x ) = =

[(

][(

)

12 x3 + 2 x ⋅ (x − 1) − 1 ⋅ 3 x 4 + x 2 + 2 3x 4 + x 2 + 2 , then f ′(x ) = x −1 (x − 1)2

)]

12 x 4 − 12 x 3 + 2 x 2 − 2 x − 3 x 4 − x 2 − 2

=

(x − 1)2

9 x 4 − 12 x 3 + x 2 − 2 x − 2

( x − 1)2

r. Given f (x ) = x −1 +

1 x −2

= x −1 + x 2 , then f ′(x ) = − x −1−1 + 2 x 2−1 = − x −2 + 2 x = −

1

+ 2x

x2

3. Compute f ′(x ) at the specified value of x . Compare your answers with the practice problem number two in Section 1.1. a. Given f (x ) = x3 , then f ′(x ) = 3x3−1 = 3x 2

at x = 1

f ′(x ) = 3 ⋅ 12 = 3

b. Given f (x ) = 1+ 2 x , then f ′(x ) = 0 + (2 ⋅1)x1−1 = 2x 0 = 2

at x = 0

f ′(x ) = 2

c. Given f (x ) = x 3 + 1 , then f ′(x ) = 3x 3−1 + 0 = 3x 2

at x = −1

f ′(x ) = 3 ⋅ (− 1)2 = 3

d. Given f (x ) = x 2 (x + 2) = x 3 + 2 x 2 , then f ′(x ) = 3x 3−1 + (2 ⋅ 2)x 2−1 = 3x 2 + 4 x at x = 2

f ′(x ) = 3 ⋅ 2 2 + 4 ⋅ 2 = 3 ⋅ 4 + 8 = 12 + 8 = 20

e. Given f (x ) = x −2 + x −1 + 1 , then f ′(x ) = − 2 x −2−1 − x −1−1 + 0 = − 2 x −3 − x −2 = − at x = 1

f ′(x ) = −

2 13

−

1

1 12

f ′(x ) =

1 2 10

=

x

3

−

1 x2

= −2 − 1 = −3

f. Given f (x ) = x + 2 = x 2 + 2 , then f ′(x ) = at x = 10

2

1 12 −1 1 1− 2 1 −1 x +0 = x 2 = x 2 = 2 2 2

1 1 2x 2

=

1 2 x

1 1 = = 0.158 2 ⋅ 3.16 6.32

4. Find f ′(0 ) and f ′(2 ) for the following functions: a. Given f (x ) = x 3 − 3 x 2 + 5 , then f ′(x ) = 3 x 3−1 − (3 ⋅ 2 )x 2−1 + 0 = 3 x 2 − 6 x

(

)

( ) b. Given f (x ) = (x + 1)(x − 1) , then f ′(x ) = [3 x ⋅ (x − 1) ]+ [1 ⋅ (x + 1) ] = 3 x − 3 x + x + 1 = 4 x − 3 x + 1 Therefore, f ′(0 ) = (4 ⋅ 0 ) − (3 ⋅ 0 ) + 1 = 0 − 0 + 1 = 1 and f ′(2 ) = (4 ⋅ 2 )− (3 ⋅ 2 )+ 1 = 32 − 12 + 1 = 21 c. Given f (x ) = x (x + 1) = x + x , then f ′(x ) = 3 x = 3x + x = 3x + 1 +x Therefore, f ′(0 ) = (3 ⋅ 0 )+ 1 = 0 + 1 = 1 and f ′(2 ) = (3 ⋅ 2 )+ 1 = 12 + 1 = 13 Therefore, f ′(0 ) = 3 ⋅ 0 2 − (6 ⋅ 0 ) = 0 − 0 = 0 and f ′(2 ) = 3 ⋅ 2 2 − (6 ⋅ 2 ) = 12 − 12 = 0 2

3

3

2

3

3

2

3

2

3

2 −1

2

1−1

0

3

3

2

2

2

2

d. Given f (x ) = 2 x 5 + 10 x 4 − 4 x , then f ′(x ) = (2 ⋅ 5)x 5−1 + (10 ⋅ 4 )x 4−1 − 4 x 1−1 = 10 x 4 + 40 x 3 − 4 x 0 = 10 x 4 + 40 x3 − 4

(

)(

)

(

)(

)

Therefore, f ′(0 ) = 10 ⋅ 0 4 + 40 ⋅ 0 3 − 4 = −4 and f ′(2 ) = 10 ⋅ 2 4 + 40 ⋅ 2 3 − 4 = 160 + 320 − 4 = 476

Hamilton Education Guides

91

Basic Differentiation

Solutions

e. Given f (x ) = 2 x −2 − 3 x −1 + 5 x , then f ′(x ) = (2 ⋅ −2)x −2 −1 + (− 3 ⋅ −1)x −1−1 + 5 x1−1 = − 4 x −3 + 3 x −2 + 5 x 0

4

= − 4 x −3 + 3 x −2 + 5 = − f ′(2 ) = −

4

+

23

3 22

x

3

+5 = −

(

)

+

3 x

+ 5 . Therefore, f ′(0 ) = −

2

4 0

+

3

4 3 + + 5 = −0.5 + 0.75 + 5 = 5.25 8 4

3 02

+ 5 which is undefined due to division by zero and

f. Given f (x ) = x −2 x 5 − x 3 + x = x 3 − x + x = x 3 , then f ′(x ) = 3 x 3−1 = 3x 2 Therefore, f ′(0 ) = 3⋅ 0 2 = 0 and f ′(2 ) = 3⋅ 2 2 = 12

x

g. Given f (x ) =

1+ x

Therefore, f ′(0 ) =

2

, then f ′(x ) = 1− 02

(1 + 0 )

2 2

[1 ⋅ (1 + x )]− [2 x ⋅ x] (1 + x ) 2

=

2 2

= 1 and f ′(2 ) =

1− 22

(1 + 2 )

(1 + x )

(1 + 4)

2

=

−3 5

1 − x2

=

2 2

1− 4

=

2 2

1 + x 2 − 2x 2

(1 + x )

2 2

= −

2

3 25

(0 ⋅ x ) − (1 ⋅1) + 3x 3−1 = − 1 + 3x 2 1 + x 3 , then f ′(x ) = x x2 x2 1 1 1 Therefore, f ′(0 ) = − 2 + 3 ⋅ 02 which is undefined due to division by zero and f ′(2 ) = − 2 + 3 ⋅ 22 = − + 12 = 11.75 4 0 2

h. Given f (x ) =

( )

ax 2 + bx , then f ′(x ) = cx − d

i. Given f (x ) =

=

acx 2 − 2adx − bd

(cx − d )

f ′(2 ) =

( )

2

[ (2ax + b ) ⋅ (cx − d ) ] − [c ⋅ (ax 2 + bx ) ] (cx − d )2

. Therefore, f ′(0 ) =

(ac ⋅ 2 )− (2ad ⋅ 2) − bd 2

(c ⋅ 2 − d )2

=

(ac ⋅ 0 )− (2ad ⋅ 0) − bd 2

(c ⋅ 0 − d )

=

2

2acx 2 − 2adx + b/ c/ x/ − bd − acx 2 − b/ c/ x/

=

(cx − d )2

0 − 0 − bd

=

(0 − d )

2

−bd/ d

/ 1 2=

= −

b and d

4ac − 4ad − bd

(2c − d )2

5. Given f (x ) = x 2 + 1 and g (x ) = 2 x − 5 find h(x ) and h ′(x ) .

(

)

a. Given h(x ) = x 3 f (x ) where f (x ) = x 2 + 1 , then h(x ) = x 3 x 2 + 1 = x 5 + x 3 and h ′(x ) = 5 x 4 + 3 x 2

(

)

b. Given f (x ) = 3 + h(x ) where f (x ) = x 2 + 1 , then h(x ) = f (x ) − 3 = x 2 + 1 − 3 = x 2 − 2 and h ′(x ) = 2 x c. Given 2 g (x ) = h(x ) − 1 where g (x ) = 2 x − 5 , then h(x ) = 2 g (x ) + 1 = 2(2 x − 5) + 1 = 4 x − 10 + 1 = 4 x − 9 and h ′(x ) = 4 d. Given 3h(x ) = 2 x g (x ) − 1 where g (x ) = 2 x − 5 , then h(x ) = h ′(x ) =

[ (8 x − 10) ⋅ 3] − [0 ⋅ (4 x 2 − 10 x − 1) ] 32

=

2 x g (x ) − 1 2 x (2 x − 5) − 1 4 x 2 − 10 x − 1 = = and 3 3 3

8 x − 10 3(8 x − 10 ) = 9 3

e. Given 3[ f (x ) ]2 − 2h(x ) = 1 where f (x ) = x 2 + 1 , then h(x ) =

(

)

(

)

)

(

2 1 3 1 3 1 − 1 + 3[ f (x ) ]2 = − + [ f (x ) ]2 = − + x 2 + 1 and 2 2 2 2 2

h ′(x ) = 3 x 2 + 1 ⋅ 2 x = 6 x 3 + 6 x

(

)

(

f. Given h(x ) = g (x ) ⋅ 3 f (x ) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = (2 x − 5) ⋅ 3 x 2 + 1 = (2 x − 5) 3 x 2 + 3

)

= 6 x 3 − 15 x 2 + 6 x − 15 and h′(x ) = 18 x 2 − 30 x + 6 g. Given 3h(x ) − f (x ) = 0 where f (x ) = x 2 + 1 , then h(x ) =

Hamilton Education Guides

f (x ) x2 + 1 2 = and h ′(x ) = x 3 3 3

92

Basic Differentiation

Solutions

(

)

h. Given 2 g (x ) + h(x ) = f (x ) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = f (x ) − 2 g (x ) = x 2 + 1 − 2(2 x − 5) = x 2 + 1 − 4 x + 10 = x 2 − 4 x + 11 and h ′(x ) = 2 x − 4

(

)

i. Given f (x ) = x 3 + 5 x 2 + h(x ) where f (x ) = x 2 + 1 , then h(x ) = f (x ) − x 3 − 5x 2 = x 2 + 1 − x 3 − 5 x 2 = − x3 + (− 5 + 1)x 2 + 1 = − x 3 − 4 x 2 + 1 and h ′(x ) = − 3 x 2 − 8 x

(

3x3 − x3 − 1

=

[

][ (

)]

3x 2 ⋅ x − 1 ⋅ x3 + 1 x3 + 1 x3 +1 − x 2 + 1 and h ′(x ) = − f (x ) where f (x ) = x 2 + 1 , then h(x ) = x x x2

j. Given h(x ) =

x2

− 2x =

2x3 − 1 x2

)

− 2x

− 2x

(

)

k. Given h(x ) = 2 f (x ) + g (x ) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = 2 x 2 + 1 + (2 x − 5) = 2 x 2 + 2 + 2 x − 5 = 2 x 2 + 2 x − 3 and h ′(x ) = 4 x + 2

[ (

l. Given [h(x )] − f (x ) = 10 where f (x ) = x 2 + 1 , then [ h(x ) ]2 = 10 + f (x ) ; h(x ) =

(

)

= x 2 + 11

1 2

and h ′(x ) =

(

(

)

)

1 −1 −1 1 2 x + 11 2 ⋅ 2/ x = x x 2 + 11 2 2/

[4 ⋅ (x + 1)]− [2 x ⋅ (4 x − 10) ] = 4 x + 4 − 8x + 20 x = − 4 x + 20 x + 4 (x + 1) (x + 1) (x + 1) 3 f (x ) 1 n. Given = where f (x ) = x + 1 , then h(x ) = 3 x f (x ) = 3 x (x + 1) = 3 x h(x ) x 2

2

2

2

2

2

2

2

2

2

2

o. Given f (x ) =

2

[0 ⋅ (x + 1)]− [2 x ⋅1] − 0 = 0 − 2 x (x + 1) (x + 1) 2

2

2

3

+ 3 x and h ′(x ) = 9 x 2 + 3

f (x ) 1 1 , which is equivalent to ; f ( x ) ⋅ [ h ( x ) + 4] = 1 ⋅ 1 ; f ( x ) h ( x ) + 4 f ( x ) = 1 = h(x ) + 4 1 h(x ) + 4 1 1 1 − 4 f (x ) 1 4 f (x ) = = −4 = 2 − − 4 and h ′(x ) f (x ) f (x ) f (x ) f (x ) x +1

; f (x )h(x ) = 1 − 4 f (x ) , and f (x ) = x 2 + 1 , then h(x ) =

=

1 2

2 g (x ) 4 x − 10 2 g (x ) 2(2 x − 5) where f (x ) = x 2 + 1 and g (x ) = 2 x − 5 , then h(x ) = = = and h ′(x ) f (x ) h(x ) x2 + 1 x2 + 1

m. Given f (x ) =

=

)]

10 + f (x ) = 10 + x 2 + 1

2

2

2

= −

2x

(x + 1)

2

2

Section 1.3 Solutions - Differentiation Rules Using the

d dx

Notation

dy for the following functions: dx dy d 5 d d d 5 a. Given y = x 5 + 3 x 2 + 1 , then = x + 3 x 2 + 1 = 5 x 4 + (3 ⋅ 2 )x + 0 = 5 x 4 + 6 x x + 3x 2 + 1 = dx dx dx dx dx dy d d d 2 2 2 b. Given y = 3 x + 5 , then = 3x + 5 = 3 x + 5 = (3 ⋅ 2 )x + 0 = 6 x dx dx dx dx dy d 3 d 3 d −1 1 d  3 1 3 1 c. Given y = x − , then = x − x −1 = x = 3 x 2 + x −1−1 = 3 x 2 + x −2 = 3 x 2 + x − x −  = x dx dx dx dx x dx  x2

1. Find

(

)

(

)

