MAT421 : Chapter 2 (Differentiation)

Chapter 2 : Differentiation

CHAPTER 2 : DIFFERENTIATION

List of Topics
1.0 The First Principle
2.0 Rules of Differentiation

2.1 Basic Rule
2.1.1 Constant Rule
2.1.2 Power Rule
2.1.3 Sum and difference Rules

2.2 Generalized Power Rule
2.3 Chain Rule
2.4 Product Rule
2.5 Quotient Rule
2.6 Trigonometric Rules
2.7 Exponential Rule
2.8 Logarithmic Rule
3.0 Implicit Differentiation
4.0 Higher Order Differentiation
5.0 Linear Approximations and Differentials

1.0 The Definition of Derivative (The First Principle)

f' (x) = lim f (x + h) − f (x)
h→0 h

Simple Examples
Find f’(x) using the definition of derivative
a) f(x) = 2x
b) f(x) = x2

c) f(x) = x

47

MAT183/ MAT421 : Calculus I

) ( ) = 2 ) ( ) = √
( + ℎ) = 2( + ℎ) = 2 + 2ℎ ( + ℎ) = √ + ℎ

Using the first principle (or the definition of derivative) Using the first principle (or the definition of derivative)

′( ) = ( + ℎ) − ( ) ′( ) = ( + ℎ) − ( )
ℎ
ℎ→0 ℎ→0 ℎ

= 2 + 2ℎ − 2 √ + ℎ − √
ℎ
ℎ→0 ℎ =

ℎ→0

= 2ℎ √ + ℎ − √ √ + ℎ + √
ℎ √ + ℎ + √
ℎ→0 ℎ = ∙

= 2 ℎ→0

ℎ→0 = (√ + ℎ)2 − (√ )2
ℎ→0 ℎ(√ + ℎ + √ )
=2 #

+ ℎ −
=

ℎ→0 ℎ(√ + ℎ + √ )

) ( ) = 2 ℎ
( + ℎ) = ( + ℎ)2 = 2 + 2 ℎ + ℎ2 =

ℎ→0 ℎ(√ + ℎ + √ )

Using the first principle (or the definition of derivative) 1
=
′( ) = ( + ℎ) − ( )
ℎ→0 √ + ℎ + √
ℎ→0 ℎ 1

= 2 + 2 ℎ + ℎ2 − 2 =
√ + 0 + √
ℎ→0 ℎ 1

= 2 ℎ + ℎ2 =
√ + √
ℎ→0 1

=#
2√

ℎ

= ℎ(2 + ℎ)

ℎ→0 ℎ

= (2 + ℎ)

ℎ→0

= 2 + 0

= 2 #

48

Chapter 2 : Differentiation

More Examples (From Previous semester papers)

Example 1/ MAR 2004/ MAT183/Q4a (ii)/ 5 marks

Use the definition of derivatives to find f' (x) for the function f(x) = x − 2x .

Solution :
) ( ) = √ − 2
( + ℎ) = √ + ℎ − 2( + ℎ) = √ + ℎ − 2 − 2ℎ

Using the first principle (or the definition of derivative)

′( ) = ( + ℎ) − ( )

ℎ→0 ℎ

= (√ + ℎ − 2 − 2ℎ) − (√ − 2 )
ℎ
ℎ→0

= √ + ℎ − 2ℎ − √ Split using theorem of limits :
ℎ
ℎ→0 [ ( ) ± ( )] = ( ) ± ( )
→ → →

= √ + ℎ − √ − 2ℎ Note : separate the term containing surd, and
ℎ the term not containing surd
ℎ→0 ℎ→0 ℎ

= √ + ℎ − √ ∙ √ + ℎ + √ − 2
ℎ √ + ℎ + √
ℎ→0 ℎ→0

= (√ + ℎ)2 − (√ )2 −2

ℎ→0 ℎ(√ + ℎ + √ )

= + ℎ − −2

ℎ→0 ℎ(√ + ℎ + √ )

= ℎ −2

ℎ→0 ℎ(√ + ℎ + √ )

= 1 −2

ℎ→0 √ + ℎ + √

1
= −2

√ + 0 + √

1 + = 2
= −2 √ + √ = 2√

√ + √
1
= −2#
2√

49

MAT183/ MAT421 : Calculus I

Example 2/ OCT 2004/ MAT183/Q2c/ 7 marks

Find the derivative of f (x) = 1 by using the definition of derivative.

2 x +1

Solution :

) ( ) = 1 Question :
2√ + 1
How to solve limits, if the function is given in
( + ℎ) = 1 fractions?

2√ + ℎ + 1

Using the first principle (or the definition of derivative)

′( ) = ( + ℎ) − ( )
ℎ
ℎ→0

′( ) = 1 ∙ [ ( + ℎ) − ( )] Answer :

ℎ→0 ℎ In that case, you better write the formula ‘per h’ as

11 − 1 multiply 1 so that it will not appear so messy like this,
= ∙ [ ]
ℎ 2√ + ℎ +1 2√ + 1 ℎ
ℎ→0 1 −1

= 1 ∙ [(2√ + 1) − (2√ + ℎ + 1)] ′( ) = [2√ + ℎ +1 2√ + 1]
ℎ (2√ +ℎ + 1)(2√ + 1) ℎ
ℎ→0 ℎ→0

