eBook Calculus Differentiation for Science And Engineering Students And

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eISBN 978-967-2897-43-9

Preface

Grateful to Allah because with His permission, the eBook Calculus - Differentiation For Science & Engineering
Students was published. This eBook is written by lecturers who have been teaching in Engineering Mathematics for more
than 15 years . This eBook can be used by all institutions of higher learning such as Polytechnics and Colleges as well as
private and public universities. The purpose for this eBook was written is to make it easier for students to gain
knowledge and review the topic of calculus in a simpler and more concise way.

Many examples in various forms of questions are included in this eBook with detailed steps of solution to make it
easier for students to quickly understand the method of its solution. In addition, students will also able to improve and
strengthen their understanding through the included practice questions. The authors hope that this eBook can benefit all
students as well as educators around the world in the field of Engineering Mathematics.

Thank You So Much.

Suhana binti Ramli
Editor

Table of Contents

Introduction Of Differentiation 1 Differentiation Of Exponential 52

Basic Rules Of Differentiation 2 Application Of Differentiation 60

a. Constant Rule 6 a. Maximum, Minimum And Point Of Inflexion 61

b. Power Rule 7 b. Rate Of Change 67

c. Sum And Difference Rule 8 Parametric Equation 72

d. Constant Multiple Rule 12 Implicit Differentiation 76

Techniques Of Differentiation 15 Partial Differentiation 82

a. Chain Rule 15 a. First Order Of Partial Differentiation 83

b. Product Rule 20 b. Second Order Of Partial Differentiation 90

c. Quotient Rule 26
Total Differentiation
Second Order Differentiation 31 95
Exercises 101
Differentiation Of Trigonometrics 102
34 References

Differentiation Of Logarithmic 43

Differentiation Of
Trigonometrics

Function

Calculus: Differentiation

Introduction Of Differentiation

Algebra Differentiation
❑ Algebra focuses on solving equations
❑ Algebra deals with operations on variables ➢ The process of finding a derivative
is called differentiation.
and numbers
➢ If given = ( ) , the differentiation
Calculus
❑ Calculus is primarily focused on is written as or ′

differentiation and integration problems.
❑ Calculus deals with operations on
➢ Application :
functions and their derivatives i) to calculate rates of change
ii) to determine where the maximum
and minimum values occur.

** Among the uses of calculus in engineering
are the study of gravity and planetary
motion, fluid flow and ship design, and
geometric curves and bridge engineering.

1

Calculus: Differentiation

Basic Rules Of Differentiation

2

Calculus: Differentiation

Basic Rules Of Differentiation

a) Constant Rule b) Power Rule

= = =

= = − = −


c) Sum & Difference Rule d) Constant Multiple Rule

= ± = ( ± )

= ′ ± ′ = ( ± ) − ( ± )


3

Calculus: Differentiation

a) Must have POWER = 4 5 5

= 4

b) POWER must be numerator 1 = 1 −5
= 3 5 3

c) =

d) Expand = 2 + 1 3 − = 3 + 5 − 2 2
= 2 − 6 + 9
= − 3 2
6 4 3 5
e) Separate 6 − 4 3 + 5 = 3 − 3 + 3
= 3 = 3 − 4 + 5 −3

5

Calculus: Differentiation Differentiate the followings:

a) Constant Rule Example 1 Example 2
4 = 5
=
= 5
=


= 0 = 0

6

Calculus: Differentiation

b) Power Function Rule

= = Differentiate the followings:

= − = −


Example 1 Example 2
= 5 = 3 4

STEPS How to
Differentiate??
1. Bring the POWER up front.
2. POWER is subtracted by 1 = 5 5−1 = 4 ∙ 3 4−1

= 5 4 = 12 3

7

Calculus: Differentiation c) Sum & Difference Rule
Differentiate the followings:
= ±
ATTENTION !
= ′ ± ′

Example 1 2 = 1 2
3 3

4 2 Example 2
3
= 4 − − 9

= 5 − 2 6 − 9
3
1
= 4 4 4−1 − 2 3 2−1 − 0

= 16 3 − 2 1 ′ = 5 5−1 − 6 2 6−1 − 9 1−1
3 3

= 16 3 − 2 = 5 4 − 4 5 − 9 0 REMEMBER !
3 = 5 4 − 4 5 − 9 0 = 1

8

Calculus: Differentiation

ATTENTION !