(

(

)

)

(

)

 3 d 2  2 d 3   1 − x dx x  −  x dx 1 − x  dy d  x 2     = d. Given y = , then = =  2 dx dx  1 − x3  1 − x3 1 − x3

x2

=

2 x − 2 x 4 + 3x 4

(1 − x )

3 2

=

x4 − 2x

(1 − x )

Hamilton Education Guides

3 2

=

(

x x3 − 2

(1 − x )

3 2

)

(

)

[ (1 − x )⋅ 2 x]−  x ⋅ (− 3x ) (1 − x ) 3

2

2

 

3 2

93

Basic Differentiation

Solutions

d   d    (x − 1) ⋅ dx 1 − 1 ⋅ dx (x − 1)  1 dy d 1 1 d d         e. Given y = 4 x 2 + , then = 4x2 +   4x2 +  =  = 8x + x −1 dx dx  x −1 dx dx  x − 1  (x − 1)2

( )

= 8x +

0 −1

= 8x −

(x − 1)

2

1

( x − 1)2

(

) (

) ( ( )

) (

) [(x + 1)⋅ (2 x + 2)] (x + 1)

d 2 d 3  3   2   x + 1 dx x + 2 x  −  x + 2 x dx x + 1  dy d  x 2 + 2 x   =    , then = = f. Given y = 3 2 3 dx dx  x3 + 1  x +1 x +1

x2 + 2x

[(

) ] = 2 x + 2 x + 2 x + 2 − 3x − 6 x = − x − 4 x + 2 x + 2 (x + 1) (x + 1) (x + 1) d dy (x + 5x − 2 x ) = dxd x = g. Given y = x (x + 5 x − 2 ) = x + 5 x − 2 x , then dx dx − x 2 + 2 x ⋅ 3x 2

4

3

2

3

4

2

5

4

5

3

)

(

=

[(x − 1)⋅ (3x

2

)] [(

4

[(

3

5

1 x

3

]

)

(

) ]

, then

(x − 1)(x + 3) x2

(

=

− 2x + 6x x

4

d d 5 x 4 − 2 x3 = 5 x 4 + 20 x 3 − 6 x 2 dx dx

dy d d 3  = x3 + 3 x 2 (x − 1) =  (x − 1) x + 3x 2 dx dx dx 

) + (x

(

3

(

=

x 2 + 3x − x − 3 x2

) (

=

−2 x/ (x − 3) x

4/ =3

+ 3x 2

) dxd (x − 1) 

)

)

dy d d −3 d 3 1  d  = 5 x − x −3 = 5x − x = 5 + 3 x −3−1 = 5 + 3 x −4 = 5 + 4  5 x − 3  = dx dx dx dx dx  x  x

=

)

x 2 + (3 − 1)x − 3 x2

=

x2 + 2x − 3 x2

, then

dy d  x2 + 2x − 3  =   dx dx  x2 

d 2  2 d 2   2 2 2  x dx x + 2 x − 3  −  x + 2 x − 3 dx x      = x ⋅ (2 x + 2 ) − x + 2 x − 3 ⋅ 2 x = x4 x4 2

+

+ 6 x + x3 + 3 x 2 ⋅ 1 = 3 x3 + 6 x 2 − 3 x 2 − 6 x + x3 + 3 x 2 = 4 x3 + 6 x 2 − 6 x = 2 x 2 x 2 + 3 x − 3

i. Given y = 5 x −

j. Given y =

2

3

h. Given y = x 2 (x + 3)(x − 1) = x3 + 3 x 2 (x − 1) , then

2

3

3

2

3

3

4

3

3

= −

[

] [(

) ] = 2x

3

+ 2 x 2 − 2 x3 − 4 x 2 + 6 x x4

2 ( x − 3) x3

(

) (

)

d   d 2   2 2 3 ⋅ dx x − x  −  x − x ⋅ dx 3 dy d  x 2 − x   x − 1  x2 − x    = [3 ⋅ (2 x − 1) ] − x − x ⋅ 0 k. Given y = x , then = =  = 9 dx dx  3  3 32  3 

[(

2x − 1 3/ (2 x − 1) 3(2 x − 1) − 0 = = 9/ = 3 9 3 dy d 2 d    d  −1 2 = l. Given y = x (x + 3) , then x (x + 3)−1 = (x + 3)−1 x 2  +  x 2 (x + 3)−1  = dx dx dx dx    

) ]

=

[

= 2 x (x + 3)−1 − x 2 (x + 3)−2 =

]

[(x + 3)

−1

][ (

⋅ 2 x + x 2 ⋅ − (x + 3)−2

2x x2 − x + 3 ( x + 3 )2

x−3 dy  x −3 d  x   x  d  x −3  x  x −3 ⋅ =  m. Given y =   , then   +    =  5 dx  5  dx  1 + x   1 + x  dy  5  1+ x   5 

(1 + x ) d

d x − x (1 + x ) dx dx (1 + x )2

d d 5 ( x − 3) − ( x − 3) 5 x − 3 [(1 + x ) ⋅ 1] − [x ⋅ 1] x [5 ⋅ 1] − [(x − 3) ⋅ 0] x x − 3 x/ + 1 − x/ x 5/ dx dx = = ⋅ ⋅ + + ⋅ ⋅ + ⋅ 25 1+ x 5 1+ x 5 52 (1 + x )2 (1 + x )2 1 + x 2/ 5/ 5

=

x−3 x 1 x−3 x 1 ⋅ = + ⋅ + 5 (1 + x )2 1 + x 5 5(1 + x )2 5(1 + x )

Hamilton Education Guides

94

)]

Basic Differentiation

Solutions

d 3  3 d   3 3  (x − 1) dx x  −  x dx (x − 1)  dy x 1 d d x       = 3x 2 +  n. Given y = x3 1 + , then = x3 +   = x3 + dx x −1 x −1 dx dx  x − 1  (x − 1)2  = 3x 2 +

[(x − 1)⋅ 3x ]− [x ⋅1] 2

3

(x − 1)2

= 3x 2 +

3x3 − 3x 2 − x3

= 3x 2 +

(x − 1)2

2 x3 − 3x 2

= 3x2 +

(x − 1)2

x 2 (2 x − 3 )

( x − 1)2

(

)

(

)

d d  2    2  3 x + x dx (2 x − 1) − (2 x − 1) dx 3 x + x  dy 2x − 1 d  2x − 1  1  2x − 1     o. Given y =  , then =   =   = 2 dx dx  3 x 2 + x  x  3x + 1  3x 2 + x 3x 2 + x

[(3x + x)⋅ 2]− [(2 x − 1)⋅ (6 x + 1) ] = 6 x (3x + x) 2

=

2

2

2

(

(

)

(3x

(bx )

2

2abx 2 + b/ 2/ x/ − abx 2 − b/ 2/ x/ − bc

=

) = 6x

+ 2 x − 12 x 2 + 2 x − 6 x − 1 2

+x

)

2

dy ax 2 + bx + c d  ax 2 + bx + c  p. Given y = , then = =  bx dx dx  bx 

− ax 2 + bx + c ⋅ b

)

(

2 2

b x

2

2

2

=

3x 6 − 9 x 2 − 4 x 6 + 8 x3

(x − 3)

2

4

r. Given y =

=

=

(

5x

(1 + x )2

b x

) (

=

(

b/ ax 2 − c

)

(1 + x )4

2

2

=

b

2/ =1 2

x

) ( ( )

) = ax

2

(bx )2

−c

bx 2

) (

) [(x − 3)⋅ 3x ]− [(x − 2)⋅ 4 x ] (x − 3) 4

3

3

2

4

2

− x6 + 8 x3 − 9 x2

(x

4

−3

)

2

 (1 + x )2 d 5 x  − 5 x d (1 + x )2     dy d  5x  dx   dx  = , then = =  dx dx  (1 + x )2  (1 + x )4

5 x 2 + 2 x + 1 − 10 x(1 + x )

− 6x2 + 6x + 1

2

d 3 d 4  4   3   x − 3 dx x − 2  −  x − 2 dx x − 3  dy d  x3 − 2   =   , then = =  q. Given y = 4 2 dx dx  x 4 − 3  x −3 x4 − 3

x3 − 2

=

(bx )2

2 2

(

+ 2/ x/ − 12 x 2 − 2/ x/ + 6 x + 1

(3x + x) (3 x + x ) d   d   bx dx (ax + bx + c ) − (ax + bx + c ) dx bx  bx ⋅ (2ax + b )    

abx 2 − bc

=

2

=

5 x 2 + 1/ 0/ x/ + 5 − 1/ 0/ x/ − 10 x 2

(1 + x )4

=

[(1 + x) ⋅ 5]− [5x ⋅ 2(1 + x) ] 2

(1 + x )4

− 5x2 + 5

(1 + x )4

2. Find the derivative of the following functions:

(

)

( )

a.

d 2 d d 3t + 5t = 3t 2 + (5t ) = (3 ⋅ 2 ) t 2 −1 + (5 ⋅ 1) t1−1 = 6t + 5t 0 = 6t + 5 dt dt dt

b.

d d d (5 x ) + d (− 2) = (6 ⋅ 3)x 3−1 + (5 ⋅1)x1−1 + 0 = 18 x 2 + 5 x 0 = 18 x 2 + 5 6x 3 + 6 x3 + 5 x − 2 = dx dx dx dx

c.

d d d 3 d 3 u + 2u 2 + 5 = 2u 2 + 5 = 3u 3−1 + (2 ⋅ 2 )u 2−1 + 0 = 3u 2 + 4u u + du du du du

(

)

( )

(

)

( )

(

 d 2 5 ⋅ dt t + 2t d  t 2 + 2t  =  d. dt  5 

(

( )

) −  (t 52

2

) dtd 5

+ 2t ⋅

=

[5 ⋅ (2t + 2)] − [(t 2 + 2t )⋅ 0] 25

=

5 (2t + 2 ) − 0 2t + 2 5/ (2t + 2 ) = = or, 25 5 2/ 5/ 5

)

1 2t + 2 1 d 2 1 d d  d  t 2 + 2t  = t + 2t =  t 2 + 2t  = (2t + 2 ) = 5 5 5 dt 5  dt dt  dt  5 

(

) (

)

d 2  2 d 3   3 2 2 3  s ⋅ dt s + 3s − 1  −  s + 3s − 1 ⋅ dt s  d  s 3 + 3s − 1   = s ⋅ 3s + 3 − s + 3s − 1 ⋅ 2 s    e. =  ds  s4 s2 s4 

Hamilton Education Guides

[ (

)] [(

) ]

=

3s 4 + 3s 2 − 2 s 4 − 6 s 2 + 2 s s4

95

Basic Differentiation s 4 − 3s 2 + 2 s s

(

)

s/ s 3 − 3s + 2 s3 − 3s + 2 = 4/ =3 s s3

=

4

Solutions

 (1 + w) ⋅ d w2  − w2 ⋅ d (1 + w) 2 2  2     d  3 w  d 3 d  w  dw   dt  = 3w2 + [(1 + w) ⋅ 2 w] − w ⋅ 1 f. = = 3w 2 +  w + w + dw  1 + w  dw dw  1 + w  (1 + w)2 (1 + w)2

[ ]

= 3w2 +

2 w + 2 w2 − w2

= 3w 2 +

(1 + w)

2

[

[(

)]

(

w 2 + 2w

(1 + w )2

)] (

)(

) (

)(

) (

) (

)(

)(

)

d d 3 2 d 2 d 2 t3 + t 2 t 2 − 3 = t 2 − 3 t + t + t3 + t 2 t − 3 = t 2 − 3 ⋅ 3t 2 + 2t + t 3 + t 2 ⋅ 2t t (t + 1) t 2 − 3 = dt dt dt dt

g.