= 1 ∙ [2√ + 1 − 2√ + ℎ − 1 ]
ℎ (2√ + ℎ + 1)(2√ + 1)
ℎ→0

= 1 ∙ [ 2√ − 2√ + ℎ + ]
ℎ (2√ + ℎ + 1)(2√ 1)
ℎ→0

= 2 1 ∙ [ √ − √ + ℎ + ]
ℎ (2√ +ℎ + 1)(2√ 1)
ℎ→0

= 2 1 ∙ [ √ − √ + ℎ + ] ∙ √ + ℎ + √
ℎ (2√ +ℎ + 1)(2√ 1) √ + ℎ + √
ℎ→0

= 2 1 ∙ [ (√ )2 − (√ + ℎ)2 ] Simplify using ( + )( − ) = −
(2√ No need to expand!
ℎ→0 ℎ + ℎ + 1)(2√ + 1)(√ + ℎ + √ )

1 − ( + ℎ)
= 2 ∙ [ ]
ℎ (2√ + ℎ + 1)(2√ + 1)(√ + ℎ + √ )
ℎ→0

1 −ℎ
= 2 ∙ [ ]
ℎ (2√ + ℎ + 1)(2√ + 1)(√ + ℎ + √ )
ℎ→0

= 2 [ −1 ]

ℎ→0 (2√ + ℎ + 1)(2√ + 1)(√ + ℎ + √ )

= 2[ −1 ]

(2√ + 0 + 1)(2√ + 1)(√ + 0 + √ )

= 2 [ −1 ] + = 2
(2√ + 1)(2√ + 1)(√ + √ ) √ + √ = 2√

−1 −1 −1
= 2( 1)2(2√ )) = 2 [ + 1)2] = #
(2√ + 2√ (2√ √ (2√ + 1)2

50

Chapter 2 : Differentiation

Example 3/ page 13/ MAR 2005/ MAT183/Q3c/ 6 marks

Find the derivative of f (x) = x + 1 from the first principle.

Solution :

) ( ) = √ + 1
( + ℎ) = √ + ℎ + 1

Using the first principle (or the definition of derivative)

′( ) = ( + ℎ) − ( )

ℎ→0 ℎ

= √ + ℎ + 1 − √ + 1

ℎ→0 ℎ

= √ + ℎ + 1 − √ + 1 ∙ √ + ℎ + 1 + √ + 1 Multiplying with their conjugate, and simplify
ℎ √ + ℎ + 1 + √ + 1 using :
ℎ→0
( + )( − ) = −
(√ + ℎ + 1)2 − (√ + 1)2
=

ℎ→0 ℎ(√ + ℎ + 1 + √ + 1)

( + ℎ + 1) − ( + 1)
=

ℎ→0 ℎ(√ + ℎ + 1 + √ + 1)

+ ℎ + 1 − − 1
=

ℎ→0 ℎ(√ + ℎ + 1 + √ + 1)

ℎ
=

ℎ→0 ℎ(√ + ℎ + 1 + √ + 1)

1
=

ℎ→0 √ + ℎ + 1 + √ + 1
1

=
√ + 0 + 1 + √ + 1
1

=
√ + 1 + √ + 1
1

=#
2√ + 1

51

MAT183/ MAT421 : Calculus I

Example 4/ page 18/ NOV 2005/ MAT183/Q2a/ 5 marks

Use the first principle of differentiation to find f' (x) for the function f (x) = x 2 − 2 .

x
Solution :

) ( ) = 2 − 2


( + ℎ) = ( + ℎ)2 − 2 ℎ = 2 + 2 ℎ + ℎ2 − 2 ℎ
+ +

Using the first principle (or the definition of derivative)

′( ) = ( + ℎ) − ( )

ℎ→0 ℎ

′( ) = 1 ∙ [ ( + ℎ) − ( )]

ℎ→0 ℎ

= 1 ∙ [( 2 + 2 ℎ + ℎ2 − 2 ℎ) − ( 2 − 2
ℎ + )]
ℎ→0

= 1 ∙ [ 2 + 2 ℎ + ℎ2 − 2 ℎ − 2 + 2
ℎ + ]
ℎ→0

= 1 ∙ [2 ℎ + ℎ2 − 2 ℎ + 2
ℎ + ]
ℎ→0

= 1 ∙ [2 ℎ + ℎ2 + 2 − 2 ℎ]
ℎ +
ℎ→0

1 ∙ [2 ℎ + ℎ2] 12 2
= ℎ + ∙ [ − ℎ]
ℎ→0 ℎ +
ℎ→0

1 1 2( + ℎ) − 2
= ∙ ℎ(2 + ℎ) + ∙ [ ]
ℎ ℎ ( + ℎ)
ℎ→0 ℎ→0

= (2 + ℎ) + 1 2 + 2ℎ − 2 ]
∙ [
ℎ→0 ℎ ( + ℎ)
ℎ→0

= (2 + ℎ) + 1 2ℎ
∙ [ ( ℎ)]
ℎ→0 ℎ +
ℎ→0

= (2 + ℎ) + ( ( 2 ℎ))

ℎ→0 ℎ→0 +

2
= (2 + 0) + ( ( + 0))

2
= 2 + 2 #

52

Chapter 2 : Differentiation

Example 5/ page 36/ APR 2007/ MAT183/Q2a/ 5 marks

Use the first principle of differentiation to find f' (x) for the function f (x) = 3 + x .

ans : f’(x) = 1
2√3+

Solution :

) ( ) = √3 +
( + ℎ) = √3 + + ℎ

Using the first principle (or the definition of derivative)