3

3 = 2

= 2 3 − 3 + 4 ℎ = 3 − 4 + 5
3
3
∴ = 2 3 − 3 −3 + 4
∴ ℎ = 2 − 4 + 5

ℎ′ = 3 23−1 − 4 4−1 + 5 1−1
= 2
3 2 3−1 − −3 3 −3−1 + 4 4−1
3 1 4 3 5 0
= 6 2 + 9 −4 + 4 3 = 2 − +
2

= 6 2 + 9 + 4 3 3 − 4 3 + 5 REMEMBER !
4 =2 0 = 1

9

Calculus: Differentiation Example 6

Example 5 = 2 − 3 2 REMEMBER !
∴ = 4 2 − 12 + 9 2 − 3 2 (expand)

= 2 − 3 2 − 3
= 4 2 − 6 − 6 + 9
= 4 2 − 12 + 9

= 3 + 1 − 2 REMEMBER ! 2 4 2−1 − 1 12 1−1 + 0
∴ = 3 2 − 5 − 2 3 + 1 − 2 (expand) =
= 3 2 − 6 + − 2
= 3 2 − 5 − 2

= 8 1 − 12 0

2 3 2−1 − 1 5 1−1 − 0 = 18 − 12
=

= 6 1 − 5 0

= 6 − 5

10

Calculus: Differentiation ATTENTION !
Example 7
5 = 5
REMEMBER !
1

= 5 2

4 + 5 3 − 5 4 + 5 3 − 5 (separate)
= 5 −12 =
= 3 + 5 2 − 4 5 3 5
= + −
= 3 + 5 2 − 5 −21

3 3−1 − 2 5 2−1 − 1 5 −12−1
= −2

= 3 2 − 10 1 + 5 −23
2

= 3 2 − 10 + 1 5
2 3

11

Calculus: Differentiation

DDiiffffeerreennttiiaatteetthhee ffoolllloowwiinnggss::

d) Constant Multiple Rule Example 1

= ( ± ) = 4 2 + 3 5

= ( ± ) − ( ± )


4 2 + 3 5−1 ∙ 8
= 5
= 40 4 2 + 3 4

STEPS How to Example 2
Differentiate??
= − 3 2 8
1. Differentiate OUTSIDE (power value).
2. Differentiate INSIDE (value in bracket)

= 8 − 3 2 8−1 ∙ 1 − 6

= 8 1 − 6 − 3 2 7 12

Calculus: Differentiation Example 5
Example 3 = 2 3 + 5 − 3 4

= 3 − 8 5

3 − 8 5−1 ∙ 3 3−1 Example 4 ′ = 4 2 3 + 5 − 3 4−1 ∙ 6 3−1 + 5 1−1
= = 4 6 2 + 5 2 3 + 5 − 3 3

5

= 15 2 3 − 8 4

1 4 + 3 3
=6

′ = 3 1 4 + 3 3−1 ∙ 4 4−1 + 3 1−1
6

1 4 3 + 3 4 + 3 2
=2

13

Calculus: Differentiation Example 7
Example 6

2 = 3 4 − 5 change the form
= 3 5 − 2 6
change the form 1

= 4 − 5 3

2 5 − 2 −6
= 3
1 4 − 5 31−1 ∙ 4 4−1
= 3

2 5 − 2 −6−1 ∙ −2 = 1 4 3 4 − 5 −32
= −6 3
3

= 8 5 − 2 −7 4 3
=
8
= 5 − 2 7 33 4 3 − 5 2

14

Calculus: Differentiation

Techniques Of Differentiation

CHAIN RULE a) Chain Rule

The chain rule has been known since Isaac Newton and
Leibniz first discovered the calculus at the end of the 17th = ×
century. The chain rule states that to compute the derivative
of composite function, f(g(x)) is f'(g(x))⋅g'(x).

15

Calculus: Differentiation DDififfeferreennttiaiatetetthheefofolllolowwininggss: :

a) Chain Rule Example 1 = 4 2 + 3 5

Step 1 = 4 2 + 3 = 5
= × Step 2
Step 3 = 5 4
STEPS = 8
1. Find u & y Step 4
o u in the bracket & y the remain
2. Differentiate u & y = ×
3. Substitute into the formula
4. Solve / simplify = 5 4 ∙ 8
5. Replace u