= 3t 4 + 2t 3 − 9t 2 − 6t + 2t 4 + 2t 3 = 5t 4 + 4t 3 − 9t 2 − 6t , or

[

[(

)]

(

)]

(

)] = (x + 5) dxd (x + 1) + (x + 1) dxd (x + 5) = [(x + 5)⋅1]+ [(x + 1) ⋅ 2 x] = x

)(

(

)

d 2 d d 5 4 d d 5 d 4 d t3 + t 2 t 2 − 3 = t (t + 1) t 2 − 3 = t + t − 3t 3 − 3t 2 = t + t − 3 t 3 − 3 t 2 = 5t 4 + 4t 3 − 9t 2 − 6t dt dt dt dt dt dt dt

[

d (x + 1) x 2 + 5 dx

h.

or,

[

(

d (x + 1) x 2 + 5 dx

2

2

2

)] = dxd (x

3

)

2

+ 5 + 2x2 + 2x = 3 x 2 + 2 x + 5

d d 3 d 2 d x + 5x + 5 = 3 x 2 + 2 x + 5 x + dx dx dx dx

+ x2 + 5x + 5 =

d   d d 2  2 d     (1 − u ) ⋅ du u  − u ⋅ du (1 − u ) (1 + u ) ⋅ du u  − u ⋅ du (1 + u ) d  u2 u  d  u 2  d  u     −    i. − −  =    = du 1 − u 1 + u  du  1 − u  du  1 + u  (1 − u )2 (1 + u )2

[ (1 − u ) ⋅ 2u ] − [u 2 ⋅ −1] [ (1 + u ) ⋅ 1 ] − [u ⋅ 1] − (1 − u )2 (1 + u )2

=

(

=

2u − 2u 2 + u 2

(1 − u )

2

) (

−

1+ u − u

(1 + u )

2

)

2u − u 2

=

1

−

(1 − u )2 (1 + u )2

=

u(2 − u )

1

−

(1 − u)2 (1 + u)2

d   d   3 3 2 2 2 3 2 r ⋅ dr 3r − 2r + 1  −  3r − 2r + 1 ⋅ dr r  d  3r 3 − 2r 2 + 1    = r ⋅ 9r − 4r − 3r − 2r + 1 ⋅ 1   j. =  dr  r r2 r2 

9r 3 − 4r 2 − 3r 3 + 2r 2 − 1

=

r2

=

[ (

)] [(

6r 3 − 2r 2 − 1 r2

(

(

)

)

d 2  2 d 3  3   s + 1 ⋅ ds 3s  − 3s ⋅ ds s + 1  d  3s 2 1 d  3s 2  d  1     − k. −   = −  =  2 3 ds  s 3 + 1  ds  s 2  ds  s 3 + 1 s 2  s +1

[(s + 1)⋅ 6s]− [3s (s + 1) 3

=

2

⋅ 3s 2

2

3

) ]

] − (s ⋅ 0)− (1 ⋅ 2s ) = 6s 2

s

4

(

4

+ 6s − 9s 4

(s + 1)

2

3

)

−

0 − 2s s

4

=

− 3s 4 + 6 s

(s + 1) 3

2

 2 d   d 2  s ⋅ ds 1 − 1 ⋅ ds s      4 s +

2 s/ s

4/ =3

= −

(

3s s 3 − 2

(s + 1) 3

2

)+ 2

s3

d 3  3 d d 2    2 d   (1 − u ) ⋅ du u  − u ⋅ du (1 − u ) u ⋅ du (u + 1) − (u + 1) ⋅ du u  d  u 3  d  u + 1  d  u3 u + 1 −      l. − − 2  =   =   du  1 − u  du  u 2  du 1 − u u4 u  (1 − u )2

= =

[(1 − u )⋅ 3u ]− [u ⋅ −1] − [u ⋅1]− [(u + 1)⋅ 2u] = 3u 2

3

2

(1 − u )2 u 2 (− 2u + 3) u/ (u + 2 ) + 4/ =3 u (1 − u )2

u4

= −

u 2 (2u − 3 )

(1 − u)

+

− 3u 3 + u 3

(1 − u )2

−

u 2 − 2u 2 − 2u u4

=

− 2u 3 + 3u 2

(1 − u )2

−

− u 2 − 2u u4

u+2

u3 3. Find the derivative of the following functions at the specified value.

a.

(

)

( )

2

2

( )

d 3 d 3 d d x + 3x 2 + 1 = x + 3 x 2 + (1) = 3 x3−1 + (3 ⋅ 2 )x 2 −1 + 0 = 3 x 2 + 6 x dx dx dx dx

Hamilton Education Guides

96

Basic Differentiation

(

)

d 3 x + 3 x 2 + 1 = 3 ⋅ 22 + 6 ⋅ 2 = 12 + 12 = 24 dx

at x = 2

[

Solutions

)] (

(

)

)

) (

(

d (x + 1) x 2 − 1 = x 2 − 1 d (x + 1)+ (x + 1) d x 2 − 1 = x 2 − 1 ⋅ 1+ (x + 1) ⋅ 2 x = x 2 − 1+ 2 x 2 + 2 x = 3x 2 + 2 x − 1 dx dx dx

b.

[

)] ( )

(

d (x + 1) x 2 − 1 = 3 ⋅ 12 + (2 ⋅ 1) − 1 = 3 + 2 − 1 = 4 dx

at x = 1

[

]

(

)

( )

(

)

d d d d 3s 3 − 3s 2 = 3s 3 + − 3s 2 = 9 s 2 − 6 s 3s 2 (s − 1) = ds ds ds ds

c.

[

] ( )

d 3s 2 (s − 1) = 9 ⋅ 02 − (6 ⋅ 0 ) = 0 ds

at s = 0

)

) (

(

d 2 d    2  (t − 1) dt t + 1  −  t + 1 dt (t − 1) d t2 +1      = d.   = dt  t − 1  (t − 1)2

=

=

2t 2 − 2t − t 2 − 1

(t − 1)2

=

t 2 − 2t − 1

(t − 1)2

(− 1)2 + (− 2 ⋅ −1) − 1 = 1 + 2 − 1 = 2 = 1 d t2 +1    = 4 4 2 dt  t − 1  (− 1 − 1)2

at t = −1

d e. du

[ (t − 1) ⋅ 2t ] − [(t 2 + 1)⋅ 1] (t − 1)2

 2 d 3  3 d 2 (u + 1) du u  − u du (u + 1)   u3      =   = (u + 1)4  (u + 1)2 

3u 4 + 6u 3 + 3u 2 − 2u 4 − 2u 3

(u + 1)

4

=

u 4 + 4u 3 + 3u 2

(u + 1)

4

[(u + 1) ⋅ 3u ]− [u 2

2

3

⋅ 2(u + 1)

(u + 1)

4

d du

at u = 1

(

] = [(u

2

) ][

+ 2u + 1 ⋅ 3u 2 − 2u 4 + 2u 3

(u + 1)

]

4

( ) ( )

 u3  1 14 + 4 ⋅ 13 + 3 ⋅ 12 8 1+ 4 + 3 = = =   = 2 16 2 24 (1 + 1)4  (u + 1) 

) (

)

(

)

d 2 d    2  2 2 2 3w du w + 1  −  w + 1 du 3w    d w w + 1 w + d 1 d  w w2 +1  /     = [3w ⋅ 2 w] − w + 1 ⋅ 3 = = = f.       dw  3w 2  dw  3w2/ =1  dw  3w  9 w2 (3w)2

(

=

g.

)

(

6 w 2 − 3w 2 − 3 9w

[(

2

=

(

)

)=w

3 w2 − 1

2

9w

3w

) ]

2

−1

) ]

3 1 22 − 1 d  w w2 +1  = = =   2 12 4 dw  3w 2  3⋅ 2

at w = 2

2

[(

( )

)

(

d 5 3 d 5 d 3 d v 2 +1 v3 = v +v = v + v = 5v 4 + 3v 2 dv dv dv dv

at v = −2

[(

) ]

(

)

d v 2 + 1 v 3 = 5 ⋅ (− 2 )4 + 3 ⋅ (− 2 )2 = (5 ⋅ 16 ) + (3 ⋅ 4 ) = 80 + 12 = 92 dv

(

)

d 3  3 d 2  2   x + 1 dx x  −  x dx x + 1  d  x 3      = = h. 2 dx  x 2 +1  x2 + 1

(

at x = 0

)

( ) ( )

[(x + 1)⋅ 3x ]− [x (x + 1) 2

2

2

2

3

⋅ 2x

]

3x 4 + 3x 2 − 2 x 4

=

(x + 1) 2

2

=

x 4 + 3x 2

(x + 1)

2

2

0 04 + 3 ⋅ 02 0+0 d  x 3  = = 2 = = 0 2 2 1 dx  x 2 +1  1 0 +1

d 5  5 d   (1 − u ) du u  − u du (1 − u ) d  3  u 2  d  u 5      = u  = = i. 2 du   1 − u  du  1 − u  ( ) − u 1  

(

[(1 − u )⋅ 5u ]− [(u ⋅ −1)] 4

5

(1 − u )

2

=

5u 4 − 5u 5 + u 5

(1 − u )

2

=

− 4u 5 + 5u 4

(1 − u )2

) ( )

0+0 0 − 4 ⋅ 05 + 5 ⋅ 0 4 d  3  u 2  u  = = = = 0 2 1 1 du   1 − u  (1 − 0)   4. Given the functions below find their derivatives at the specified value.

at u = 0

a.

(

)

(

)

ds 2 ds ds ds ds given s = t 2 − 1 + (3t + 2 )2 , then = t − 1 + (3t + 2 )2 = 2t + 2(3t + 2 )2 −1 ⋅ (3t + 2 ) = 2t + 2(3t + 2 ) ⋅ 3 dt dt dt dt dt

Hamilton Education Guides

97

Basic Differentiation

Solutions

ds = (20 ⋅ 2 ) + 12 = 40 + 12 = 52 dt d 3 d t + 3t 2 + 1 − t 3 + 3t 2 + 1 2t 2t t 3 + 3t 2 + 1 dy dy 2t ⋅ 3t 2 + 6t − t 3 + 3t 2 + 1 ⋅ 2 dt dt b. given y = , then = = 2 2t dt dt 4t 2 (2t )

= 2t + 18t + 12 = 20t + 12

at t = 2

=

6t 3 + 12t 2 − 2t 3 − 6t 2 − 2 4t

=

2

4t 3 + 6t 2 − 2 4t

(

) = 2t

2/ 2t 3 + 3t 2 − 1

=

2

)

)(

(

4/ t

2

3

(

) (

)

+ 3t 2 − 1 2t 2

2

4 2 + 3 −1 dy 2 ⋅ 13 + 3 ⋅ 12 − 1 = = = = 2 dt 2 2 2 ⋅ 12 2 2 dw dw d 2 c. given w = x 2 + 1 + 3 x , then = x + 1 + 3x = 2 x 2 + 1 dx dx dx

at t = 1

(

)

(

)

(

)

2 −1

(

)

(

)

d 2 d x + 1 + 3x = 2 x 2 + 1 ⋅ 2 x + 3 dx dx

dw = 4 ⋅ (− 1)3 + (4 ⋅ −1) + 3 = −4 − 4 + 3 = −5 dx 2 2 2 dy dy d  2 3 d d  2 3  d. given y = x 2 x3 + 2 x + 1 + 3 x , then = x x + 2 x + 1  + 3x x x + 2 x + 1 + 3x  =   dx dx dx  dx    dx

= 4 x3 + 4 x + 3

at x = −1

(

(

)

= x3 + 2 x + 1

2

(

)

(

)

d 3 d 2 x + x2 x + 2x + 1 dx dx

) [ (

(

2

(

(

)

(

)

)

2    + 3 =  x3 + 2 x + 1 ⋅ 2 x  +  x 2 ⋅ 2 x3 + 2 x + 1   

)]

)(

2

)

= 2 x x3 + 2 x + 1 + 2 x 2 x3 + 2 x + 1 3x 2 + 2 + 3

2 −1

(

)

= 6x2 −

2

d 3  x + 2x + 1  + 3 dx 

⋅

dy = 0+0+3 = 3 dx

at x = 0

Section 1.4 Solutions - The Chain Rule 1. Find the derivative of the following functions. Do not simplify the answer to its lowest term.

( ) ( ) ⋅ 2 x = 6 x (x + 2) b. Given y = (x + 1) , then y ′ = − 2(x + 1) ⋅ 2 x = − 4 x (x + 1) 5−1 c. Given y = (x − 1) , then y ′ = 5(x 3 − 1) ⋅ 3 x 2 = 15 x (x − 1) 3

a. Given y = x 2 + 2 , then y ′ = 3 x 2 + 2 2

−2

3

5

3−1

−2 −1

2

2

1 3x

2

4

 1 + x2   1 + x2  f. Given y =  3  , then y ′ = 4  3   x   x      ⋅

− x 4 − 3x 2 x

6

2 −1

, then y ′ = (2 ⋅ 3)x 3−1 +

3

(

4 −1

−3

2

2

  1  1   , then y ′ = 21 −  d. Given y = 1 − 2  2  x x    

e. Given y = 2 x 3 +

2

2

⋅

2x x4

3

4

 4  1  1  x/ ⋅ = 41 − = 3  1 − 2  2  4/ =3 x x x x    

(0 ⋅ 3x )− (6 x ⋅1) = 6 x + 0 − 6 x = 6 x − 6 x/ 9x 9x (3x ) [2 x ⋅ x ]− [3x (1 + x )] = 4  1 + x  ⋅ 2 x − 3x ⋅ 2

2

2 2

3

2

2

2

)

3

3

4/ =3

4

 x3   

x6

 1 + x2   1 + x 2  − x 2/ x 2 + 3  = 4 3  ⋅ = − 4 / = 6 4 3   x   x    x 

2

4

2

x6

− 3x 4

3x3  1 + x2    x3   

3

= 4

 x2 + 3     x4   

3 2 3−1 3 3   1  x +1  x +1  x +1    x +1  x +1 g. Given y = x 2   + x2   ⋅ ⋅ x2  = 2 x    + 3  , then y ′ = 2 x 2 −1 ⋅  3  3     3   3   3   3   

[

h. Given y = x (x + 1)2 + 2 x

(

) (

] = [x (x 3

⋅ 3x 2 + 4 x + 3 = 3 x 3 + 2 x 2

x  i. Given y =  − 2 x3  3  

−1

) ] = (x + 3 x ) (3 x + 4 x + 3 ) 2

+ 2x + 1 + 2x 2

3

+ 2x2 + x + 2x

) = (x 3

3

)

3

(

+ 2 x 2 + 3 x , then y ′ = 3 x3 + 2 x 2 + 3 x

)

2

x  , then y ′ = −  − 2 x3  3  

Hamilton Education Guides

3

−1−1

−2

x  1  1  ⋅  − (2 ⋅ 3)x3−1  = −  − 2 x 3   − 6 x 2  3 3 3      

98

3−1

Basic Differentiation

Solutions

(

)

(

4

) ⋅ (3x + (3 ⋅ 2)x + 0) = 4(x + 3 x + 1) (3 x + 6 x ) [2t ⋅ (1 + t )]− [2t ⋅ t ] = 3 t  ⋅ 2t + 2/ t/ − 2/ t/ = 3t ⋅ 2t ⋅ 1+ t  (1 + t ) (1 + t ) (1 + t ) (1 + t )   4 −1

j. Given y = x3 + 3 x 2 + 1 , then y ′ = 4 x3 + 3 x 2 + 1 3

 t2   2   , then y ′ = 3  t  k. Given y =  2 2  1+ t  1+ t    3t 4 ⋅ 2t

=

6t 4 +1

=

3−1

2 2

− 2 −1−1

, then y ′ =

x3

(

⋅ −2 x − 2 −1 = 2 x −3 1 + x − 2

[− 2 (x + 1)