′( ) = ( + ℎ) − ( )

ℎ→0 ℎ

= √3 + + ℎ − √3 +

ℎ→0 ℎ

= √3 + + ℎ − √3 + ∙ √3 + + ℎ + √3 + Multiplying with their conjugate, and simplify
ℎ √3 + + ℎ + √3 + using :
ℎ→0
( + )( − ) = −
(√3 + + ℎ)2 − (√3 + )2
=

ℎ→0 ℎ(√3 + + ℎ + √3 + )

(3 + + ℎ) − (3 + )
=

ℎ→0 ℎ(√3 + + ℎ + √3 + )

3 + + ℎ − 3 −
=

ℎ→0 ℎ(√3 + + ℎ + √3 + )

ℎ
=

ℎ→0 ℎ(√3 + + ℎ + √3 + )

1
=

ℎ→0 (√3 + + ℎ + √3 + )

1
=

√3 + + 0 + √3 +
1

=
√3 + + √3 +
1

=#
2√3 +

53

MAT183/ MAT421 : Calculus I

Example 6/ page 54/ OCT 2008/ MAT183/Q2a/ 5 marks

Use the first principle of differentiation to find f' (x) for the function f (x) = 2x .

3−x

Ans : 6
(3− )2

Solution :

) ( ) = 2 Question :
3 −
How to solve limits, if the function is given in
( + ℎ) = 2( + ℎ) = 2 + 2ℎ fractions?
3 − ( + ℎ) 3− − ℎ

Using the first principle (or the definition of derivative)

′( ) = ( + ℎ) − ( )
ℎ
ℎ→0

′( ) = 1 ∙ [ ( + ℎ) − ( )]

ℎ→0 ℎ Answer :

1 2 + 2ℎ 2 In that case, you better write the formula ‘per h’ as

= ℎ ∙ [3 − −ℎ − 3 − ] multiply 1 so that it will not appear so messy like this,
ℎ
ℎ→0

1 −1
1 (2 + 2ℎ)(3 − ) − 2 (3 − − ℎ) ′( ) = [2√ + ℎ +1 2√ + 1]
= ∙ [ ] ℎ
ℎ (3 − − ℎ)(3 − ) ℎ→0
ℎ→0

1 6 − 2 2 + 6ℎ − 2ℎ − 6 + 2 2 + 2ℎ
= ∙ [ ]
ℎ (3 − − ℎ)(3 − )
ℎ→0

1 6ℎ )]
= ∙ [(3
ℎ − − ℎ)(3 −
ℎ→0

= [(3 6 )]

ℎ→0 − − ℎ)(3 −

6
= (3 − − 0)(3 − )

6
= (3 − )(3 − )

6
= (3 − )2 #

54

Chapter 2 : Differentiation

2.0 Rules of Differentiation

2.1 Basic Rule

2.1.1 Constant Rule

If y = a, where a is a constant
Then dy = 0

dx
Or

d (a) = 0 , where a is a constant

dx

2.1.2 Power Rule

If y = axn, where a is a constant
Then dy = nax n−1

dx
Or

( )d ax n = nax n−1, where a is a constant

dx

2.1.3 Sum and Difference Rules

If y = f(x)  g(x)

Then dy = f' (x)  g' (x)

dx

55

MAT183/ MAT421 : Calculus I

Examples

Differentiate with respect to x.

a) y = 2x 3 − x + 3 + 4
x

b) f (x) = 1 + 3x − 2 + 5 x 2

x4

c) g(x) = 2 − 4x2 + 1 +
3x 3 3x

d) h(x) = 5x 4 + 2x − 1 + e

3x

Solution :

) = 2 3 3 ) ( ) = 1 + 3 − √2 + 5√ 2
− √ + √ + 4 4

= 2 3 − 1 + 3 −12 + 4 ( ) = −4 + 3 − √2 + 2
2
5

differentiate differentiate

= 6 2 − 1 −21 − 3 −32 + 0 ′( ) = −4 −5 + 3 − 0 + 2 52−1
2 2 5

= 6 2 − 1 − 3 −23 # ′( ) = −4 −5 + 3 + 2 −35 #
2√ 2 5

#

Before differentiate,

 move up variable x from denominator

using the properties of indices. 1 = −


 change the surd (@ radical) function

5√ 2 to exponent (power) 2 so that we

5

can differentiate using power rule

 no need to change surd constant √2
to 221, because differentiate any

constant will become zero.

56

Chapter 2 : Differentiation

) ( ) = 2 − 4 2 + 1 + ) ℎ( ) = 5 4 + √2 − 1 +
3 3 3√ √3

( ) = 2 −3 − 4 2 + −31 + ℎ( ) = 5 4 + √2√ − 1 +
3 √3√

ℎ( ) = 5 4 11
+ √2 2 − +
√3 1
differentiate 2

2 1 −34 ℎ( ) = 5 4 + √2 1 − 1 −12 +
3 3 2
′( ) = ∙ −3 −4 − 8 − + 0 √3

′( ) = −2 −4 − 8 − 1 −43
3
differentiate
2 1
′( ) = − 4 − 8 − 3 −34 # ℎ′( ) = 20 3 1 −12 1 1 −21−1
2 − 2
+ √2 ∙ ∙ − +0
√3

