CHAPTER 2 DIFFERENTIATION

Asma Asdayana binti Ibrahim
Hanim binti Yahaya

‘DBM 20023 Engineering Mathematics 2: Differentiation’ has been designed to comply the Engineering
Mathematics 2 syllabus used in Polytechnic.

This book explains on the techniques of differentiation. It covers the basic rules of differentiation,
parametric equation, implicit differentiation, second order differentiation and derivatives of
trigonometric, logarithmic and exponential functions. It also explains application of differentiation in
calculating the rates of change in addition solving maximum and minimum problems.

We acknowledge that there are elements to this book that can be improved which we will strive
continually address in upcoming editions. We would like to give our highest appreciation to all relevant
individuals involved either directly or indirectly in the creation of this book. We hope that this book can
help to make the learning of Engineering Mathematics 2: Differentiation become an enjoyable and
beneficial experience.

Asma Asdayana Ibrahim
Hanim Yahaya

Summary ………………………………………………………………………………………………….………..ii
Learning Outcomes………………..…………………………………………………………………………..iii
Subtopics…………………………………………………………………………………………………………….iv
Formula……………………………………………………………………………………………………………….v

Describe Rules of Differentiation .................................................1
Use Trigonometric, Logarithmic and Exponential Function........23
Apply Second Order Differentiation ...........................................57
2.4 Demonstrate the Application of Differentiation ........................63
Understand Parametric Equation...............................................79
Apply Implicit Differentiation .....................................................85
Construct Partial Differentiation ................................................86
2.8 Apply The Technique of Total Differentiation ............................86

i

DIFFERENTIATION

TOPIC SUMMARY

This topic explains more on the
techniques of differentiation. It covers

the basic rules of differentiation,
parametric equation, implicit
differentiation, second order

differentiation and derivatives of
trigonometric, logarithmic and

exponential functions. It also explains
application of differentiation in
calculating the rates of change in

addition solving maximum and minimum
problems. This topic also helps to solve

paramateric equation, implicit
differentiation, partial differentiation

and total of differentation.

ii

DIFFERENTIATION

SUBTOPICS

Basic Rules - Techniques of Trigonometric,
Constant, Power, Differentiation - Logarithmic and
Product, Quotient
Sum and and Chain Rule Exponential
Difference Rule Functions

Second Order Gradient of the Maximum,
Differentiation Curve, Stationary Minimum and
Point of Inflexion
Points

Rate of Change Parametric Implicit
Equation Differentiation

Partial Techniques of
Differentation Total

Differentiation

iii

FORMULA FORMULA

BASIC DERIVATIVES
(where u is a function of x)

d k   0; k = constant

dx

 d xn  nxn1

dx

d ln x   1

dx x

 d ex  e x

dx

d cos u = – sin u du

dx dx

d sin u= cos u du

dx dx

d tan u= sec2 u du

dx dx

d cotu = – cosec2 u du

dx dx

d sec u= sec u tan u du

dx dx

d cosec u= – cosec u cot u du

dx dx

iv

FORMULA

DIFFERENTIATION OF PRODUCT RULE


( ) = +
DIFFERENTIATION OF QUOTIENT RULE

= −
( )



DIFFERENTIATION OF PARAMETRIC EQUATION


= ×
DIFFERENTIATION OF INVERSE FUNCTIONS

(where u is a function of x)

 d sin1 u = 1 du , u  1

dx 1 u2 dx

 d cos1 u =  1 du , u  1

dx 1 u2 dx

 d 1 du
1 u2 dx
dx
tan1 u =

 d = 1 du
1 u2 dx
dx
cot1 u

 d sec1 u = 1 du , u  1

dx u u2 1 dx

 d cosec1u =  1 du , u  1
dx u u2 1 dx

v

Describe Rules of Differentiation

Explanation: The process of finding a derivative.

If given = ( ) , then the first derivative by with respect to x is or ′( ) or ′.