][

⋅ 1 ⋅ x3 − 3 x3−1 ⋅ (x + 1)− 2 x6

(

=

6x2

(x + 2)

2

3

x3

(

6x2 − 2x x3 + 2 2x = 2 1 x3 + 2

−

(

)

3

)

−2

=

2

(

x 3 1 + x −2

4

2 2

2 2

)

2

] = [− 2 x (x + 1) ]− [3 x (x + 1) ] −3

3

(

)

2

3

2

3

)

  2  − 1 =  1  ⋅ 6 x 2  1 − x3  1 − x3 x2  

[3x ⋅ (x + 2)]− [3x ⋅ x ] − 2 x (x + 2) 3−1

− x 2 , then y ′ =

x3 + 2

3/

2 2

−2

2

x6

2   1  2 −1 3x 2 1  1  ′ = 2  n. Given y =  , then y ⋅ +       1 − x3  x  1 − x3  1 − x3 

o. Given y =

3/

2

2 2

− 2 −1

(x + 1)−2

2

2

2

2 4

−2 −1

m. Given y =

2

3

2

6t 5

=

2 2+ 2

3

2 −1

2

(1 + t ) ⋅ (1 + t ) (1 + t ) (1 + t ) l. Given y = (1 + x ) , then y ′ = − (1 + x ) 2 2

3−1

2 −1

=

 6x2 − 1 = 2 x2 1 − x3 

)

( ) (x + 2)

3x 2 x3 + 2 − 3x5 2

3

(

− 2x =

)

3

−

1

x2

3/ x/ 5/ + 6 x 2 − 3/ x/ 5/

(x + 2) 3

2

− 2x

2

2. Find the derivative of the following functions at x = 0 , x = 1 , and x = −1 .

(

)

(

5

)

a. Given y = x 3 + 1 , then y ′ = 5 x3 + 1

( )( ) = 0 ⋅1 y′(1) = (15 ⋅ 1 )(1 + 1) = 15⋅ 2 4

y′(0 ) = 15 ⋅ 02 03 + 1 2

[

4

3

y′(−1) = 15 ⋅ (− 1) 2

(

] [(− 1) + 1] 3

)

5 −1

(

(

(

= 0 ⋅1 = 0

4

= 15⋅ 24 = 15⋅ 16 = 240 and

4

= 15 ⋅ (− 1 + 1) 4 = 15 ⋅ 04 = 15 ⋅ 0 = 0

(

4

)

4

4

) (3x

b. Given y = x3 + 3 x 2 − 1 , then y ′ = 4 x3 + 3 x 2 − 1

) (0 + 2) = 0.(− 1)

y′(0 ) = (12 ⋅ 0 ) 03 + 3 ⋅ 02 − 1

)

⋅ 3 x 2 = 15 x 2 x3 + 1 . Therefore,

3

3

4 −1

2

(

)

) (x + 2) . Therefore,

+ 6 x = 12 x x3 + 3 x 2 − 1

3

⋅2 = 0

3

y′(1) = (12 ⋅ 1) 13 + 3 ⋅ 12 − 1 (1 + 2 ) = 12 ⋅ (1 + 3 − 1) 3 ⋅ 3 = 12 ⋅ 27 ⋅ 3 = 972 and

[

] (− 1 + 2) = − 12 (− 1 + 3 − 1)

y′(−1) = (12 ⋅ −1) (− 1) 3 + 3 ⋅ (− 1)2 − 1 2

3

 x   x  c. Given y =    , then y ′ = 2  x +1  x +1

2⋅0

y′(0 ) = y′(1) =

=

(0 + 1) 3 2 ⋅1

(1 + 1) 3

=

2 ⋅ −1

y′(−1) =

(− 1 + 1) 3

(

0

13 2

23

=

)

2 −1

⋅

=

0 = 0 1

=

2 1 = = 0.25 and 8 4

−2 03

= −

1 ⋅ (x + 1) − 1 ⋅ x

(x + 1)2

(

)

2

(

= − 12 (− 2 + 3) 3 = − 12 ⋅ 13 = −12 ⋅ 1 = −12

1 2x  x   x  x/ + 1 − x/ = 2 = 2 = . Thus, ⋅ ⋅  x + 1  (x + 1)2  x + 1  (x + 1)2 (x + 1)3

2 which is undefined due to division by zero 0

(

)

(

)

2 2   d. Given y = x x 2 + 1 , then y ′ = 1 ⋅ x 2 + 1  + 2 x 2 + 1   

y′(0 ) = 02 + 1

3

)

2 −1

(

)

 ⋅ 2 x = x2 + 1 

2

(

)

+ 4 x x 2 + 1 . Therefore,

+ (4 ⋅ 0 ) 02 + 1 = 12 + 0 = 1

Hamilton Education Guides

99

Basic Differentiation

(

)

y′(1) = 12 + 1

2

y′(−1) =

2

Solutions

(

)

+ (4 ⋅ 1) 12 + 1 = 22 + 4 ⋅ 2 = 4 + 8 = 12 and

[(− 1) + 1]

2

[

]

+ (4 ⋅ −1) (− 1)2 + 1 = (1 + 1) 2 − 4 ⋅ (1 + 1) = 22 − 4 ⋅ 2 = 4 − 8 = −4

(

)

(

3

)

e. Given y = x3 + 2 x 2 + 1 , then y ′ = 3 x3−1 + 2 ⋅ 3 x 2 + 1

(

)

y′(0 ) = 3 ⋅ 02 + (12 ⋅ 0 ) 02 + 1

(

)

y′(1) = 3 ⋅ 12 + (12 ⋅ 1) 12 + 1

2

2

=

3x 4

[

]

=

(1 + x ) (1 + x ) 6 ⋅ 05

y′(0 ) = y′(1) =

(1 + 0 ) (1 + 1 )

2 4

0

=

2 4

6 ⋅ 15

14 6

=

2

(3 ⋅ 2)x 4+1

6 ⋅ (− 1)

(1 + x )

[1 + (− 1) ]

3−1

(

[2 x ⋅ (1 + x )]− [2 x ⋅ x ] ⋅ (1 + x ) 2

2

2 2

6 x5

(1 + x )

2 4

4+ 2

2

6 ⋅ −1

(1 + 1)

= −

4

6 2

4

= −

3 6 = − = −0.375 8 16

[1 ⋅ (x + 1)]− [2 x ⋅ x] = 5 x  ⋅ x + 1 − 2 x    x +1 (x + 1) (x + 1) 2

5 −1

⋅

4

2

2

y′(1) =

6

6

2

2

(

2

) (

) (3x

2 x2 + 1

=

3

4

2

2

) ( 2

1

2

) = 2⋅2

0

2

)

− x2 − 1

=

(

)

) (2 x − 1)

2 x2 + 1

2

x3

(

)

(

)

3

2

. Therefore,

2

(− 1)3

Hamilton Education Guides

0

0

⋅ (2 − 1) 2 ⋅ 4 ⋅ 1 8 = = = 8 and 1 1 1

] ⋅ [2 ⋅ (− 1) − 1]

2 (− 1) + 1 2

2

(

3   2 ⋅ 2 x ⋅ x 2  − 2 x ⋅ x 2 + 1  2 3 2    = 6x x + 1 − 2x x + 1 4 4 x x

2

2 1 + 1 ⋅ 2 ⋅1 − 1

y′(−1) =

2

) = 2 ⋅1 ⋅ (0 −1) = 2 ⋅ −1 = − 2 which is undefined due to division by zero

2

3

2

x3

2 02 + 1 ⋅ 2 ⋅ 0 2 − 1 0

3−1

2

x2

x2

x 4/ =3

[

(x + 1) (x + 1)

6

3

2

3

) [3x − (x + 1)]

(

1 − x2

2

2

2

⋅

6

2

(

2

5x4

2

4

y′(0 ) =

2

=

6

6

2

(

2

2

2

4

2 x/ x 2 + 1

2

6

6

2

=

)

2

2

4

(

. Therefore,

) = 5x (1 − x ) . Therefore, (x + 1) (x + 1) 0 5 ⋅ 0 ⋅ (1 − 0 ) 0 ⋅1 = = = 0 y′(0 ) = 1 1 (0 + 1) 5⋅0 0 5 ⋅ 1 ⋅ (1 − 1 ) 5 ⋅ (1 − 1) = = = = 0 and y′(1) = 64 64 2 (1 + 1) 0 5 ⋅ 1 ⋅ (1 − 1) 5 ⋅ (− 1) ⋅ (1 − (− 1) ) 5⋅0 = = = = 0 y′(−1) = 64 2 (1 + 1) ((− 1) + 1)  3(x + 1) ( x + 1) 1 h. Given y = (x + 1) ⋅ = y= , then y ′ =  4

2

 x 2  2 x + 2 x3 − 2 x3  ⋅ = 3 2  1 + x2    1 + x2

6 3 = = 0.375 and 16 8

5

5x4 1 − x2

. Therefore,

= 3 ⋅ 1 − 12 ⋅ (1 + 1)2 = 3 − 12 ⋅ 22 = 3 − 12 ⋅ 4 = 3 − 48 = −45

 x   x   , then y ′ = 5 g. Given y =   2  x2 + 1   x +1 

=

2

0 = 0 1

=

=

2 4

=

2 2+ 2

=

4

5

y′(− 1) =

2

3

2 2

)

= 3 + 12 ⋅ 22 = 3 + 12 ⋅ 4 = 3 + 48 = 51 and

 2    , then y ′ = 3 x    1 + x2    

2x

⋅

2 2

(

⋅ 2 x = 3 x 2 + 12 x x 2 + 1

= 3 ⋅ 0 + 0 ⋅12 = 0 + 0 = 0

y′(−1) = 3 ⋅ (− 1) 2 + (12 ⋅ −1) (− 1) 2 + 1

 x2 f. Given y =   1+ x2 

3−1

2

=

2(1 + 1)2 ⋅ (2 ⋅ 1 −1) 8 2 ⋅ 22 ⋅ (2 −1) 2 ⋅ 4 ⋅1 = = = − = −8 −1 −1 −1 1

100

Basic Differentiation

Solutions

2

 x3   3   + 5 x , then y ′ = 2 x  i. Given y =     x −1  x −1   

[3x

2 −1

⋅

2

] [ ] + 5 = 2

⋅ (x − 1) − 1 ⋅ x3

(x − 1)

x3   3 x3 − 3 x 2 − x3  ⋅ +5  x −1  (x − 1)2    

2

(

)

2 x3 x 2 (2 x − 3) 2 x3 ⋅ x 2 2 x3 − 3x 2 2 x3 2 x3 − 3x 2 2 x3+ 2 (2 x − 3) 2 x5 (2 x − 3) ⋅ +5 = +5 = +5= + 5 . Therefore, +5 = ⋅ 2 2 2 2 1 + x − 1 (x − 1) x − 1 (x − 1) (x − 1) (x − 1) (x − 1) (x − 1) 3

=

2 ⋅ 05 ⋅ (2 ⋅ 0 − 3)

y′(0 ) = y′(1) =

(0 − 1)

2 ⋅ 15 ⋅ (2 ⋅ 1 − 3)

(1 − 1) 3

0 ⋅ (0 − 3)

+5 =

3

+5 =

1

2 ⋅ (2 − 3)

2 ⋅ (− 1) 5 ⋅ [(2 ⋅ −1) − 3]

y′(−1) =

(− 1 − 1)

3

+5 =

03

+5 =

0 +5 = 0+5 = 5 1

+5 =

3

2 2 ⋅ −1 + 5 = − + 5 which is undefined due to division by zero and 0 0

2 ⋅ −1 ⋅ [− 2 − 3]

+5 =

(− 2) 3

−2 ⋅ −5 10 5 +5 = + 5 = + 5 = 1.25 + 5 = 6.25 8 4 −8

3. Use the chain rule to differentiate the following functions. 2 d 3  3 d 2 2 3 2 t ⋅ dt (t + 1)  − (t + 1) ⋅ dt t   (t + 1)3      = t ⋅ 3(t + 1) − (t + 1) ⋅ 2t   = t4 t4  t 2 

[

d a. dt

(t + 1)2 (3t − 2t − 2)

=

t

3

][

= d c. dx = d.

(

)

2

18u 5 u 2 + 1 − 12u 3

d dx

9u

8

3t 2 (t + 1)2 − 2t (t + 1)3 t

4

=

t/ (t + 1)2 [3t − 2(t + 1)] t 4/ =3

t3

) ( ) ( ) (u + 1) = 6u (u + 1) [3u (

)

=

(t + 1) 2 (t − 2)

=

3  3 d  4 d 2  u +1  −  u 2 +1 3u 3u 4  3  2  du du d  u +1      = = b. du  3u 4  4 2 3u  

(

]

2

3

3/

2

2

9u

2

(

(

)

(

(

)(

)

3−1 3   4   2 ⋅ 2u  −  u 2 + 1 ⋅ (3 ⋅ 4 )u 4−1 ⋅1 3u ⋅ 3 u + 1    9u 8

)]

− 2 u2 + 1

8/ =5

2

2 u 2 + 1 3u 2 − 2 u 2 − 2

=

3u

5

) = 2(u + 1) (u − 2) 2

2

3u

2

5

 3 d 2 2 d 3  2 3−1 3 2 −1 (1 − x ) dx (2 x + 1)  − (2 x + 1) dx (1 − x )   (2 x + 1)3   = (1 − x ) ⋅ 3(2 x + 1) ⋅ 2 − (2 x + 1) ⋅ 2(1 − x ) ⋅ −1    =   (1 − x )4 (1 − x )4  (1 − x )2 

][

[

6(1 − x ) 2 (2 x + 1) 2 + 2(2 x + 1) 3 (1 − x )

=

(1 − x )4

(

)

2(1/ − x/ )(2 x + 1) 2 [3(1 − x ) + (2 x + 1) ]

(1 − x ) 4/ =3

(

)

(

=

]

2 (2 x + 1) 2 (4 − x )

(1 − x )3

(

)

)

2  3 2 d 2 2 −1    3  3 d 3 3 3 3 3 ⋅ 3x 2   x − 1 (2 x + 1)  = (2 x + 1) dx x − 1  +  x − 1 dx (2 x + 1)  = (2 x + 1) ⋅ 2 x − 1     

(

[

)

(

)] (

(

)

)[

(

)]

2 2     +  x3 − 1 ⋅ 3(2 x + 1)3−1 ⋅ 2 = 6 x 2 (2 x + 1)3 x3 − 1 + 6 x3 − 1 (2 x + 1)2  = 6(2 x + 1)2 x3 − 1 x 2 (2 x + 1) + x3 − 1    

(

)(

(

)

)(

)

= 6(2 x + 1)2 x3 − 1 2 x3 + x 2 + x3 − 1 = 6(2 x + 1) 2 x 3 − 1 3 x 3 + x 2 − 1 e.