ℎ′( ) = 20 3 11 + 1 −32
+ √2 ∙ ∙ 2√3
2 1
2

ℎ′( ) = 20 3 + 11 + 1 −32
√2 ∙ 2 ∙ √ 2√3

ℎ′( ) = 20 3 + √2 ∙ 11 + 1 −23
∙ 2√3
2 √

(√2)

ℎ′( ) = 20 3 + 11 + 1 −23
∙
√2 √ 2√3

ℎ′( ) = 20 3 + 1 + 1 −32 #
√2 2√3

57

MAT183/ MAT421 : Calculus I

2.2 Generalized Power Rule

If = ( )

= • ( ) −1 • ′


Examples

Use Generalized Power Rule to find the derivative of
) = ( 3 − 2)6
1
) =
2√ + 1
) = √ + 1

Solution : 1
) = ( 3 − 2)6 ) =

differentiate 2√ + 1
= (2√ + 1)−1

= 6( 3 − 2)5 ∙ (3 2 − 0) differentiate


= 6( 3 − 2)5 ∙ 3 2 = −(2√ + 1)−2 ∙ (2 ∙ 1 + 0)
2√

= 18 2( 3 − 2)5 # 1 1
= − (2√ + 1)2 ∙ √

) = √ + 1 1 #
= ( + 1)21 = −
√ + 1)2
(2√

differentiate

= 1 ( + 1)−21 ∙ (1 + 0)
2

1 1
= 2 ∙ √ + 1

1
= 2√ + 1 #

58

Chapter 2 : Differentiation

2.3 Chain Rule


= •

= • •

= • • •

Examples

Use Chain Rule to find the derivative of
) = ( 3 − 2)6
1
) =
2√ + 1
) = √ + 1

Solution :

) = ( 3 − 2)6 ℎ = 3 − 2
= 6

Differentiate with respect to u = 3 2


= 6 5

Using Chain Rule,


= ∙

= 6 5 ∙ 3 2


= 18 2 5


= 18 2( 3 − 2)5 #


59

MAT183/ MAT421 : Calculus I

1 ) = √ + 1
) =
= ( + 1)21
2√ + 1
= (2√ + 1)−1 = 2√ + 1 = 21 ℎ = + 1

= −1 ℎ = 1

1

= 2 2 + 1

= 2 ∙ 1 −12 + 0
2

= −21 Differentiate with respect to u


1 = 1 −12
= √ 2

Differentiate with respect to u 1 1
= 2 ∙ √

= − −2 1
= 2√

1 Using Chain Rule,
= − 2

Using Chain Rule, = ∙
1
= 2√ ∙ 1
= ∙ 1
= 2√ + 1 #
1 1
= − 2 ∙ √

1
= − √ 2

1
= − #
√ 1)2
(2√ +

60

Chapter 2 : Differentiation

2.4 Product Rule

=

= ′ +


2.5 Quotient Rule


=

′ −
= 2

Examples

Differentiate with respect to x.

)    =   (3 − 3)(2 + 5)

)    =   (5 2 + 3)4(1 − )3

)    =    ( 3 3)2
+

)    =    2 − 4
( 2 + 2)3

Solution :

)    =   (3 − 3)(2 + 5) )    =   (5 2 + 3)4(1 − )3

   =   3 − 3 = 2 + 5    =   (5 2 + 3)4 = (1 − )3
′   =   3 − 3 2 ′ = 5 4 ′   =  4 (5 2 + 3)3 ∙ 10 ′ = 3(1 − )2 ∙ −1
′   =  40  (5 2 + 3)3 ′ = −3(1 − )2

Differentiate using product rule,

Differentiate using product rule,

= ′ + ′

= ′ + ′

= (2 + 5)(3 − 3 2) + 5 4 (3 − 3) = 40  (5 2 + 3)3(1 − )3 − 3(5 2 + 3)4(1 − )2


= 6 − 6 2 + 3 5 − 3 2 + 15 5 − 5 7 =  (5 2 + 3)3(1 − )2(40 (1 − ) − 3(5 2 + 3))


− 9 2 + 18 5 − 5 7 # = (5 2 + 3)3(1 − )2(40 − 40 2 − 15 2 − 9)
= 6

= (5 2 + 3)3(1 − )2(40 − 55 2 − 9) #


61

MAT183/ MAT421 : Calculus I

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

   =    = ( 3 + 3)2    =    2 − 4 = ( 2 + 2)3
′   =   1 ′ = 2( 3 + 3) ∙ 3 2 ′   =  2 ′ = 3( 2 + 2)2 ∙ 2
′ = 6 2( 3 + 3) ′ = 6 ( 2 + 2)2

Differentiate using quotient rule, Differentiate using quotient rule,
′ − ′ ′ − ′
= 2 = 2

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

( 3 + 3)[( 3 + 3) − 6 3] =   2  ( 2 + 2)2[( 2 + 2) − 3( 2 − 4)]
( 2 + 2)6
= ( 3 + 3)4

[( 3 + 3) − 6 3] =   2  [( 2 + 2) − 3( 2 − 4)]
= ( 3 + 3)3 ( 2 + 2)4

3 − 5 3 =   2  ( 2 +2 − 3 2 + 12)
= ( 3 + 3)3 # ( 2 + 2)4

=   2  (−2 2 + 14)
( 2 + 2)4

=   −4 3 + 28 #
( 2 + 2)4

62

Chapter 2 : Differentiation

2.6 Trigonometric Rules

( ) ( )
① = ① = ∙ ′

( ) ( )
② = − ② = − ∙ ′

③ ( ) = 2 ③ ( ) = 2 ∙ ′


( ) ( )
④ = ④ = ∙ ′

( ) ( )
⑤ = − ⑤ = − ∙ ′

⑥ ( ) = − 2 ⑥ ( ) = − 2 ∙ ′


Examples
Differentiate with respect to x.