IF THEN





( ) ′( )

FORMULA 2.1 Rules Of
Differentiation
2.1.1 Basic rules
2.1.2 Techniques
Constant rule
Power rule Product rule
Quotient rule
Constant multiple rule
Sum and difference rule Chain rule

1

FORMULA Identify basic rules of differentiation

DIFFERENTIATION IF THEN
RULES

Constant Rule =
where is constant = 0

= = −1
=

Constant Multiple
Rule

= −1


= ( ) + ( ) = ′ ( ) + ′ ( )

Sum rule

= ( ) − ( ) = ′ ( ) − ′ ( )

Difference rule

= ( + ) = ( + ) −1 . ( + )

Power rule

2

Steps To Differentiate Constant And Simple Function

Step Explanation Example

Step 1 Identify the varible terms Find the variable and constant terms
and constant terms in in this function:
equation
= 5 3 + 9 2 − 7 + 3
A variable term is any term that includes a
variable and a constant term is any term that  The variable terms are
has only a number without a variable. 5 3, 9 2, − 7

 The constant term is 3

Step 2 Multiply the coefficients Here's how you do it:
of each variable term by  5 3 = 5 × 3 = 15
their respective  9 2 = 9 × 2 = 18
exponents.  −7 = −7 × 1 = −7

Their products will form the new coefficients of
the differentiated equation. Once you find their
products, place the results in front of their
respective variables.

Step 3 Lower each exponent by Here's how you do it:
one degree.  5 3 = 5 2
 9 2 = 9 1
To do this, simply subtract 1 from each  −7 = −7
exponent in each variable term.

Step 4 Replace the old The derivative of constants is zero so
coefficients and old you can omit 3, the constant term,
exponents with their new from the final result.
counterparts.
 5 3 becomes 15 2
To finish differentiating the equation, simply  9 2 becomes 18x
replace the old coefficients with their new
coefficients and replace the old exponents with  −7 becomes−7
their values lowered by one degree.
 The derivative of the

function
= 5 3 + 9 2 − 7 + 3

is = 15 2 + 18 − 7

3

Differentiate the following functions:

Example: 1 Example: 2

= =


= 0 = 1

Example: 3 Example: 4

= = −

= 4 3 = −4 −5


Example: 5 Example: 6

= = − −

= 15 2 = 20 −6


4

Example: 7 Example: 8

= − = − +

= 6 − 6 2 = −5 −6 + 25 4


= −5 + 25 4
6

Example: 9 Example: 10

= ( − )
= ( + )

= 5(4 − 4)5(4)


= 20(4 − 4)5 = 4(7 + 3)−4

= (−4)(4)(7 + 3)−4(7)

= −112(7 + 3)−4

= −112
(7 + 3)4

5

EXERCISE 1

Differentiate the following functions:

a) = 4 b) = 7

Ans: 4 3 Ans: 0

c) = 3 2 d) = 5

Ans: 6

Ans: 5 4

e) = 7 f) ( ) = 5 6
3 4
3

Ans: − 28 −5 Ans: 10 5

3

6

g) = 4 + 4 2 h) = 15 + 2 4 + 4 −3 + + 6

Ans: 4 3 + 8 Ans: 15 14 + 8 3 − 12 −4 + 1

i) = 3 + 2 5 + 7 j) = 7 + 1 7
2 3 5 4
14

Ans: − 9 −4 + 2 4 Ans: 1 6
2 2

k) = (2 + 3)2 l) = (1 + 2)2

Ans: 8 + 12 Ans: 4 + 4 3
7

m) = 5 4 − 2 n) = 2 6+4 5+3



Ans : 20 3 − 2 Ans : 10 4 + 16 3

o) ( ) = 3 − 1 5 − 7 p) ( ) = 15 − 3 4 − 1 − 5 − 15
4 5 4 2 3

Ans : −12 −5 − 4 − 7 Ans : 15 14 − 6 3 + 3 −4 − 5
4 r) = (1 − 2)2

q) = ( − 3)2

Ans : 2 − 6 Ans : −4 + 4 3
8

FORMULA Describe techniques of differentiation
(Product and Quotient Rule)