2

d  3 1  1   3 s − 2  = 2 s − 2  ds  s + 6 s + 6 

(

)

(

2 −1

d d 2  2  3−1 s + 6 ⋅ ds (1) − 1 ⋅ ds s + 6 ⋅ 3s − 2  s2 + 6 

(

(

)

) (

) (

⋅

1  d  3 1  1  d 3 d 1   3  3 s − 2  = 2 s − 2  = 2 s − 2   ⋅  s − ds 2 ds  s + 6 s + 6 s + 6 s + 6   ds  

)

  3 1   2 2s 1   2 0 − 2 s   3 = − ⋅ 2 s 3s +  = 2 s − 2    ⋅ 3s − 2 2 2  s + 6  s + 6   s +6  s2 + 6   

( )

(

) (

)

(

)

) (

(

) (

(

)

  2 

)

3  3 d 2 3−1 3 d 2  2   2    t +1 t −1  −  t2 −1 t +1  3 t +1 ⋅ 3 t2 −1 ⋅ 2t  −  t 2 − 1 ⋅ 2t   2   d  t −1  dt dt         f. = = 2 2 2 2 dt  t 2 + 1  t +1 t +1  

(

)

Hamilton Education Guides

)

101

Basic Differentiation

(

Solutions

)(

)

(

)

) [( ) ( ( )

(

)]

2  3  2 2 2 2 6t t + 1 t − 1  − 2t t − 1  2t t 2 − 1 3 t 2 + 1 − t 2 − 1 = = 2 2 t2 +1 t2 +1

(

)

=

)[ ] = 4t (t − 1) (t (t + 1) (t + 1) d (u + 1)  − (u + 1) dud (u + 1)  du

(

2

2t t 2 − 1 3t 2 + 3 − t 2 + 1

(

)

(

(

)

 2 2 (u + 1) ⋅ 3 u + 1 =

=

)[

(

=

(

2θ 2 + 3

(

3 −1

)

  ⋅ 2u  −  u 2 + 1   (u + 1)4

3

(u + 1)

(θ − 1)

(

3

2

 θ2 +3  = 2  3  (θ − 1) 

(θ − 1)

3

2 −1

] [(

(θ − 1) 2

)]

(θ − 1)

(

(

)

(

(

7 r 6 r 2 + 2r

=

(

)

(

) = 2(u + 1) (2u

3

2

2

+ 3u − u 2 − 1

(

]

=

)

2

)

+ 3u − 1

(u + 1)

3

) (

7

(

)

)(

(

(θ − 1)

(

)

2

6

2

)

(

)

=

[

)

(

3+ 4

) = − 2(θ

2

)(

+ 3 ⋅ θ 2 + 2θ + 9

(θ − 1) 7

)

(

)

(

(

) − [r ⋅ 3r (r + 2)⋅ 2(r + 1) ] = 7r (r (r + 2r )

(

7 r 6 r 2 + 2r

3

)

2 −1

6

6

2

(

)

⋅

d 2  r + 2r  dr 

2

+ 2r

)

7

)

) − 6r (r + 2)(r + 1) (r + 2r ) 3

8

6

2

4. Given the following y functions in terms of u , find y ′ . a. Given y = 2u 2 − 1 and u = x − 1 , then y = 2(x − 1)2 − 1 and y ′ = 2 ⋅ 2(x − 1)2 −1 − 0 = 4( x − 1) b. Given y = =

[

(x − 1) 3

c. Given y =

)] [ ( )

(

3 x3−1 ⋅ x3 − 1 − 3 x3−1 ⋅ x3 x3 u and u = x 3 , then y = 3 and y ′ = 2 u −1 x −1 x3 − 1

3/ x5/ − 3 x 2 − 3/ x5/ 2

u

1 + u2

(

= −

3x2

(x − 1) 3

and u = x + 1 , then y =

) [ ( ( )

)(

)]

x2 + 1

2

3

2

3

3

2

(

(

)

1 + x2 + 1

2

(

)

(

)

(

)

2  2 −1   ⋅ 2x ⋅ x2 + 1  2 x ⋅ 1 + x 2 + 1  − 2 x 2 + 1     and y ′ = 2 2  2 1 + x + 1 

(

2  2 2 2 2 2 2 x + 2 x x + 1  − 4 x x + 1 ⋅ x + 1 2x + 2x x2 + 1 − 4x x2 + 1 = = = 2 2 2 2   2 2 1 + x + 1  1 + x + 1  1 1 d. Given y = u 2 − and u = x 4 , then y = x8 − and y ′ = 8 x8−1 − 0 = 8x 7 2 2

Hamilton Education Guides

] = [3x (x − 1)]− [3x ⋅ x ] (x − 1)

2

2

)]

2/ 2 2 θ 2 + 3 (θ − 1) 2θ (θ − 1) − 3 θ + 3 2 θ 2 + 3 2θ (θ − 1)3 − 3 θ 2 + 3 (θ − 1)2 = ⋅ ⋅ (θ − 1) 6/ = 4 (θ − 1) 3 (θ − 1)3 (θ − 1)6

2 θ 2 + 3 ⋅ 2θ 2 − 2θ − 3θ 2 − 9

) − [r ⋅ 3(r + 2r )⋅ (2r + 2) ] (r + 2r ) 3

2

2  3   2 2 2 6u (u + 1) u + 1  − 2 u + 1 (u + 1) (u + 1)4

3 3 d 7 3  2 7 −1   7 2 d 2 r 2 + 2r r − r7 r + 2r   r + 2r ⋅ 7 r  − r ⋅ 3 r + 2r  dr dr  = = 6 6 3 2 r + 2 r r 2 + 2r 

)

)

(u + 1)4

(

6

+3

3

2

+2

 (θ − 1)3 d θ 2 + 3  −  θ 2 + 3 d (θ − 1)3  2 2      d θ +3 θ +3  dθ dθ    ⋅   = 2 ⋅ 3 dθ  (θ − 1) 3  (θ − 1)6  (θ − 1) 

)

4

2

(u + 1)

⋅ 2θ − θ 2 + 3 ⋅ 3(θ − 1) 2

) ⋅ [2θ (θ − 1) − 3(θ

2θ 2 + 3 3

4/ =3

2

2

2

2

3

2

)

 ⋅ 2(u + 1) 2 − 1   =

)] = 2(u + 1) (3u

(

2

) ⋅ [(θ − 1)

 d  r7 i. dr  r 2 + 2r  =

)

2(u/ + 1/ ) u 2 + 1 3u (u + 1) − u 2 + 1

d θ 2 +3  h.   dθ  (θ − 1)3  =

(

2

2

 2 3 2  (u + 1) 3 1   3 d  u2 + 1  d  2 d  2 1    u +1  g. = =   =  u +1 2 du  u +1   du  (u + 1)2  du   ) ( u 1 +      

2 2

2

(

)

)

(

)

(

)

2

)

2x − 2x x + 1

(

)

2  2 1 + x + 1 

2

2

(

)

2  2 x 1 − x 2 + 1    = 2 2   2 1 + x + 1   

(

)

102

Basic Differentiation

e. Given y = u

=

(

8x 1 − x

)

2 3/

(1 − x )

and u = 8x

=

2 8/ =5

f. Given y =

=

4

Solutions

1 1− x 2

(

(1 − x )

2 4

 2 0 ⋅ 1 − x and y ′ =

)  − 4(1 − x ) (1 − x )

2 4 −1

4

(u + 1)

3

 ⋅ −2 x ⋅ 1 2  = 0 + 8x 1 − x 8 1 − x2

(

2 8

(1 − x )

u2

(

)

)

3

2 5

and u = x − 1 , then y =

2 x3 (x − 1) − 3 x 2 (x − 1)2 x

, then y =

1

x 2/ (x − 1) [2 x − 3(x − 1) ]

=

6

(x − 1)2 (x − 1 + 1)3 =

x 6/ = 4

=

(x − 1)2 x

3

and y ′ =

(x − 1) (2 x − 3x + 3) x4

=

[2(x − 1)

2 −1

][

⋅ x3 − 3 x 2 ⋅ (x − 1) 2 x

]

6

( x − 1)(− x + 3) x4

Section 1.5 Solutions - Implicit Differentiation Use implicit differentiation method to solve the following functions. a. Given x 2 y + x = y , then ; y′ =

2x y + 1

)

(

(

(

)

d 2 d ( y ) ; 2 x ⋅ y + x 2 ⋅ y′ + 1 = y′ ; 2 x y + 1 = y′ − x 2 y′ ; 2 x y + 1 = y′ 1 − x 2 x y+x = dx dx

1 − x2

(

)

)

d d (0) ; (1 ⋅ y + x ⋅ y′) − 6 x + y′ = 0 ; y − 6 x = − x y′ − y′ ; y − 6 x = − y′(x + 1) x y − 3x 2 + y = dx dx y − 6x y − 6x 6x − y ; ; y′ = = − y′ ; y′ = − x +1 x +1 x +1

b. Given x y − 3 x 2 + y = 0 , then

c. Given x 2 y 2 + y = 3y 3 , then

(

)

( ) (

)

d d 2 2 x y +y = 3y 3 ; 2 x ⋅ y 2 + 2 y y′ ⋅ x 2 + y′ = 9 y 2 y′ ; 2 x y 2 + 2 x 2 y y′ = 9 y 2 y′ − y′ dx dx

(

)

; 2 x y 2 = 9 y 2 y′ − y′ − 2 x 2 y y′ ; 2 x y 2 = y′ 9 y 2 − 1 − 2 x 2 y ; y ′ =

(

(

)

9 y2 − 1 − 2 x2 y

d d (5 x ) ; (1 ⋅ y + y′ ⋅ x ) + 3 y 2 ⋅ y′ = 5 ; y + y′x + 3 y 2 y′ = 5 ; x y′ + 3 y 2 y′ = 5 − y x y + y3 = dx dx 5− y

d. Given x y + y 3 = 5 x , then ; y′ x + 3 y 2 = 5 − y ; y′ =

)

2 x y2

x + 3 y2

e. Given 4 x 4 y 4 + 2 y 2 = y − 1 , then

(

(

)

)

d d ( y − 1) ; 4 4 x3 ⋅ y 4 + 4 y 3 y′ ⋅ x 4 + 4 y y′ = y′ ; 16 x3 y 4 + 16 x 4 y 3 y′ 4x4 y 4 + 2 y 2 = dx dx

(

)

+4 y y′ = y′ ; 16 x3 y 4 = y′ − 16 x 4 y 3 y′ − 4 y y′ ; 16 x3 y 4 = y′ 1 − 16 x 4 y 3 − 4 y ; y′ =

(

(

)

(

)

+ 2 x 2 y y′ = 0 ; y + 2 x y 2 = − y′x − 2 x 2 y y′ ; y + 2 x y 2 = − y′ x + 2 x 2 y ; y′ = −

(

h. Given x y 3 + x 3 y = x , then

( ) (

)

y + 2 x y2 x + 2x2 y

)

(

(

)

) (

)

d d (x ) ; 1 ⋅ y 3 + 3 y 2 y′ ⋅ x + 3 x 2 ⋅ y + y′ ⋅ x 3 = 1 ; y 3 + 3 x y 2 y′ + 3 x 2 y + x 3 y′ = 1 x y 3 + x3 y = dx dx

(

)

; 3 x y 2 y′ + x 3 y′ = 1 − y 3 − 3 x 2 y ; y′ 3 x y 2 + x 3 = 1 − y 3 − 3 x 2 y ; y ′ =

Hamilton Education Guides

)

d 2 d 2 x − y2 x ; 1 ⋅ y 2 + 2 y y′ ⋅ x + y′ = 2 x ; 2 y y′x + y′ = 2 x − y 2 ; y′ = x y2 + y = dx dx 2x y + 1

g. Given x y 2 + y = x 2 , then

1

1 − 16 x 4 y 3 − 4 y

d d (0) ; (1 ⋅ y + y′ ⋅ x ) + 2 x ⋅ y 2 + 2 y y′ ⋅ x 2 = 0 ; y + y′x + 2 x y 2 x y + x 2 y 2 − 10 = dx dx

f. Given x y + x 2 y 2 − 10 = 0 , then

i. Given y 2 + x 2 y = x , then

16 x 3 y 4

(

1 − y3 − 3x2 y 3x y2 + x3

)

 d 1 12 −1 1 − 12 d  12 2  y + x2 y  = ′ ′ ( ) ; ; y y x y y x ⋅ + ⋅ + ⋅ = 2 1 y y′ + 2 x y + x 2 y′ = 1 x  dx 2 2 dx  