= 40 4

Step 5 = 40 4 2 + 3 4

16

Calculus: Differentiation

Example 2 Example 3 1 4 + 3 3
= 6
Step 1
= 2 3 + 5 − 3 4 Step 2 = 4 + 3 = 1 3
Step 3 6
= 2 3 + 5 − 3 = 4
Step 1 Step 4 = 4 3 + 3 = 1 2
Step 2 Step 5 2
Step 3
= 6 + 5 = 4 3
Step 4 = ×
Step 5
= × = 1 2 ∙ 4 3 + 3
2

= 4 3 ∙ 6 + 5 1 4 3 + 3 2
=2
= 4 6 + 5 3
1 4 3 + 3 4 + 3 2
=2
= 4 6 2 + 5 2 3 + 5 − 3 3

17

Calculus: Differentiation

Example 5

Example 4 = 3 4 − 5

2 change the form 1 change the form
= 3 5 − 2 6 1
= 4 − 5 3
2 5 − 2 −6 Step 1 = 4 − 5 = 3
= 3 Step 2
Step 3 1 −23
Step 1 = 5 − 2 = 2 −6 = 4 3 = 3
Step 2 Step 4
Step 3 3 Step 5

Step 4 = −4 −7
Step 5 = −2 = ×

= 1 −23 ∙ 4 3
= × 3

= −4 −7 ∙ −2 = 1 4 3 −23

3

= −8 −7 4 3
= 33 4 3 − 5 2
8
= − 5 − 2 7 18

Calculus: Differentiation

CONCLUSION:

If given question = 7 − 6 2 8 therefore it can be solved either by using Composite Function or Chain Rule.

** Refer to what method the question want !!! Chain Rule

Composite Function = 7 − 6 2 8

= 7 − 6 2 8 = 7 − 6 2 = 8

= 8 7
= −6

= 8 7 − 6 2 7∙ −12 = ×

= −96 7 − 6 2 7 = 8 7 ∙ −6


= −48 7

= −48 7 − 6 2 7

19

Calculus: Differentiation

Techniques Of Differentiation

PRODUCT RULE b) Product Rule

If = . is the product of a function, then to obtain
the derivative of the product of that function, the = +
product rule must be used.

20

Calculus: Differentiation

ATTENTION:

If given questions as followings:

= 3 7 − 6 2 Expand Bracket has a POWER
= 7 3 − 6 5
= 3 7 − 6 2 3
bracket has = 2 + 3 4 7 − 6 2 3
NO POWER !
Differentiate using
= + 2 7 − 6 2 Expand PRODUCT RULE !
= 7 − 6 3 + 14 − 12 2
21
= 7 − 6 2 2 Expand bracket has the
= 7 − 6 2 7 − 6 2 POWER of 2 !
= 47 − 12 2 + 36 4

Calculus: Differentiation Example 1

b) Product Rule Differentiate = 2 3 2 + 3 5

Step 1 = 2 3 = 2 + 3 5
= + Step 2
Step 3 = 6 2 = 10 2 + 3 4
STEPS
1. Find u & v Step 4
o u in front & v at the back Step 5
2. Differentiate u & v = +
3. Substitute into the formula
4. Factorize 2 + 3 5 ∙ 6 2 + 2 3 ∙ 10 2 + 3 4
5. Solve / simplify =

= 6 2 2 + 3 5 + 20 4 2 + 3 4

= 2 2 2 + 3 4 3 2 + 3 + 10 2

= 2 2 2 + 3 4 3 2 + 9 + 10 2
= 2 2 2 + 3 4 13 2 + 9

22

Calculus: Differentiation

Example 2

Differentiate = 2 3 + 1 1 − 4 3

Step 1 = 2 3 + 1 = 1 − 4 3
Step 2
Step 3 = 6 2 = 3 1 − 4 2 −4

Step 4
Step 5 = −12 1 − 4 2


= +

1 − 4 3 ∙ 6 2 + 2 3 + 1 ∙ −12 1 − 4 2
=

= 6 2 1 − 4 3 − 12 2 3 + 1 1 − 4 2

= 6(1 − 4 )2[ 2 1 − 4 − 2 2 3 + 1 ]

= 6 1 − 4 2 2 − 4 3 − 4 3 − 2 23
= 6 1 − 4 2 2 − 8 3 − 2

Calculus: Differentiation Differentiate = − 1 3 3 + 1 5
Example 3

Step 1 = − 1 3 = 3 + 1 5
Step 2
Step 3 = 3 −1 2 1 = 5 3 + 1 4 3

Step 4
= 3 − 1 2 = 15 3 + 1 4


= +

3 + 1 5 ∙ 3 − 1 2 + − 1 3 ∙ 15 3 + 1 4
=

= 3 3 + 1 5 − 1 2 + 15 − 1 3 3 + 1 4

= 3 3 + 1 4 − 1 2 3 + 1 + 5 − 1

Step 5 = 3 3 + 1 4 − 1 2 3 + 1 + 5 − 5
= 3 3 + 1 4 − 1 2 8 − 4
= 12 3 + 1 4 − 1 2 2 − 1 24