PRODUCT RULE

To differentiate a function in the form
=

You should use the formula

= +

QUOTIENT RULE

To differentiate a function in the form


=

You should use the formula

= −


2

DIFFERENTIATION RULES IF THEN
Product rule =
where = ( )and = ( ) = +

= −
= 2
Quotient rule where = ( )and = ( )

9

Steps To Differentiate Using Product Rule

Step Explanation Example

Step 1 Learn the Product Rule. The function:
It is as follows; = ( + 2)( 3)

If
=

then


= +

where and are function of

Step 2 Work out all the terms you Here's how you do it:
will need. = ( + 2) ,

It is useful to write them down in a list = ( 3)


= 1

= 3 2


Step 3 Substitute these values of Here's how you do it:

, , and back into the


original formula : = ( + 2)(3 2) + ( 3)(1 )



= +

Step 4 Simplify your answer Here's how you do it:
where possible.
= 3 3 + 6 2 + 3
10

= 4 3 + 6 2


Differentiate the following functions:

Example: 1 Example: 2

= ( + ) ( − ) = ( + )( − )

= ( + 2)3 = ( − 5) = ( − 3)5

= 1 = (3 + 2)

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

= (3) = 5( −


= + = +
= ( + 2)3(1) + ( − 5)3( + 2)2 = (3 + 2) 5( − 3)4 + ( − 3)5(3)
= ( + 2)3 + 3( + 2)2( − 5) = ( − 3)4[5(3 + 2) + 3( − 3)]
= ( + 2)2[( + 2) + 3( − 5)] = ( − 3)4[15 + 10 + 3 − 9]
= ( + 2)2[4 − 13] = ( − 3)4[18 + 1]

11

Steps To Differentiate Using Quotient Rule

Step Explanation Example

Step 1 Learn the Quotient Rule. The function:
3
It is as follows;
= ( + 4)2

If


=
then

= −
2

where and are function of

Step 2 Work out all the terms Here's how you do it:
you will need. = 3

It is useful to write them down in a list = ( + 4)2

= 3 2



= 2 ( + 4)

Step 3 Substitute these values of Here's how you do it:

, , and back into


the original formula : (( + 4)2)(3 2) − ( 3)(2 ( + 4) )
= (( + 4)2)2
−
= 2

Step 4 Simplify your answer Here's how you do it:
where possible.
2( + 4)[3( + 4) − 2 ]
=
( + 4)4

2( + 4)[ + 4]
= ( + 4)4

12

Differentiate the following functions:

Example: 1 Example: 2

( − ) +
= = −

= (3 2 − 3 )2 = 4 = 2 2 + 3 = 3 − 2

= 2(3 2 − 3 )(6 − 3) = 4 = 4 = 3 2 − 2


= (12 − 6)(3 2 − 3 )

= − −

=
2 2

(4 )((12 − 6)(3 2 − 3 )) − ((3 2 − 3 )2)(4) ( 3 − 2 )(4 ) − (2 2 + 3)(3 2 − 2)
= (4 )2 = ( 3 − 2 )2

(3 2 − 3 )[(4 )(12 − 6) − (3 2 − 3 )(4)] 4 4 − 8 2 − 6 4 + 4 2 − 9 2 + 6
= 16 2 = ( 3 − 2 )2

(3 2 − 3 )[28 2 − 24 − 12 2 + 12 ] −2 4 − 13 2 + 6
= 16 2 = ( 3 − 2 )2

(3 2 − 3 )[16 2 + 12 ]
= 16 2

13

EXERCISE 2

Solve the following functions using product or quotient
rules:

a) = ( 2 + 2)(3 − 5)

Ans : 9 2 − 10 + 6

b) = (2 − 4)(3 + )

Ans: 4 + 2

c) = (4 3 + 3)3(3 − 2)