103

Basic Differentiation

Solutions

   1 −1   1  1 − 2 xy ; y′ y 2 + x 2  = 1 − 2 xy ; y′ 1 + x 2  = 1 − 2 xy ; y′ = 1 2  2y 2    + x2   2 y

(

)

(

)

1 −3 1 1 −1 d 2 d  14   y  ; 2 x ⋅ y + y′ ⋅ x 2 + 2 y y′ = y 4 y′ ; 2 x y + x 2 y′ + 2 y y′ = y 4 y′ x y + y2 =   4 4 dx dx  

1

j. Given x 2 y + y 2 = y 4 , then

 1 − 34 1 −3 −2 x y −2 x y y y′ = −2 x y ; y′ x 2 + 2 y − y 4  = −2 x y ; y′ = ; y′ = 3 1 − 1 4 4 2 2   x + 2y − x + 2y − y 4 4 3 4 4 y

; x 2 y′ + 2 y y′ −

k. Given x + y 2 = x 2 − 3 , then l. Given x 4 y 2 + y = −3 , then

(

)

(

)

d d 2 2x − 1 x + y2 = x − 3 ; 1 + 2 y y′ = 2 x ; 2 y y′ = 2 x − 1 ; y ′ = dx dx 2y

(

(

)

)

d 4 2 d (− 3) ; 4 x3 ⋅ y 2 + 2 y y′ ⋅ x 4 + y′ = 0 ; 4 x3 y 2 + 2 x 4 y y′ + y′ = 0 x y +y = dx dx

(

)

− 4 x3 y2

; 2 x 4 y y′ + y′ = −4 x3 y 2 ; y′ 2 x 4 y + 1 = −4 x3 y 2 ; y′ =

(

2 x4 y + 1

)

(

)

d 7 d (8) ; 7 y 6 y′ − 2 x ⋅ y 4 + 4 y 3 y′ ⋅ x 2 − 1 = 0 ; 7 y 6 y′ − 2 x y 4 − 4 x 2 y 3 y′ = 1 y − x2 y 4 − x = dx dx

m. Given y 7 − x 2 y 4 − x = 8 , then

(

)

; 7 y 6 y′ − 4 x 2 y 3 y′ = 1 + 2 x y 4 ; y′ 7 y 6 − 4 x 2 y 3 = 1 + 2 x y 4 ; y′ =

[

]

(

1 + 2 x y4

7 y6 − 4 x 2 y3

)

n. Given (x + 3)2 = y 2 − x , then

d (x + 3)2 = d y 2 − x ; 2(x + 3) = 2 y y′ − 1 ; 2 x + 6 + 1 = 2 y y′ ; y′ = 2 x + 7 dx dx 2y

o. Given 3 x 2 y 5 + y 2 = − x , then

d d (− x ) ; 3 2 x ⋅ y 5 + 5 y 4 y′ ⋅ x 2 + 2 y y′ = −1 ; 6 x y 5 + 15 x 2 y 4 y′ + 2 y y′ = −1 3x 2 y 5 + y 2 = dx dx

(

(

)

(

)

)

; 15 x 2 y 4 y′ + 2 y y′ = −1 − 6 x y 5 ; y′ 15 x 2 y 4 + 2 y = −1 − 6 x y 5 ; y′ = −

1 + 6 x y5

15 x 2 y 4 + 2 y

Section 1.6 Solutions - The Derivative of Functions with Fractional Exponent 1. Find the derivative of the following exponential expressions. 1 1 1 −1 1 1−5 1 −4 a. Given y = x 5 , then y′ = x 5 = x 5 = x 5 5 5 5

( )

b. Given y = 4 x3

1 2

3

= 4 x 2 , then y′ =

2 1− 3 1 1 (2 x + 1)3 −1 ⋅ 2 = 2 (2 x + 1) 3 = 2 (2 x + 1) − 3 3 3 3

1

c. Given y = (2 x + 1)3 , then y′ =

(

)

1

d. Given y = 2 x 2 + 1 8 , then y′ =

(

e. Given y = 2 x 3 + 3 x

(

f. Given y = x 3 + 8

( )

g. Given y = x3 =

1 2

)

2 3

)

3 5

(

)

(

(

3 3 2 x + 3x 5

2 3 x +8 3

, then y′ = 3

(

(

)

)

1−8 1 −1 4x 1 2 x 2x2 + 1 8 = 2x + 1 8 ⋅ 4x = 2x2 + 1 8 2 8

, then y′ =

1

1 3 −1 3 2 3− 2 3 ⋅ 4 x 2 = ⋅ 4/ x 2 = 6 x 2 2/ 2

)

) ⋅ (6 x + 3) = 53 (2 x 3 −1 5

2 −1 3 ⋅ 3x 2 1

2

=

(

2 3 x +8 3/

− (3 x − 1)3 = x 2 − (3 x − 1)3 , then y′ =

)

2 −3 3

3

+ 3x

)

−2 5

−7

8

(

)

⋅ 3 2x2 + 1 =

(

⋅ 3/ x 2 = 2 x 2 x 3 + 8

)

(

)(

9 2x2 + 1 2x3 + 3x 5

)

−2 5

−1

3

1− 3 1 3 3− 2 1 3 32 −1 1 x − (3 x − 1)3 −1 ⋅ 3 = x 2 − (3 x − 1) 3 ⋅ 3/ 2 3 2 3/

2 3 12 x − (3 x − 1) − 3 2 1

1

h. Given y = x 2 (x + 1) 8 , then y′ = 2 x ⋅ (x + 1)8 +

Hamilton Education Guides

2 2 1 7 1−8 1 1 1 (x + 1)8 −1 ⋅ x 2 = 2 x (x + 1)8 + x (x + 1) 8 = 2 x ( x + 1)8 + x ( x + 1) − 8 8 8 8

104

Basic Differentiation

(

Solutions

)

2 5

i. Given y = x 3 + 1

)

(

)

(

)

(

2 −5 2 −1 6 2 3 1 1 −1 1 1− 2 6 x + 1 5 ⋅ 3x 2 + x 2 = x 2 x3 + 1 5 + x 2 = x 2 x 3 + 1 5 5 2 2 5

1

+ x 2 , then y′ =

−3 5

+

1 − 12 x 2

2    2 2 2 2 −1 2 −1 2 2 −3 1 ⋅ x 3  −  x 3 ⋅ (x + 1) x 3 − x 3 ( x + 1) x 3 − x 3 (x + 1) 3 x +1  =   3 3 j. Given y = 2 , then y′ =  = 4 4 4 3 3 3 x x x3 x

(x + 1) 2

k. Given y = =

(

x3 x2 + 1

)

1 2

x2

−1

(

x4

)

)

(

(

)

(

)

)

1− 2   1  2/ x3 2 x + 1 2  − 2 x ⋅ x 2 + 1 2    2/    =  x4

x4

1 2

1   1 1 1 −1 2 2 −1 1 − 23 3 ( x + 1) − 2(x + 1) ⋅ x 3  −  x 3 ⋅ (x + 1)  2 x x ( x + 1) 2 3     3 , then y′ = = 2 2 x3 x3

1

x3

2. Use the

, then y′ =

− 2x x2 + 1

2

(x + 1)2

l. Given y =

(

1 −1 1 1 2 2  2  x + 1 2 ⋅ 2 x ⋅ x  − 2 x ⋅ x + 1 2     2

d notation to find the derivative of the following exponential expressions. dx

d  15 x a. dx 

2

 1   = 2  x 5    

2 −1

⋅

1 2 1−4 1 1 −1 2 1 −4 d 15 2 −3 x = 2 x 5 ⋅ x 5 = x 5 ⋅ x 5 = x 5 5 = x 5 , or 5 5 5 5 dx

2

 d 52 2 2 −1 2 2 −5 2 −3  = x = x5 = x 5 = x 5 5 5 dx 5  1 1 1 −1 d 1 1 1 d (x − 1) 2 = (x − 1)2 ⋅ (x − 1) = (x − 1) − 2 ⋅ 1 = 1 ( x − 1) − 2 b. dx 2 2 2 dx d  15 x dx 

(

)

(

)

(

)

(

)

(

(

)

)

−2

c.

1− 3 1 −1 1 1 1 d 2 2x 2 d 2 x + 1 3 = x2 + 1 3 ⋅ x +1 x + 1 = x2 + 1 3 ⋅ 2x = dx 3 dx 3 3

d.

−1− 4 − 1 −1 d −1 1 1 3x2 3 d 3 x + 1 4 = − x3 + 1 4 ⋅ x3 + 1 = − x3 + 1 4 ⋅ 3x 2 = − x +1 dx 4 dx 4 4

(

1 d  (x − 1)2 e. dx  x 2 

(

)

1 8

d 3 x + 2x dx

g.

d  3 2  x +1 x dx  = 3x

(

(

1 3 x + 2x 8

=

)

1 −1 8 ⋅

(

1 3

3

2 3

(

)

)

−5 4





 d  3 1 h. x ⋅ dx  x 2 +1 

(

)

)

1 2

  3  = d  x 1  dx  2  x + 1 2 

(

(

)

)

1  1− 2   2 2 4 2 3 x x + 1 2  −  x x + 1 2     = =  x2 + 1

Hamilton Education Guides

(

(

−1

)

(

)

) ⋅ (3x −7 8

) (

(

)

 3x2 + 2  3  x + 2x +2 =    8  

2

)

(

)

−7

8

)

 2 d 3     2   d 2 2 2 −1  x + 1  +  x3 + 1 ⋅ x 3  =  x 3 ⋅ 3x 2  +  x3 + 1 ⋅ x 3   = x 3 ⋅ 3 dx dx         

8 −1 2 3 2x 3 + x +1 ⋅ x 3 = 3x 3 + 3 3

(

(

1 d 3 x + 2 x = x3 + 2 x 8 dx

)( )  = dxd (x + 1)x

2+ 2

)

1  1 d 1 −1   1  2 1  2 d 1 1  2 x2 ( x − 1) − 2 − 2 x ( x − 1) 2   x ⋅ 2 (x − 1)2  − (x − 1)2 ⋅ 2 x   x dx (x − 1)2  − (x − 1)2 dx x        2  = = =  x4 x4 x4 

f.

(

)

3

3

(

)

(

)

⋅ x3 + 1

(

)

(

)

(

)

1 1 1 1 −1  2   2 d 3  3 d 2 2  3 1 2 x +1 2   x + 1 2 ⋅ x  − x ⋅  x + 1 2 ⋅ 3x  −  x ⋅ x + 1 2 ⋅ 2 x   2 dx   dx  =      =  2 2  x +1 x +1 

(

)

(

1   2 2 4 2 3 x x + 1 2  −  x x + 1    x2 + 1

)

−1  2

 

105

Basic Differentiation

 d  x5 i. dx  3  x + 1

)

(

(

=

j.

Solutions

(

2 3

)

)

(

(

(

)

)

2 2 2 2 −1  3  3 d 5  5 d 3 4  5 2 3 2 x +1 3    x + 1 3 ⋅ x  − x ⋅  x + 1 3 ⋅ 5 x  −  x ⋅ x + 1 3 ⋅ 3x  dx   dx 3  =   =     4 4  3 3 3 3 x +1 x +1 

(

)

(

)

)

2 −3  2   4 3 2 x7 3 x +1 3  5 x x + 1 3  −     3 

(x + 1)

4 3

3

(

(

)

(

)

2   4 3 7 3 5 x x + 1 3  −  2 x x + 1     = 4 x3 + 1 3

(

)

−1  3

)

 

1 1 1 1 1 1 1 1 1 1 d  (x − 1) 2 (x + 1) 3  = (x + 1)3 d (x − 1)2  + (x − 1)2 d (x + 1)3  = (x + 1)3 ⋅ 1 (x − 1)2 −1  + (x − 1)2 ⋅ 1 (x + 1)3 −1   dx dx dx  2 3          2 1 1 1 1− 3  1 1− 2  1 1 1 1 1 =  (x + 1)3 ⋅ (x − 1) 2  +  (x − 1)2 ⋅ (x + 1) 3  =  ( x + 1)3 ( x − 1) − 2  +  ( x − 1) 2 ( x + 1) − 3  2  3   2  3

k.

d dx

(

)

(

)

(

(

)

(

)

(

)

(

)

(

)

(

)

1− 2  1   = 3 x 2 x 2 + 1 2  +  x 4 x 2 + 1 2  = 3 x 2 x 2 + 1    

l.

)

(

)

(

)

1 1 1 −1 1 1      2  2  3 2 2  3 1 2 2 2 d x3 + x3 d x 2 + 1 2  =  x + 1 2 ⋅ 3x  +  x ⋅ x + 1 2 ⋅ 2 x    x x +1  =  x + 1 dx dx 2          

1 2

)

(

+ x4 x2 + 1

(

)

−1

2

(

)

(

)

 2     −1  − 1 d 3  3 d −1  −1 − 1 −1 d  3 2 1 x  + x x 2 + 1 3  =  x 2 + 1 3 ⋅ 3x 2  +  x3 ⋅ − x 2 + 1 3 ⋅ 2 x  x x + 1 3  =  x + 1 3 dx   dx 3 dx        

(

(

)

−1− 3   − 1   2x4 2 = 3 x 2 x 2 + 1 3  + − x + 1 3  = 3x2 x2 + 1     3

)

−1

3

−

(

2 x4 2 x +1 3

)

−4

3

Section 1.7 Solutions - The Derivative of Radical Functions 1. Find the derivative of the following radical expressions. Do not simplify the answer to its lowest term.