Calculus: Differentiation

Find the derivative for = 2 + 3 2 − 3 2 5

Step 1 = 2 + 3 = 2 − 3 2 5
Step 2
Step 3 1

Step 4 = 2 + 3 2
Step 5
1 2 + 3 −12 2 = 5 2 − 3 2 4 −6
= 2
= 2 + 3 −21
= −30 2 − 3 2 4


= +

2 − 3 2 5 ∙ 2 + 3 −12 + 1
=
2 + 3 2 −30 2 − 3 2 4

= 2 − 3 2 5 ∙ 2 + 3 −21 − 30 2 + 3 1 2 − 3 2 4
2

= 2 − 3 2 4 2 + 3 −21 2 − 3 2 − 30 2 + 3

= 2 − 3 2 4 2 + 3 − 1 2 − 3 2 − 60 2 − 90
2

= 2 − 3 2 4 2 + 3 − 1 2 − 63 2 − 90
2

25

Calculus: Differentiation

Techniques Of Differentiation

QUOTIENT RULE c) Quotient Rule

The quotient rule is a method of finding the derivative of = −
a function that is the ratio of two differentiable functions. 2
Its function is in the form



26

Calculus: Differentiation

Example 1 Differentiate = 2 3
2+3 5
Step 1
Quotient Rule Step 2 = 2 3 = 2 + 3 5
Step 3
= 6 2 = 5 2 + 3 4 2
Step 4
− Step 5
= = 10 2 + 3 4

2 −

=
2

STEPS 2 + 3 5 ∙ 6 2 − 2 3 ∙ 10 2 + 3 4
=
1. Find u & v 2 + 3 5 2
o u is nominator & v is denominator
6 2 2 + 3 5 − 20 4 2 + 3 4
2. Differentiate u & v = 2 + 3 10
3. Substitute into the formula
4. Factorize 2 2 2 + 3 4 3 2 + 3 − 10 2
5. Solve / simplify = 2 + 3 10

2 2 3 2 + 9 − 10 2 2 2 9 − 7 2
= 2 + 3 6 = 2 + 3 6

27

Calculus: Differentiation

Find the derivative for = 2 3+1
1−4 3

Step 1 = 2 3 + 1 = 1 − 4 3
Step 2
Step 3 = 6 2 = 3 1 − 4 2 −4

Step 4
Step 5 = −12 1 − 4 2

= −


2

1 − 4 3 ∙ 6 2 − 2 3 + 1 ∙ −12 1 − 4 2
=
1 − 4 3 2

6 2 1 − 4 3 + 12 2 3 + 1 1 − 4 2
= 1 − 4 6

6 1 − 4 2 2 1 − 4 + 2 2 3 + 1
= 1 − 4 6

= 6 2 − 4 3 + 4 3 + 2 6 2 + 2
= 1 − 4 4
1 − 4 4 28

Calculus: Differentiation

Example 3 Find the derivative for = −1 3
3 +1 5
Step 1
Step 2 = − 1 3 = 3 + 1 5
Step 3
= 3 − 1 2 1 = 5 3 + 1 4 3
Step 4
Step 5
= 3 − 1 2 = 15 3 + 1 4

= −


2

3 + 1 5 ∙ 3 − 1 2 − − 1 3 ∙ 15 3 + 1 4
=
3 + 1 5 2

3 3 + 1 5 − 1 2 − 15 − 1 3 3 + 1 4
= 3 + 1 10

3 3 + 1 4 − 1 2 3 + 1 − 5 − 1
= 3 + 1 10

3 − 1 2 3 + 1 − 5 + 5 3 − 1 2 6 − 2 6 − 1 2 3 −
= 3 + 1 6 = 3 + 1 6 = 3 + 1 6

29

Calculus: Differentiation

Find the derivative for = 7 +2 2
4 3+1 3

Step 1 = 7 + 2 2 = 4 3 + 1 3
Step 2
Step 3 = 12 3 +1 2 3 2
= 2 7 + 2 7
Step 4 = 36 2 3 + 1 2
Step 5 = 14 7 + 2