) = 2
) = 3 2
) = √
) = 5
) = 3 4
) = 3 3

Solution : ) = 3 2 ) = √ 1

) = 2 = √ = 2
Differentiate wrt x
Differentiate wrt x Differentiate wrt x ′ = 1 −12
2
= 2 ∙ 2
= − 3 2 ∙ 6 = 2 √ ∙ 1 11
= 2 2 # 2√ ′ = 2 ∙ √

= −6 3 2 # = 2 √ # 1
2√ ′ =

2√

63

MAT183/ MAT421 : Calculus I

) = 5 ) = 3 4 ) = 3 3

= 5 5 ∙ 5 = ( 4 )3 3
= 3 ∙ 3

= 3( 4 )2 ∙ 4 ∙ 4 = 3

= 5 5 5 #

= 12 4 2 4 # = − 3 ∙ 3


= −3 3 #


for example (f), we can also differentiate in the following way,

) = 3 3 = 3 = 3

= ′ = 3 ∙ 3 ′ = − 2 3 ∙ 3

= ′ + ′ ′ = 3 3 ′ = −3 2 3


= ( 3 )(3 3 ) + ( 3 )(−3 2 3 )


= 3 3 3 − 3 3 2 3


= 3 3 ∙ 3 − 3 3 ∙ 1 3
3 2

= 3 2 3 − 3
3 3

3 2 3 − 3
= 3

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

−3 2 3
= 3

= −3 3 #


64

Chapter 2 : Differentiation

2.7 Exponential Rule

( ) = ( ) = • ′


Examples
Differentiate with respect to x.

) = 2
) = 3 2

) = √
) = 4
) = 3−2

Solution :

) = 2 ) = 3 2 ) = √ 1

= √ = 2

Differentiate wrt x Differentiate wrt x Differentiate wrt x ′ = 1 −21
2

= 2 ∙ 2 = 3 2 ∙ 6 = √ ∙ 1 11
2√ ′ = 2 ∙ √

= 2 2 # = 6 3 2 # √ 1
= 2√ # ′ =

2√

) = 4 ) = 3−2

Differentiate wrt x Differentiate wrt x

= 4 ∙ − 4 4 ∙ 4 = 3−2 ∙ (3 2 − 2)


= −4 4 4 4 # = (3 2 − 2) 3−2 #


65

MAT183/ MAT421 : Calculus I

2.8 Logarithmic Rule

( ) 1 ( ) = 1 • ′
=

Hot Tips !!
 If = ( )

Write = + (using the properties of logarithm)

 If = ( )



Write = − (using the properties of logarithm)

Examples
Differentiate with respect to x.
) = (3 )
) = (2 3)
) = ( 4 )
) = (3 3 )

2 + 1
) = ( 2 − 1)

Solution :

) = (3 ) ) = (2 3) ) = ( 4 )
Differentiate wrt x
Differentiate wrt x Differentiate wrt x
1
= 3 ∙ 3 = 1 ∙ 6 2 = 1 ∙ 2 4 ∙ 4
1 2 3 4
= #
3 = 4 4 2 4
= #

66

Chapter 2 : Differentiation

) = (3 3 ) ) = (3 3 )

using the properties of logarithms = ( ) = 3 = 3

= (3 ) + ( 3 ) Differentiate wrt x ′ = 3 ′ = 3 3 ∙ 3

Differentiate wrt x = 1 ∙ ( ′ + ′) ′ = 3 3 3

1 1
= 3 ∙ 3 + 3 ∙ 3 3 ∙ 3 1
= 3 3 ∙ (3 3 + 9 3 3 )

1 = 3 1 3 ∙ 3 3 (1 + 3 3 )
= + 3 3 #

= 1 (1 + 3 3 )


1
= + 3 3 #

2 + 1 2 + 1 = 2 + 1 = 2 − 1
) = ( 2 − 1) ) = ( 2 − 1) ′ = 2 ′ = 2

using the properties of logarithms
= ( 2 + 1) − ( 2 − 1) = ( )

Differentiate wrt x

Differentiate wrt x 1 ′ − ′
= ( ) ∙ ( 2 )
1 1 ′ − ′
= 2 + 1 ∙ 2 − 2 − 1 ∙ 2 = ( 2 )

2 2
= 2 + 1 − 2 − 1 #

Or… 2 − 1 2 ( 2 − 1) − 2 ( 2 + 1) )
= 2 + 1 (
2 ( 2 − 1) − 2 ( 2 + 1) ( 2 − 1)2
= ( 2 + 1)( 2 − 1)
2 3 − 2 − 2 3 − 2 1 2 ( 2 − 1) − 2 ( 2 + 1) )
= ( 2 + 1)( 2 − 1) = 2 + 1 (
−4 ( 2 − 1)
= ( 2 + 1)( 2 − 1) #
2 3 − 2 − 2 3 − 2
= ( 2 + 1)( 2 − 1)

−4
= ( 2 + 1)( 2 − 1) #

67

MAT183/ MAT421 : Calculus I

More Examples (From Previous Semester Paper)

Example 1/ MAR 2005/ MAT183/ Q1c (5 marks)

Find dy if y =  2 + cos 2x 
ln 
( )dx  2 + x 2 3 

Solution :
2 + cos 2

= [ (2 + 2)3 ]
Using the properties of logarithm

= (2 + cos 2 ) − (2 + 2)3
= (2 + cos 2 ) − 3 (2 + 2)
Differentiate with respect to x,

1 3
= 2 + cos 2 ∙ − 2 ∙ 2 − 2 + 2 ∙ 2

−2 2 6
= 2 + cos 2 − 2 + 2 #

Example 2/ MAR 2005/ MAT183/ Q2c (4 marks)

Given f (x) =  x 3 , find f' (x) .