(

)

1

a. Given y = x 2 + 1 = x 2 + 1 2 , then y′ =

b. Given y = x 3

(

(

)

)

1 −1 −1 1 2 2/ x 2 x + 1 2 ⋅ 2x = x +1 2 = 2/ 2

x

(x + 1) 1 1 + 3 x − 5 = (x + 3 x − 5) , then y′ = (x + 3 x − 5) ⋅ (3 x + 3) = (x + 3 x − 5) (3 x + 3) = 2 2 1 2

3

1 −1 2

3

2

1

1

1 2

2

−1

3

d. Given y =

e. Given y =

f. Given y =

g. Given y =

(x + 1) x +1 = x x x

2

x 2 −1

=

x3 + 3x 2 x2 + 3 x +1

x

(

1

2

Hamilton Education Guides

)

3

= x 2 + 3 x 2 , then y′ = 1

=

2 x3 + 3x − 5

)

1 2

1

( x − 1) 2

(

)

1 −1 1   1 2 2 2 1 2 x ⋅ x − 1 2  −  x − 1 2 ⋅ 2/ x ⋅ x  2/ 2x x2 − 1 2 − x3 x2 − 1     , then y′ = = x2 − 1 x2 − 1

(x − 1) = (x ) + 3 x 1 2

3 2

(

)

1 −1 1 1   1 1 x ( x + 1) − 2 − ( x + 1) 2  2 (x + 1)2 ⋅ x  − 1 ⋅ (x + 1)2    2 , then y′ = = x2 x2

2

2

(

3 x2 + 1

1

1

c. Given y = x 2 + x − 1 = x 2 + (x − 1)2 , then y′ = 2 x 2 −1 + (x − 1) 2 −1 = 2 x + (x − 1) − 2 = 2 x +

1 2

2

2

(

)

(

)

−1

2

3 1 3 32 − 1 3 3− 2 x + (3 ⋅ 2 )x 2 −1 = x 2 + 6 x = x 2 + 6 x 2 2 2 1 −1

 x2 + 3  2  2 2 x2 + 3  , then y′ = 1  x + 3  =   x +1  x +1 2  x + 1   

⋅

(

) = 1  x

2 x ⋅ (x + 1) − 1 ⋅ x 2 + 3

(x + 1)

2

2

+ 3  2  x + 1 

−1

2

⋅

x2 + 2x − 3

( x + 1) 2

106

Basic Differentiation

4

h. Given y =

i. Given y =

2. Use the

Solutions

x 3 −1

x2 x

3

1

x

x3

(x − 1) =

1 4

(

)

1 3  x −1 4 , then y′ = 

1 −1 4 ⋅ 3x 2

x2 =

x3/ =1

=

1

x 2/ ⋅ x 2

x 1

= x⋅x

−1

2

1− 1

= x

2

(

)

5

1 1  1 1 −1 3x 2 3 ⋅ x 2  −  x 2 ⋅ x3 − 1 4  x −1 2    = 4 x

1

= x 2 , then y′ =

x2

(

)

−3 4

− x

(

1 − 12 3 x x −1 2

)

1 4

1 1− 2 1 12 − 1 1 1 −1 1 −1 x = x2 1 = x 2 = x 2 2 2 2 2

d notation to find the derivative of the following radical expressions. dx

(0 ⋅ x ) − (1 ⋅ 1) = 1 − 1 d d 1 d  1 d  2 1 x +   = 1+ x+  =  x +  = dx  x dx dx  x  dx  x x2 x2

a.

 1 −1 1 −1 1 (x − 1) d x  −  x d (x − 1)    2 2 1 x  d  x  d  x  dx   dx  = 1  x  2 ⋅ [1 ⋅ (x − 1) ] − [1 ⋅ x ] b. = ⋅    =  2  x −1 2  x −1 dx  x − 1  dx  x −1  (x − 1) 2 (x − 1) 2 1 x  =   2  x −1

1− 2 2

x/ − 1 − x/ 1 x  = −  ⋅  2 2  x −1 (x − 1)

d  x3 d  x 3  c. = dx  (x + 1)12 dx  x +1  

−1

2

⋅

1

( x − 1) 2

1 d 1  (x + 1) 12 ⋅ 3 x3 −1  −  x3 ⋅ 1 (x + 1) 12 −1  3  3 d  (x + 1) 2 dx x  −  x dx (x + 1) 2      2    =    =   x +1 x +1 

3 3(x + 1) 12 x 2  −  x (x + 1) 1−22      2   = = x +1

d  (x + 5) 2 d  x + 5  d. = dx  x dx  x   1

 1   x3  2 −1 3 x ( x + 1) 2  −  2 ( x + 1) 2    x +1

1  1 d 1 1 −1    d   1 1 1  x ( x + 5) − 2 − ( x + 5) 2   x (x + 5) 2  − (x + 5) 2 dx x   x ⋅ 2 (x + 5) 2  − (x + 5) 2 ⋅ 1  =  dx    =   = 2  x2 x2 x2 

1   1 1 −1   −1 d 3 d − 12 d  3 d  3 d  3 x2  d  3 d  3 x   x + x 2 ⋅ x −1  = x + x2  = x + x 2  = = x + x e. x + x +  =         dx dx dx  x  dx  dx  dx  dx  x      

= 3 x3 −1 −

1 − 12 − 1 1 −1− 2 1 −3 = 3x 2 − x 2 = 3 x 2 − x 2 x 2 2 2

d  2 x f. 1+ dx  x3 =

1  2    = d 1 + 2 x 3  dx  x  

 1− 6  1 −3 1   −5 d  d  d  d  2  = 1 + 2 x 2  1 + 2 x 2  = 1 + 2 x 2 x −3  = + 1 2 x  =     dx  dx  dx  dx       

−5 − 2  −5  −7 − 5 −1 d (1) + d  2 x 2  = 0 − 5 x 2 ⋅ 2 = − 5 x 2 ⋅ 2/ = − 5 x 2 2/ 2 dx dx  

3. Find the derivative of the following radical expressions. 3− 2 3 −1 3 1 −1 1 1   3 −1 d  3  d (x ) ; d  x 2 + y 2  = 1 ; d x 2 + d y 2 = 1 ; 3 x 2 + 1 y 2 y′ = 1 ; 3 x 2 + 1 y 2 y′ = 1  x + y= 2 2 dx dx 2 2 dx  dx   dx 

a. ;

b.

1 −1 3 1 3 12 1 − 12 x + y y′ = 1 ; y 2 y′ = 1 − x 2 ; 2 2 2 2 d dx

(

)

x + y3 =

y′ 1 2y 2

= 1−

1 1 1 3 1 3 12 x ; y′ = 2 y 2 1 − x 2  ; y′ = 2 y 2 − 3( xy ) 2 2 2  

1 −1 1  1  −1 −1 d (2) ; d  x 2 + y3  = 0 ; d x 2 + d y 3 = 0 ; 1 x 2 + 3 y 2 y′ = 0 ; 1 x 2 + 3 y 2 y′ = 0 ; 3 y 2 y′ = − 1 x 2 dx 2 2 dx dx 2 dx  

Hamilton Education Guides

107

Basic Differentiation

; 3 y 2 y′ = −

1

d (x y ) = d dx dx

d.

d dx

(

y′

;

1 2y 2

1

6 x 2 y2

( x ) ; dxd (x y ) = dxd x

)

1 2

; y

  d d d 12 1 1 −1 1 −1 1 1  x+x y= x ; y + x y′ = x 2 ; x y′ = x 2 − y ; y′ =  1 − y  2 2 dx dx dx x 2   2x 

 d 12 d 3 1 12 − 1 1 − 12 d  12 2  y + x3  = 0 ; ′ ; ; y + x = 0 y y + 3 x = 0 y y′ = −3 x 2 ;  2 2 dx dx dx  

y + x3 = 0 ;

=

1

; y′ = −

1 2x 2

c.

Solutions

y′ 1 2y 2

= −3 x 2

1 − 3x 2 ; y′ = −6 x 2 y 2 1

(

)

e.

d  4 d 2 ( x ) ; d x 2 + y 2 = 1 ; d x 2 + d y 2 = 1 ; 2 x + 2 y y′ = 1 ; 2 y y′ = 1 − 2 x ; y′ = 1 − 2 x  x +y = dx dx dx dx  2y  dx

f.

d dx

(

)

( )

1 1 1 d 1 3 y 2 y′ d 3 (x + 1)2 = 3 y 2 d y ; 1 (x + 1) 2 − 1 = 3 y 2 y′ ; 1 (x + 1) − 2 = 3 y 2 y′ ; y ; = 1 dx dx dx 2 2 1 2(x + 1) 2

x +1 =

1

1

; 6 y 2 y′(x + 1) 2 = 1 ; y′ =

(

1

6 y ( x + 1) 2 2

)

1 −1 1   −1 d d (2) ; d  x y 2 + x 2  = 0 ; y 2 d x + x d y 2 + 1 x 2 = 0 ; y 2 ⋅ 1 + x ⋅ 2 y y′ + 1 x 2 = 0 x y2 + x = dx dx 2 dx dx 2 dx  

g.

−1

 1 −1 1 1  1 − 12 y y2 x 2 y2 1  − x − y 2  ; y′ = − ′ ; 2 x y y′ = − x 2 − y 2 ; y′ = ; ; y′ = − 3 − − y = − − 1   1 + 2 2x y  2 2x 4 xy 2 xy 2 xy  4x 2 y 4x 2 y 3 1 1 3 −1 d  3 d (x y ) = 0 ; d x 2 + d (x y ) = 0 ; 3 x 2 +  y d x + x d y  = 0 ; 3 x 2 + ( y + x y′) = 0 ; x y′ = − 3 x 2 − y  x + 2 dx dx 2 2 dx  dx   dx  dx

h.

1

; y′ =

i.

d dx

(

 1  3 12 y 3 3x 2 y 3 y y 3  − x − y  ; y′ = − ; y′ = − 1 − − ; y′ = − − ; y′ = − − 1 1   − − 1 x 2 2 x x x x x  2x ⋅ x 2 2x 2 2x 2

)

x + 3y =

1 −1 1  1  −1 −1 d ( y ) ; d  x 2 + 3 y  = y′ ; d x 2 + d 3 y = y′ ; 1 x 2 + 3 y′ = y′ ; 1 x 2 + 3 y′ − y′ = 0 ; 1 x 2 + 2 y′ = 0 dx dx  2 2 2 dx dx 

1 −1 1 1 ; 2 y′ = − x 2 ; 2 y′ = − 1 ; y′ = − 1 2 2x 2 4x 2 4. Evaluate the derivative of the following radical expressions for the specified value of x . 3

2

a. Given y = 3 x + x =

( )

1

3x3 2

2

+x =

2

+ x , then y′ =

1 32

3

3

3 3 −1 3 2 12 3 2 3−2 2 ⋅ x 2 + 2 x 2 −1 = x + 2x x + 2x = 2 2 2

5.196 32 3 2 12 y′ = + 2 = 4.598 +2 = ⋅ 1 + (2 ⋅ 1) = 2 2 2

at x = 1

)

(

(

b. Given y = x 2 + 1

x2 − 1 4x2

)

1

x = x2 + 1 x 2 = x

5 23 1 − 12 5 3 1 x + x = x2 + 1 2 2 2 2x 2

c. Given y =

3 x2

3

3

=

1 32

=

2+ 1

2

1

+ x2 = x

at x = 0

[ (

4 +1 2

y′ =

)]

[2 x ⋅ 2 x] − 2 ⋅ x 2 − 1 x2 − 1 , then y′ = 2x (2 x )2

Hamilton Education Guides

5

1

1

+ x 2 = x 2 + x 2 , then y′ = 5 32 ⋅0 + 2

=

1 1 2⋅ 02

= 0+

4x2 − 2x2 + 2 4x2

=

1 0

5 5 − 2 1 1− 2 5 52 − 1 1 12 − 1 x + x = x 2 + x 2 2 2 2 2 is undefined due to division by zero

2x2 + 2 4x2

=

(

)=

2 x2 + 1 4x2

x2 + 1 2x2

108

Basic Differentiation

Solutions 22 + 1

y′ =

at x = 2

d. Given y =

=

2 ⋅ 22

e. Given y =

2

⋅

− 12 + 1

=

11   22

−1

2

⋅

2

1 2

3

(

2

)

3 ⋅ 02 3 ⋅ 0 +1 2

−1

2

2

3

3 ⋅ 02 ⋅ 2

+ 12 ⋅ 02 =

1

(0 + 1) 3

1 2

11   22

=

2

2

2

2

−1+1

3

−1

2

⋅

2 (1 + 1) 1 = (x + 1) + 4 x , then y′ = (x + 1) 2

x3 + 1 + 4 x3 y′ =

at x = 0

−1

[1 ⋅ (x + 1)]− [2x ⋅ x] = 1  x  ⋅ − x + 1 2  x + 1  (x + 1) (x + 1)

1 −1

1

1 x  2  x 2 =  2   , then y′ =  2 2 2  x + 1   x +1 x +1 x

1 1  y′ =  2  2 1 +1

at x = 1

4 +1 5 = = 0.625 2⋅4 8

−1

2

1 −1 2 ⋅ 3x 2

2

2

2

⋅0 = 0

+ (4 ⋅ 3)x3 − 1 = 1

+ 12 ⋅ 02 = 0 ⋅

(

1 12

(

)

3x 2 3 x +1 2

−1

2

+ 12 x 2

+0 = 0+0 = 0

)

3    5 3 3 −1 2 3 1 2 x ⋅ x 2  −  x 2 ⋅ x + 1  2x 2 − x 2 x2 + 1 2 x +1 x +1 x +1     2 f. Given y = = = , then y′ = = 3 1 3 x3 x 3 2 2 x3 x x

2

2

2

(

)