= −
2

4 3 + 1 3 ∙ 14 7 + 2 − 7 + 2 2 ∙ 36 2 3 + 1 2
=
(4 3 + 1 3)2

56 3 + 1 3 ∙ 7 + 2 − 36 2 3 + 1 2 7 + 2 2
= 16 3 + 1 6

4 3 + 1 2 7 + 2 14 3 + 1 − 9 2 7 + 2
= 16 3 + 1 6

4 3 + 1 2 7 + 2 14 3 + 14 − 63 3 − 18 2 30
= 16 3 + 1 6

7 + 2 14 − 18 2 − 49 3
= 4 3 + 1 4

Calculus: Differentiation

Second Order Of Differentiation

➢ The second order of differentiation is

when the differentiation made twice

➢ The second derivatives of a function is
usually denoted as ′′( ) and for =

( ), then the second order written as 2
2
➢ Among the uses of the second order of

differentiation is to determine the optimum

value.

differentiate second time

31

Calculus: Differentiation

Example 1 Find the derivative for the followings:

= (5 + 3 4)5

= 4 3 + 2 = (2 + 3 )( − 1)

change the form = 5(5 + 3 4)4 ∙ 12 3 For 2nd order derivative, must

= 3 2 − − 2 use product rule :

= 12 2 + 2 = 6 − 1 = 60 3(5 + 3 4)4 = +
2
2 = 6
2 2 = 60 3 ∙ 4(5 + 3 4)312 3 + (5 + 3 4)4180 2
2 = 24 2

= 2880 6(5 + 3 4)3 + 180 2(5 + 3 4)4

= 20 2(5 + 3 4)3[144 4 + 9 (5 + 3 4)]

= 20 2(5 + 3 4)3 144 4 + 45 + 27 4

32

Calculus: Differentiation

Example 2 Find the value of 2 when = 3 for each of the following
2

= (2 − 1)4 Must use product rule for 1st order and 2nd

6 order derivative :
2
= 5 3 − + 9 + 3 change = +
the form

= 5 3 − 6 −2 + 9 + 3 = ∙ 4(2 − 1)3 ∙ 2 + 1 (2 − 1)4 ∙ 2


= 8 (2 − 1)3 + 2(2 − 1)4

= 15 2 + 12 −3 + 9 2
2 =
2 8 ∙ 3(2 − 1)2 ∙ 2 + (2 − 1)3 ∙ 8 + 4 ∙ 2(2 − 1)3 ∙ 2
2
= 30 − 36 −4 = 48 (2 − 1)2 + 8(2 − 1)3 + 16(2 − 1)3

36 = 48 (2 − 1)2 + 24(2 − 1)3
= 30 − 4
= 6(2 − 1)2[8 + 4(2 − 1)]
Substitute = 3 into 2 = 6(2 − 1)2 16 − 4
2

2 36 806 Substitute = 3 into 2
2 = 30 3 − 34 = 9 2 2
2 = 6(2 3
− 1)2 16 3 −4 = 6600 33

Calculus: Differentiation

Differentiation of Trigonometric Function

34

Calculus: Differentiation

Trigonometry Function

Basic Derivatives

Reciprocal Identities
=

= − = Trigonometric Identities
2 + 2 = 1
= 2 1 1 + 2 = 2
= 2 + 1 = 2

= − 1 35
=

= 1
= =

= − 2

Calculus: Differentiation

How To Derive Trigonometry Function

Method Method

(trigonometry with power 1) (trigonometry with power > 1)

STEPS STEPS

1. Differentiate trigonometry function 1. Differentiate Power
2. Differentiate the angle (in bracket) 2. Differentiate trigonometry function
3. Differentiate the angle (in bracket)
Examples :
= 5 , Examples :
= sin 5 + 4 = 25 ,
= 4(5 + 4)

36

Calculus: Differentiation

Trigonometry Function

STEPS STEPS
1) Differentiate Power
1) Differentiate trigonometry function 2) Differentiate trigonometry function
2) Differentiate the angle (in bracket) 3) Differentiate the angle (in bracket)

= 2 = 1 − 2 = 32

1 2 1 2 1 2 3
= 1 − 2 ∙ −2
= 2 ∙ 2 = 3 22 ∙ 2 ∙ 2
= 2 2
= −2 1 − 2 = 6 22 2

37

Calculus: Differentiation

Example 1 Find the derivative for the followings: 32
= 8 cos 4
change the form
3
2 = 8 cos(2 −4)
= 3
= 3 2 2 3 − 2 −4 ∙ −8 −5
= 3 ∙ 3 = 8