1− x 
Solution :

3
( ) = (1 − )
Using generalized power rule (combine with quotient rule)

2 ′ − ′ = = 1 −
′( ) = 3 (1 − ) ( 2 )

2 (1 − ) − (−1) ′ = 1 ′ = −1
′( ) = 3 ∙ (1 − )2 ( (1 − )2 )

3 2 1 − + 1
′( ) = (1 − )2 ∙ (1 − )2

3 2 2 −
′( ) = (1 − )2 ∙ (1 − )2

′( ) = 3 2(2 − ) #
(1 − )4

68

Chapter 2 : Differentiation

Example 3/ NOV 2005/ MAT183/ Q1c (6 marks)

Differentiate the following functions with respect to x :

i) y = sin 3x
1+ cos 3x

ii) y = ln x 2 2x 
+ 3x + 1

Solution :

3 = 3 = 1 + 3
) = 1 + 3
′ = 3 3 ′ = −3 3

=

Using quotient rule,

′ − ′
= 2

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

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

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

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

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

3
= 1 + 3 #

2
) = ( 2 + 3 + 1)
Using the properties of logarithms,

= (2 ) − ( 2 + 3 + 1)

Differentiate wrt x,

1 1
= 2 ∙ 2 − 2 + 3 + 1 ∙ (2 + 3)

1 2 + 3
= − 2 + 3 + 1 #

69

MAT183/ MAT421 : Calculus I

Example 4/ OCT2006/ MAT183/ Q5a (5 marks)

Find dy if y = x 4 (sin 2x)3 .

dx
Solution :

= 4( 2 )3 = 4 = ( 2 )3
′ = 4 3 ′ = 3( 2 )2 ∙ 2 ∙ 2
= ′ = 6 2 2 2

Differentiate wrt x (using Product Rule)

= ′ + ′


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


= 4 3 3 2 + 6 4 2 2 2


= 2 3 2 2 (2 2 + 3 2 ) #


Example 5/ APR 2009/ MAT183/ Q2c (3 marks)

Find dy if y = tan 1− x  .
dx 1+ x 

Solution :

1 − = 1 − = 1 +
= (1 + ) ′ = −1 ′ = 1


= ( )

Differentiate wrt x

(using Trigo-Rule & Quotient Rule)

= 2 ∙ ′ − ′
( ) ( 2 )

= 2 1 − ∙ −(1 + ) − (1 − )
(1 + ) ( (1 + )2 )

= 2 1 − ∙ −1 − − 1 +
(1 + ) ( (1 + )2 )

= 2 1 − ∙ ((1 −2
(1 + ) + )2)

= −2 2 1 − #
(1 + )2 (1 + )

70

Chapter 2 : Differentiation

Example 6 APR 2007/ MAT183/ Q1c (5 marks)

( )Given y = e−x , show that 1+ x 2 dy + (1+ x)2 y = 0 .
( )1+ x 2 dx

Solution :

− = − = 1 + 2
= (1 + 2) ′ = − − ′ = 2


=

Differentiate wrt x (using Quotient Rule)

′ − ′
= 2

− − (1 + 2) − 2 −
=
(1 + 2)2

− − − 2 − − 2 −
=
(1 + 2)2

− − (1 + 2 + 2 )
=
(1 + 2)2

− − ( 2 + 2 + 1)
=
(1 + 2)2

− − (1 + )2
= (1 + 2)2

Hence,

(1 + 2) + (1 + )2 = (1 + 2 ) ∙ − − (1 + )2 + (1 + )2 ∙ −
(1 + 2)2 (1 + 2)

(1 + 2) + (1 + )2 − (1 + )2 + − (1 + )2
= − (1 + 2) (1 + 2)

(1 + 2) + (1 + )2 = 0 # ( ℎ )


71

MAT183/ MAT421 : Calculus I

Example 7/ APR 2011/ MAT183/ Q2a (4 marks)

Find dy if y = x 2 ln(x + 1).

dx
Solution :

= 2 ( + 1)
=

Differentiate wrt x (using Product Rule) = 2 = ( + 1)
′ = 2
= ′ + ′ ′ = 1 1
+

2
= 2 ( + 1) + + 1 #

. Example 8/ OCT2009/ MAT183/ Q1c (5 marks)

Find dy if y = ln e3x +4  .
dx x2 −1

Solution :

3 + 4 3 + 4 = 3 + 4 = 2 − 1
= ( 2 − 1 ) = ( 2 − 1 ) ′ = 3 3 ′ = 2
Using the properties of logarithms,

= ( )

Differentiate wrt x

= ( 3 + 4) − ( 2 − 1) 1 ′ − ′
= ( ) ∙ ( 2 )
Differentiate wrt x ′ − ′
= ( 2 )
= 1 4 ∙ 3 3 − 1 1 ∙ 2
3 + 2 −