Ans : 3(4 3 + 3)3[40 3 − 24 2 + 3]
14

d) = ( 2 + 2)3(4 − 3 2)2

Ans : ( 2 + 2)2(4 − 3 2)(−30 3)

e) = 4+5
2−2

Ans : 2 5−8 3−10
( 2−2)2

f) = 9−

2 +3

Ans : −21
(2 +3)2

15

g) = ( +1)(2 2−3)

+2

Ans : 4 3+14 2+12 −3
( +2)2

h) = (3 +1)8
3 3

Ans: (3 +1)7(45 3−9 2)
(3 3)2

i) = 3
(5+2 4)3

Ans : (5+2 4)2(15 2−18 6)
(5+2 4)6

16

2.1.2 Describe techniques of differentiation
(Chain rule)

FORMULA CHAIN RULE

To differentiate two function f (x) and g (x)
If = ( ) and = ( )

You should use the formula

= ×

DIFFERENTIATION RULES IF THEN
Chain Rule =
where = ( ) = ×

17

Steps To Differentiate Using Chain Rule

Step Explanation Example

Step 1 Learn the Quotient Rule. The function:
It is as follows; = (2 4 − 1)3

If
=

then


= ×

where is function of

Step 2 Work out all the terms you Here's how you do it:
Step 3 will need. = 3

It is useful to write them down in a list = 2 4 − 1

= 3 2


= 8 3


Substitute these values of Here's how you do it:

, , and back into the


original formula : = (3 2) × (8 3)



= ×

Step 4 Simplify your answer Here's how you do it:
where possible.Then,
r e p l a c e w i t h t h e f u n c t i o n = 24 3 2
o f .

= 24 3(2 4 − 1)2


18

Differentiate the following functions:

Example: 1 Example: 2

= ( + ) = ( − )

= 5 = 2 + 2 = 4 3 = − 3

= 5 4 = 2 = 12 2 = 1



= × = ×

= 5 4 × 2 = 12 2 × 1

= 10 4 = 12 2
= 12( − 3)2
= 10 ( 2 + 2)4

Example: 3 Example: 4

= ( − )−
= ( − )

= −4 = 3 − 11 = 2 −6 = 2 − 5 2

= −4 −5 = 3 = −12 −7 = −10



= × = ×

= −4 −5 × 3

= −12 −5 = −12 −7 × −10

= −12(3 − 11)−5 = 120 −7
= 120 (2 − 5 2)−7

19

EXERCISE 3

Solve the following functions using chain rule:
a) = (4 + 2 3)5

Ans : 30 2(4 + 2 3)4

b) = 3(2 5 + )3

Ans : 9(10 4 + 1)(2 5 + )2
20

c) = 5
(3 2−4)3

Ans : −90 (3 2 − 4)−4

d) = (4 − 3 2)−7

Ans : 42 (4 − 3 2)−8
21

e) = 10
(5+2 4)6

Ans : −480 3(5 + 2 4)−7

f) = 1
√(3 4+2)

Ans : −6 3(3 4 + 2)−23
22

Use Trigonometric, Logarithmic and
Exponential Function

FORMULA Derivatives of Trigonometric Function

TRIGONOMETRIC FUNCTION DIFFERENTIATION
=
= cos

=
= − sin

= = sec2


=
= sec tan

=
= −cosec cot

= = −cosec2


=
== (cos ) ( )

= (cos )( )
= cos

=
= (− sin ) ( )

= (− sin )( )
= − sin

23

TRIGONOMETRIC FUNCTION DIFFERENTIATION
=
= (sec2 ) ( )

= (sec2 ) ( )

= sec2

= ( + )
= cos ( + ) ( + )

= cos ( + ) ( )
= cos ( + )

= ( + )
= −sin ( + ) ( + )

= − sin( + ) ( )
= − sin( + )

= ( + ) = sec2( + ) ( + )

= sec2( + ) ( )

= sec2( + )