( ) 5

2 ⋅ 32 −

y′ =

at x = 3

(

)

3 12 2 ⋅3 3 +1 (2 ⋅ 15.58) − (1.5 ⋅ 1.732 ⋅ 10) = 31.18 − 25.98 = 0.193 2 = 27 27 33

Section 1.8 Solutions - Higher Order Derivatives 1. Find the second derivative of the following functions. a. Given y = x3 + 3 x 2 + 5 x − 1 , then y ′ = 3 x 3−1 + (3 ⋅ 2 )x 2 −1 + 5 x1−1 − 0 = 3 x 2 + 6 x + 5 x 0 = 3 x 2 + 6 x + 5 and y ′′ = (3 ⋅ 2 )x 2 −1 + 6 x1−1 + 0 = 6 x + 6 x 0 = 6 x + 6

b. Given y = x 2 (x + 1)2 , then y ′ = 2 x ⋅ (x + 1)2 + 2(x + 1)2 −1 ⋅ x 2 = 2 x (x + 1)2 + 2 x 2 (x + 1) = 2 x (x + 1)2 + 2 x3 + 2 x 2 and y ′′ =

[2 ⋅(x + 1) + 2(x + 1)

2 −1

2

]

(

)

⋅ 2 x + (2 ⋅ 3)x3 −1 + (2 ⋅ 2 )x 2 −1 = 2 (x + 1)2 + 4 x(x + 1) + 6 x 2 + 4 x = 2 x 2 + 2 x + 1 + 4 x 2 + 4 xx

(

)

+ 6 x 2 + 4 x = 2 x 2 + 4 x + 2 + 10 x 2 + 8 x = 12 x 2 + 12 x + 2 = 2 6 x 2 + 6 x + 1

c. Given y = 3 x 3 + 50 x , then y′ = (3 ⋅ 3)x3−1 + 50 x1−1 = 9 x 2 + 50 and y ′′ = (9 ⋅ 2 )x 2 −1 + 0 = 18 x d. Given y = x5 +

e. Given y =

y ′′ =

=

[(6x

2

1

x2

= x5 + x −2 , then y′ = 5 x5−1 − 2 x −2 −1 = 5 x 4 − 2 x −3 and y ′′ = (5 ⋅ 4 )x 4 −1 + (− 2 ⋅ −3)x −3−1 = 20 x 3 + 6 x −4

] [ ] − (5 ⋅ 2)x

[

3 x 2 ⋅ (x + 1) − 1 ⋅ x3 x3 − 5 x 2 , then y′ = x +1 (x + 1) 2

][

)

(

+ 6 x ⋅ (x + 1) 2 − 2(x + 1) ⋅ 2 x3 + 3 x 2

(x + 1) 4

(x/ + 1/ )[6 x(x + 1) 2 − 2(2 x3 + 3x 2 ) ] (x + 1) 4/ =3

(

− 10 =

)]

− 10 x1 − 1 =

2 −1

=

3x 3 + 3x 2 − x 3

[6x(x + 1)(x + 1) ]− [2(x + 1)(2x 2

3

( x + 1) 3

2 x 3 + 3x 2

(x + 1)2

+ 3x 2

(x + 1) 4

6 x ( x + 1) 2 − 2 x 2 (2 x + 3 )

)

(x + 1)2

− 10 x =

)]

− 10 x and

− 10

− 10

f. Given y = x 3 x 2 − 1 = x5 − x3 , then y′ = 5 x5−1 − 3 x3−1 = 5 x 4 − 3 x 2 and y ′′ = (5 ⋅ 4 )x 4 −1 − (3 ⋅ 2 )x 2 −1 = 20 x 3 − 6 x g. Given y = x 4 +

(

)

(

)

1 x 8 − 7 x 5 + 5x 1 , then y′ = 4 x 4−1 + 8 x 8−1 + (− 7 ⋅ 5)x 5−1 + 5 x 1−1 = 4 x 3 + 8 x 7 − 35 x 4 + 5 and 10 10 10

y ′′ = (4 ⋅ 3)x3−1 +

[

]

(

)

1 (8 ⋅ 7 )x7 −1 + (− 35 ⋅ 4)x 4−1 + 0 = 12 x 2 + 1 56 x6 − 140 x3 = 5.6 x 6 − 14 x 3 + 12 x 2 10 10

Hamilton Education Guides

109

Basic Differentiation

Solutions

1 = x 2 − (x + 1)−1 , then y′ = 2 x 2−1 + (x + 1)−1−1 = 2 x + (x + 1)−2 and y ′′ = 2 x1−1 − 2(x + 1) − 2 −1 x +1 2 = 2 x 0 − 2(x + 1)−3 = 2 − 2(x + 1)−3 = 2 − ( x + 1)3 1 i. Given y = 2 − 3 x = x −2 − 3 x , then y′ = − 2 x −2−1 − 3 x 1−1 = − 2 x −3 − 3 and y ′′ = (− 2 ⋅ −3)x −3−1 − 0 = 6 x −4 x

h. Given y = x 2 −

2. Find y′′′ for the following functions. a. Given y = x5 + 6 x3 + 10 , then y ′ = 5 x 5−1 + (6 ⋅ 3)x 3−1 + 0 = 5 x 4 + 18 x 2 , y ′′ = (5 ⋅ 4 )x 4−1 + (18 ⋅ 2 )x 2−1 = 20 x 3 + 36 x , and y ′′′ = (20 ⋅ 3)x 3−1 + 36 x 1−1 = 60 x 2 + 36 b. Given y = x 2 +

1 = x 2 + x −1 , then y ′ = 2 x 2−1 − x −1−1 = 2 x − x −2 , y ′′ = 2 x1−1 + 2 x −2 −1 = 2 + 2 x −3 , and y ′′′ = − 6 x −4 x

c. Given y = 4 x3 (x − 1)2 , then y′ =

[(4 ⋅ 3)x

][

]

(

)

⋅ (x − 1)2 + 2(x − 1)2 −1 ⋅ 4 x3 = 12 x 2 (x − 1)2 + 8 x 3 (x − 1) = 12 x 2 x 2 − 2 x + 1

3−1

+ 8 x 4 − 8 x3 = 12 x 4 − 24 x 3 + 12 x 2 + 8 x 4 − 8 x 3 = 20 x 4 − 32 x 3 + 12 x 2 y ′′ = (20 ⋅ 4 )x 4−1 − (32 ⋅ 3)x 3−1 + (12 ⋅ 2 )x 2−1 = 80 x 3 − 96 x 2 + 24 x and y ′′′ = (80 ⋅ 3)x 3−1 − (96 ⋅ 2 )x 2−1 + 24 x 1−1 = 240 x 2 − 192 x + 24 x 0 = 240 x 2 − 192 x + 24

[1 ⋅ (x + 1) ] − [1 ⋅ x] = x + 1 − x = 1 = (x + 1)−2 , y ′′ = − 2(x + 1)−2−1 = − 2(x + 1) − 3 and x , then y ′ = x +1 (x + 1)2 (x + 1)2 (x + 1)2

d. Given y =

6

y ′′′ = (− 2 ⋅ −3)(x + 1) − 3 −1 = 6 (x + 1) − 4 =

( x + 1)4

e. Given y = x8 − 10 x5 + 5 x − 10 , then y ′ = 8 x8−1 + (− 10 ⋅ 5)x5−1 + 5 x1−1 − 0 = 8 x 7 − 50 x 4 + 5 x 0 = 8 x 7 − 50 x 4 + 5 , y ′′ = (8 ⋅ 7 )x 7 −1 − (50 ⋅ 4 )x 4 −1 + 0 = 56 x 6 − 200 x3 and y ′′′ = (56 ⋅ 6 )x 6 −1 − (200 ⋅ 3)x3−1 = 336 x 5 − 600 x 2 f. Given y =

=

=

=

x −1 x

2

+ 5 x 3 , then y ′ =

[1 ⋅ x ]− [2 x ⋅ (x − 1) ] + (5 ⋅ 3)x 2

x

4

[

3−1

][

− 1 ⋅ x3 − 3 x 2 ⋅ (− x + 2 ) x/ (− x + 2 ) −x + 2 2 2 ′′ = = , y + x 15 + 15 x x 4/ =3 x3 x6

2 x3 − 6 x 2 x

6

+ 30 x =

− 6 x 4 + 24 x3 x

8

x 2/ (2 x − 6 )

+ 30 =

x

6/ = 4

− 6 x3/ (x − 4 ) x

8/ =5

2x − 6

+ 30 x =

x

4

+ 30 = −

x2 − 2x2 + 2x x

x5

4

] + (15 ⋅ 2)x

[2 ⋅ x ]− [4x 4

+ 30 x and y ′′′ =

6( x − 4 )

3. Find f ′′(0 ) and f ′′(1) for the following functions.

=

2 −1

3

x

+ 15 x 2 =

− x2 + 2x x4

− x3 + 3x3 − 6 x 2

=

x6

⋅ (2 x − 6 )

8

] + 30x

1 −1

=

+ 15 x 2

+ 30 x 2 x 4 − 8 x 4 + 24 x3 x8

+ 30

+ 30

a. Given f (x ) = 6 x5 + 3 x3 + 5 , then f ′(x ) = (6 ⋅ 5)x5−1 + (3 ⋅ 3)x3−1 + 0 = 30 x 4 + 9 x 2 and f ′′(x ) = (30 ⋅ 4 )x 4 −1 + (9 ⋅ 2 )x 2 −1 = 120 x3 + 18 x . Therefore, f ′′(0 ) = 120 ⋅ 03 + 18 ⋅ 0 = 0 and f ′′(1) = 120 ⋅ 13 + 18 ⋅ 1 = 120 + 18 = 138 b. Given f (x ) = x3 (x + 1)2 , then f ′(x ) =

[3x

3−1

][

]

(

)

⋅ (x + 1)2 + 2(x + 1)2 −1 ⋅ x3 = 3 x 2 (x + 1)2 + 2 x3 (x + 1) = 3 x 2 x 2 + 2 x + 1

+ 2 x 4 + 2 x3 = 3 x 4 + 6 x3 + 3 x 2 + 2 x 4 + 2 x3 = 5 x 4 + 8 x3 + 3 x 2 and f ′′(x ) = (5 ⋅ 4 )x 4 −1 + (8 ⋅ 3)x3−1 + (3 ⋅ 2 )x 2 −1

(

) (

)

(

) (

)

= 20 x3 + 24 x 2 + 6 x . Therefore, f ′′(0 ) = 20 ⋅ 03 + 24 ⋅ 02 + (6 ⋅ 0 ) = 0 and f ′′(1) = 20 ⋅ 13 + 24 ⋅ 12 + (6 ⋅ 1) = 50

Hamilton Education Guides

110

Basic Differentiation

Solutions

c. Given f (x ) = x + (x − 1) 2 , then f ′(x ) = 1 + 2(x − 1) 2 −1 = 1 + 2(x − 1) = 1 + 2 x − 2 = 2 x − 1 and f ′′(x ) = 2 x1−1 − 0 = 2 . Therefore, f ′′(0 ) = 2 and f ′′(1) = 2 d. Given f (x ) = (x − 1) − 3 , then f ′(x ) = − 3(x − 1) − 3 −1 = − 3(x − 1) − 4 and f ′′(x ) = (− 3 ⋅ −4 )(x − 1) − 4 −1 = 12 (x − 1) − 5 = Therefore, f ′′(0 ) =

12

(0 − 1)

5

(

=

12

(x − 1) 5

12 12 12 = −12 and f ′′(1) = = which is undefined due to division by zero. 5 −1 0 (1 − 1)

)

e. Given f (x ) = (x − 1) x 2 + 1 , then f ′(x ) =

[1 ⋅ (x + 1)]+ [2 x ⋅ (x − 1) ] = x 2

2

+ 1 + 2 x 2 − 2 x = 3 x 2 − 2 x + 1 and

f ′′(x ) = (3 ⋅ 2 )x 2 −1 − 2 x1−1 + 0 = 6 x − 2 . Therefore, f ′′(0 ) = (6 ⋅ 0 ) − 2 = 0 − 2 = −2 and f ′′(1) = (6 ⋅ 1) − 2 = 6 − 2 = 4

(

)

f. Given f (x ) = x3 − 1

2

(

)

+ 2 x = x3 − 1

[ (

)

2

1

(

)

+ (2 x ) 2 , then f ′(x ) = 2 x3 − 1

2 −1

⋅ 3x 2 +

(

)

1 1 1 (2 x ) 2 − 1 ⋅ 2 = 6 x 2 x3 − 1 + (2 x ) − 2 2

]

3 3 1 1 (2 x ) − 2 − 1 ⋅ 2/ = 12 x 4 − 12 x + 18 x 4 − (2 x ) − 2 = 30 x 4 − 12 x − (2 x ) − 2 . Thus, f ′′(0) 2/ 1 1 = − which is undefined, and f ′′(1) = 30 ⋅ 14 − 12 ⋅ 1 − = 30 − 12 − 0.35 = 17.65 3 0 (2 ⋅ 1) 2

and f ′′(x ) = 12 x ⋅ x3 − 1 + 3 x 2 ⋅ 6 x 2 − = 30 ⋅ 04 − 12 ⋅ 0 −

1 3

(2 ⋅ 0) 2

Hamilton Education Guides

111

About the Author Dan Hamilton received his B.S. degree in Electrical Engineering from Oklahoma State University and Master's degree, also in Electrical Engineering from the University of Texas at Austin. He has taught a number of math and engineering courses as a visiting lecturer at the University of Oklahoma, Department of Mathematics, and as a faculty member at Rose State College, Department of Engineering Technology, at Midwest City, Oklahoma. He is currently working in the field of aerospace technology and has published several books and numerous technical papers.

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