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

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

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

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

3 3 1 2
= 3 + 4 − 2 − 1 #

72

Chapter 2 : Differentiation

Example 9/ OCT 2008/ MAT183/ Q2b (6 marks)

Differentiate the following functions with respect to x :

( )i) f (x) = 4 7 − x3

ii) f (x) = ln 2x −5 
cos 2 3x 

Solution :

) ( ) = 4√(7 − 3) 2 − 5
) ( ) = [ 2 3 ]
( ) = (7 − 3)41 Using the properties of logarithms,

Using generalized power rule, ( ) = (2 − 5) − ( 2 3 )

′( ) = 1 (7 − 3 )−34 ∙ −3 2 ( ) = (2 − 5) − ( 3 )2
4
( ) = (2 − 5) − 2 ( 3 )
3 3 )−34
′( ) = − 4 2 (7 − # Differentiate wrt x,

′( ) = 1 − 1
2 − 5 ∙ 2 2 ∙ 3 ∙ − 3 ∙ 3

′( ) = 2 + 6 3
2 − 5 3

′( ) = 2 + 6 3 #
2 − 5

Example 10/ MAR 2012/ MAT183/ Q2a (6 marks)

Find dy for the following :
dx

i) y = 1 + 3x 2 − 4 x (2 marks)
x

ii) y = 2 x−2 (4 marks)

x

Solution :

) = 1 + 3 2 − 4√ ) = 2 √ − 2
=

= −1 + 3 2 − 1 = 2 = √ − 2 = ( − 2)12
′ = 2
4

Differentiate wrt x, 1
2
= − −2 + 6 − 1 −34 # ′ = ( − 2)−12
4

Differentiate wrt x (using Product Rule)

= ′ + ′


= 2 √ − 2 + 1 2 ( − 2)−21
2

2
= 2 √ − 2 + 2√ − 2 #

73

MAT183/ MAT421 : Calculus I

3.0 Implicit Differentiation

Simple Examples Procedure to solve implicit differentiation
a) if A = y
 Differentiate each expression one by one
 A’ = dy (when we differentiate the terms containing y, it must has )
dx

b) if B = 2y
 B’ = 2 dy  Separating the terms containing and not containing
dx
 Factorise

 Solve for (write as a subject)


c) if C = 3y2
 C’ = 6y dy
dx

d) If D= y = 1

y2

 D’ = 1 • y − 1 • dy = 1 dy
2

2 dx 2 y dx

e) If A = xy

u=x v=y
u’ = 1 v’ = dy

dx

Using product rule
A’ = vu' + uv'

A’ = y + x dy
dx

74

f) If B = x Chapter 2 : Differentiation
y
75
u=x v=y

u’ = 1 v’ = dy
dx

Using quotient rule
A’ = vu' − uv'

v2

y − x dy
A’ = dx

y2

g) If B = sin3y
 B’ = 3 cos3y dy
dx

h) If C = cosec 3y2
 C’ = − cosec 3y2 cot 3y2• 6y• dy
dx
C’ = −6y cosec 3y2 cot 3y2 dy
dx

i) If A = e4y
 A’ = e4y • 4 dy
dx
A’ = 4 e4y dy
dx

j) If B = etany
 B’ = etany •sec2y• dy
dx

k) If C = ln 2y
 C’ = 1 • 2 dy = 1 dy
2y dx y dx

MAT183/ MAT421 : Calculus I

More Examples (From Previous Semester Papers)

Example 1/ MAR 2004/ MAT183/ Q4a (i) (5 marks)

Given y 2 = sin3 2x + tan(xy ) , find dy by implicit differentiation.

dx

Solution :

2 = 3(2 ) + = =
Let = +
Differentiate : ′ = ′ + ′ − − − − − 
′ =
Where, ′ = 1

= 2 = 3(2 ) =
= ( (2 ))3
Using generalized power rule, ′ = 2 [ ′ + ′]
Using generalized power rule,
′ = 2 ′ = 2 [ +
′ = 3( (2 ))2 ∙ (2 ) ∙ 2 ]
′ = 6 3(2 ) (2 )
′ = 2 + 2


By substituting A’, B’ and C’ into 

6 3(2 ) (2 ) + 2 + 2
2 =

Separating the term containing and not containing


2 − 2 = 6 3(2 ) (2 ) + 2

Factorize


(2 − 2 ) = 6 3(2 ) (2 ) + 2

Write as a subject

6 3(2 ) (2 ) + 2
= #
2 − 2

76

Chapter 2 : Differentiation

Example 2/ OCT 2004/ MAT183/ Q2b (7 marks)

Let y be a function of x. If (y + 1)2 = 3 sec 3 (2x) − x dy by using implicit differentiation.

e y , find dx

Solution :

= =

( + 1)2 = 3 3(2 ) −
Let = +
Differentiate : ′ = ′ + ′ − − − − − 

Where, ′ = 1
′ =

= ( + 1)2 = 3 3(2 )
= 3( (2 ))3
= −
Using generalized power rule,
Using generalized power rule, ′ − ′
′ = 9( (2 ))2 ∙ (2 ) (2 ) ∙ 2 [ 2 ]
′ = 2( + 1) ′ = 18 3(2 ) (2 ) ′ = −


′ = − ቏
቎ 2
−

′ =

− +

2

By substituting A’, B’ and C’ into 

+
2( + 1) =
−
18 3(2 ) (2 ) +
2

Multiplying by 2

2 2( + 1) = 18 2 3(2 ) (2 ) +
)
− +
2 (
2

2 2( + 1) = +

18 2 3(2 ) (2 ) −

2 2( + 1) − =

18 2 3(2 ) (2 ) −

(2 2( + 1) − =

) 18 2 3(2 ) (2 ) −


18 2 3(2 ) (2 ) −
= #


2 2( + 1) −

77

MAT183/ MAT421 : Calculus I

Example 3/ MAR 2005/ MAT183/ Q4a (5 marks)

Let y be a function of x. If e xy − x 2 + 2y = 1, find dy by using implicit differentiation.
dx

Ans : = 2 −
2+

Solution :

Try This !