= = sin −1 ( )
(sin )
= sin −1 (cos )(1)

= sin −1 cos

= = cos −1
(cos ) ( )
= cos −1 (−sin )(1)

= − cos −1 sin

= = tan −1
(tan ) ( )
= tan −1 (sec2 )(1)

= tan −1 sec2

24

Steps To Differentiate Trigonometric Function

Step Explanation Example

Step 1 Set up the equation: The function:
It is as follows; = ( + )
= ( + )

S t e p 2 I f + i s a d i f f e r e n t i a b l e Here's how you do it:
function then,
= ( + ) ( )

= cos( + ) . ( + )

Which is the derivative of sin ( + ) is The derivative of Times the
cos ( + ) times the derivative of ( + ) ( + )
derivatives of

( + )

Step 3 Rewritten your answer Here's how you do it:
where possible.

= 8 cos (4 2 + 3)

25

Differentiate each if the following functions

Example: 1 Example: 2

= = ( + )

= co s(2 2 + 1) (2 2 + 1)
= (−sin 8 ) (8 )
= cos (2 2 + 1) (4 )
= (− sin 8 )(8)
= −8 sin 8 = 4 cos (2 2 + 1)

Example: 3 Example: 4

= ( − ) =

= 3 sec2(2 − 2) (2 − 2) 5−1
(sin ) ( )
= 3 sec2(2 − 2) (2) = 5 sin

= 6 sec2(2 − 2) = 5 sin 5−1 (cos )(1)

= 5 sin 4 cos

26

EXERCISE 4

Solve the following trigonometric functions using suitable
method:

a) = sin(2 − 2)

Ans : 2 cos(2 − 2)

b) = cos 7

Ans : −7 sin 7
27

c) = sin 2 + sin2

Ans : 2 cos 2 + 2 sin cos

d) = tan6

Ans: 6 tan5 sec2
28

e) 3 sin − 4 cos

Ans: 3 cos + 4 sin

f) 3 3 + tan (6 − 3)

Ans: 9 2 + 6 sec2(6 − 3)
29

FORMULA Derivatives of Logarithmic Function

LOGARITHMIC FUNCTION
= (ln )
=
= ( + ) 1
=
=

= ln( + ) . ( + )

= 1 ) . ( )
( +


= ( + )

Apply the chain rule

1
= .

DO YOU REMEMBER? (LAW OF LOGARITHMIC)

= ( + ) ( + )

= +

= −



30

Steps To Differentiate Logarithmic Functions

Step Explanation Example

Step 1 Set up the equation: The function:
It is as follows; = ( + )

= ( + )

S t e p 2 I f + i s a d i f f e r e n t i a b l e Here's how you do it:
function then, 1
= (2 + 3) × 2
= 1 ) . ( + )
( +

One over Times the
logarithmic derivatives
function of function

Step 3 Rewritten your answer Here's how you do it:
where possible. 2
= (2 + 3)

31

Differentiate each if the following functions

Example: 1 Example: 2

= ln 8 x = ln ( 2 x + 6 )

STEP 1 STEP 2
= ln (2 + 6). (2 + 6)

= ln (8 ) . (8 ) 1
= 2 + 6 . 2
1
= 8 . 8 2
= 2 + 6
=8
2
8 = 2( + 3)

= 1 1
= ( + 3)

Example: 3 Example: 4

= ln (5 x + 7) = ( + )



=2 (5 + 7) . (5 + 7) 3
= 2 (4 + 5)

1 3 . 1 5) (4 + 5)
= 2 . (5 + 7) . 5 = 2 (4 +

10 12
= (5 + 7) = 2(4 + 5)

6
= (4 + 5)

32

Example: 5 Example: 6

= ( 2 + 2 + 1) = ( − )

= ( 2 + 1 + 1) ( 2 + 2 + 1) 1 (8 − 3)
= 7 −
2 3)
(8
2 +2
= 7 (8−3 2)
( 2+2 +1)
= (8 − 3)
= 2( +1) 56 − 21 2
( +1)( +1)
= (8 − 3)
= 2
( +1)