= − 3 ∙ 3 22 32
= 3 3 = 5 4
= −3 3

1 = change the form
= 2 4
1

= 2

= 1 24 ∙ 4 ∙ 1 −12
2 = − 2

= 2 2 1 − 4 2
=−

2

38

Calculus: Differentiation

Example 2 Find the derivative for the followings: = 4 1 −

1 change the
form
= 4 1 − 2

= 5 + 3 1 −21
= 4 2
= 3 + 2 − 2 1− ∙ −

= − 5 + 3 ∙ 5 + 3 2 2 2 1 −
= − 3 + 2 ∙ 3 =

= −3 3 + 2 = − 5 + 3 2 5 + 3

= 3 1 − 4 3 composite function 2 change the
8 = 7 − 4 form

= 7 − 2 −4

= 3 1 − 4 3 ∙ 3 1 − 4 2 −4 = 2 7 − 2 −4 ∙ −8 −5
8
8 2
9 = − 5 2 7 − 4
= −2
1 − 4 2 1 − 4 3

39

Calculus: Differentiation

Example 3 Find the derivative for the followings:

= 34 = 2 2 5 + 3

24 = 2 ∙ 2 5 + 3 ∙ − 5 + 3 ∙ 5 + 3 2

= 3 ∙ ∙ −sin 4 4

= −12 24 4 = −4 5 + 3 2 5 + 3 5 + 3

3 = 4 1
5 7 − 4
= 5 1 − 4 change the form

= 4 7 − −4

= 5 ∙ 3 4 1 − 4 ∙ 1 − 4 ∙ −4 = 4 ∙ 3 7 − −4 ∙ 2 7 − −4 ∙ 4 −5
5

= −12 4 1 − 4 ∙ 1 − 4 = 16 3 7 − −4 2 1
5 7 − 4

40

Calculus: Differentiation

Example 4 Find the derivative for the followings:

Chain Rule Product Rule

= 3 3 = 4 5 − 4 2

= 3 = 3 = 4 = 5 − 4 2

= 2 3(3 2) = 3 2 = 4 3 = 5 − 4 2 −8


= 3 2 2 3 = −8 5 − 4 2


= × = +

= 3 2 ∙ 3 2 2 3 = 5 − 4 2 ∙ 4 3 + 4 ∙ −8 5 − 4 2


= 9 2 2 3 2 = 4 3 5 − 4 2 − 8 5 5 − 4 2

= 9 2 2 3 2 3 3 = 4 3 5 − 4 2 − 2 2 5 − 4 2

41

Calculus: Differentiation

Quotient Rule

2 + 5
= 3 + 2 2

= 2 + 5 = 3 + 2 2

= − 2 +5 ∙ 2 = 3 2 + 4


= −2 2 + 5

= −
2

3 + 2 2 ∙ −2 2 + 5 − 2 + 5 ∙ 3 2 + 4
=
( 3 + 2 2)2

−2 3 + 2 2 2 + 5 − 3 2 + 4 2 + 5
= ( 3 + 2 2)2

42

Calculus: Differentiation

Differentiation Of Logarithm Function

➢ The expression for the derivative of
the natural logarithm function is

= = ( )′ = 1.



➢ It can be solved either using the
logarithm differentiation formula
or a chain rule.

➢ The law of logarithms should be
applied first if necessary before
differentiation.

➢ Logarithm differentiation can
also involve the product rule or
quotient rule.

43

Calculus: Differentiation

Logarithmic Function

Formula Differentiation of Law of Logarithm
Logarithm Function
+ =
a) = 1

− =
=
b) + = 1 ( + )
= 1
+
=
c) ln = 1 ∙



where is any value in bracket

d) ln = ′



44

Calculus: Differentiation

How To Derive Logarithmic Function

= ln 2 = ln 1 − 2

STEPS 1 1 2

1) Write 1 1 2 1
= 1 − 2 ∙ (−2)
+ = 2 ∙ 2 Don’t forget to
follow the steps
2) Differentiate + 1 u = 1 - 2x
OR = 45
u = 2x
1) Write 1 STEPS

STEPS 1) Write 1

2) Differentiate u 1) Write 1
(u is any value in bracket )
2) Differentiate u

2) Differentiate u OR ′
ln =