Example: 7 Example: 8

= ( + )
=
= 3 (6 + 5)
1 = 2 − 7
= 3 (6 + 5) (6 + 5) = 2 . − 7
= 2 . 1 − 0
18
= (6 + 5)

2
=

Example: 9 Example: 10

= √ + ( + )
= ( − )

1 = (2 + ) − (2 − )

= (5 + 3)2

1 = 1 . (1) − 1 ) . (−1)
= 2 (5 + 3) (2 + ) (2 −

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


= 5 = (2− )+(2+ )
2 (5 +3) (2+ )(2− )

= 4
4− 2

33

EXERCISE 5 Ans : 1
F i n d f o r t h e f u n c t i o n b e l o w .


Ans : −2
a) = ln 9
1−2x
b) = ln ( 1 − 2x )
Ans : 27
c) = 9 ln (3 x + 2)
3x+2
34

d) = 5 ln (9 x +4) Ans : 15

3 9x+4

e ) = ln(2 2 + 5)3 Ans : 12
f ) = 9 + ln 2 (2 2+5)

Ans : 9 − 1

35

g ) = ln 9 Ans : 3

(4−3 ) 4−3

h ) = ln

sin

Ans : sin − cos

sin

36

i ) = ln √1 − 3

Ans : −3

2(1−3 )

j) = ln 3
(5−3 )4

Ans: 12

5−3

37

FORMULA Derivatives of Exponential Function

EXPONENTIAL FUNCTION DIFFERENTIATION

= = ( x)
= +

= =

= ( + ) .
( + )

= + .

= +

Apply the chain rule

= .


DO YOU REMEMBER? (LAW OF EXPONENT)

× +
−
×


( )



−


38

Steps To Differentiate Exponential Functions

Step Explanation Example

Step 1 Set up the equation: The function:
It is as follows; = (2 −4)

= +

Step 2 I f + i s a d i f f e r e n t i a b l e Here's how you do it:
function then,

= ( + ) . = (2 −4) ×2
( + )

Which is the derivative of to the ( + ) is The exact Times the
an exact copy of to the ( + ) times the copy of derivatives
devivative of ( + ) the of exponent
exponential
function

Step 3 Rewritten your answer Here's how you do it:
where possible.

= 2 (2 −4)

39

Differentiate each if the following functions

Example: 1 Example: 2
= + = − −

STEP 1

STEP 2 ( 5−7 ) (5 − 7 )
. .
= ( + ) (9 + 2) = −2

= + . 9 = −2 − . (−7)

= 9 + = 14 5−7

Example: 3 Example: 4
= ( + )
= + +



= 5 9 +4 +x = +2 + + 4
3 = 3 + 5

= 5 ( 9 +4) . (9 + 4) + 1
3

= 5 . 9 9 +4 +1 = ( 3 ). (3 ) + ( 5 ). (5 )

3

= 15 9 +4 +1 = 3 3 + 5 5

40

Example: 5

= +


= 4 −3 + 5 − 3
= + 2

( ) . ( ) + ( 2 ) (2 )
= .

= − 2 2

Example: 6

= ( − ) + − −



= 7 ( 3− 2). (3 − 2) + −8 (−8 ) − 9

( 5 ) . 5

= 7 3− 2 (−2 ) + −8 (−8 ) − 9
5
5

−8

= −14 3− 2 − 8 5 − 9

5

41

EXERCISE 6 Ans: 5 5 +1
F i n d f o r t h e f u n c t i o n b e l o w .



a) = 5 +1

b) = 2 3 + 8 9−2

Ans: 6 3 − 16 9−2

c)

= 2 7 − 4



Ans: 2 7 −

74

42

d) = 3 2 +5 +

8

Ans: 3 2 +5 + 1

4

e) = 2 ( 3 − 5 )

Ans: 5 5 − 7 7
43