78

Chapter 2 : Differentiation

Example 4/ NOV 2005/ MAT183/ Q2b (6 marks)

( )Let y be a function of x. If y 4 = tan x 2 + y 2 + e xy , find dy by using implicit differentiation.
dx

Ans : = 2 2( 2+ 2)+
4 3−2 2( 2+ 2)−

Solution :

Try This !

79

MAT183/ MAT421 : Calculus I

Example 5/ APR 2006/ MAT183/ Q2b (7 marks)

Find dy for xy + 3x 2 = 2 cos(x + y ) − e3y by using implicit differentiation.

dx
Solution :

xy + 3x2 = 2 cos(x + y ) − e3y

Let A + B = C + D

Differentiate :
A' + B' = C' + D' −−−−−−−−− 

Where,

= = 3 2 = 2 ( + ) = − 3

Using product rule, Using Power rule, Using tigonometric rule, Using exponential rule,
′ = 6
′ = + ′ = 2 • − ( + ) • (1 + ′ = − 3 • 3
)

′ = −2 ( + ) (1 + ) ′ = −3 3

Expand,

′ = −2 ( + ) − 2 ( + )


By substituting A’, B’, C’ and D’ into ,

y + x dy + 6x = − 2 sin(x + y ) − 2 sin(x + y ) dy − 3e3y dy

dx dx dx

Separating the terms containing dy not containing dy
dx dx

x dy + 3e3y dy + 2sin(x + y )dy = − 6x − y − 2sin(x + y )
dx dx dx

Factorize dy
dx

( )dy x + 3e3y + 2sin(x + y ) = − 6x − y − 2sin(x + y )

dx

Thus, dy = − 6x − y − 2sin(x + y ) #
dx x + 3e3y + 2sin(x + y )

80

Chapter 2 : Differentiation

Example 6/ OCT 2006/ MAT183/ Q5b (6 marks)

Find dy for 3x 2 y + 4y 3 = tan x  by using implicit differentiation.
dx y

Solution :

3 2 + 4 3 =
( )

Let + =

Differentiate : ′ + ′ = ′ − − − − − 

Where, = 3 2 = 4 3
= ( )
Using product rule,
′ = 12 2 ′ −
′ = ′ + ′ [ 2 ′
′ = 2 ( ) ]

′ = 6 + 3 2 = = −
′ = 1 ( ) ቎ 2 ቏
′ = 2
′ =
= 3 2 =
′ = 6 2 ( ) − 2 ( )
2
′ = ′ =

By substituting A’, B’ and C’ into 

2 ( ) − 2 ( )

6 + 3 2 + 12 2 =
2

Multiplying by 2

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

6 3 + 3 2 2 + 12 4 = 2 − 2
( ) ( )

3 2 2 + 12 4 + 2 = 2 − 6 3
( ) ( )

(3 2 2 + 12 4 + 2 = 2 − 6 3
( )) ( )

2 ( ) − 6 3 #
= 3 2 2 + 12 4 + 2 ( )

81

MAT183/ MAT421 : Calculus I

Example 7/ OCT 2007/ MAT183/ Q2b (6 marks)

( )Find dy for 3 ln x 2y − 2x + y 2 = 2 by using implicit differentiation.
dx

Solution : = √2 + 2
= (2 + 2)1⁄2
( )3ln x2y − 2x + y 2 = 2

Let A − B = 2
Differentiate :

A' − B' = 0
A' = B' −−−−−−−−− 

Where,
= 3 ( 2 )

Using the properties of log, Using Generalized Power rule,
= 3( 2 + )
= 3(2 + ) ′ = 1 (2 + 2 )−1⁄2 (2 + 2
2 )
= 6 + 3

Differentiate ′ = 1 • 1 • 2 (1 +
2 √2 + )
′ = 6 • 1 + 3 • 1 2


6 3 ′ = 1 +

′ = + √2 + 2

By substituting A’ and B’ into , Expand :

6+ 3 dy 1+ y dy 6 √2 + 2 + 3 √2 + 2 = + 2
dx
=
x y dx 2x + y2
Separating the terms containing not containing

6y + 3x dy 1+ y dy 3 √2 + 2 − 2 = − 6 √2 + 2
dx = dx

xy 2x + y 2 Factorize



(3 √2 + 2 − 2) = − 6 √2 + 2

Cross multiply
2x + y 2 6y + 3x dy  = xy 1+ y dy  Thus, = −6 √2 + 2 #
 dx   dx  3 √2 + 2− 2

82

Chapter 2 : Differentiation

Exercise (Implicit Differentiation)

1. Find dy for 4 tan 2 x − xy 2 = sin 2y by using implicit differentiation.
dx

8 2 − 2 Page 48/ APR 2008/ MAT183/ Q2b (5 marks)
= 2 2 + 2

2. Find dy for xy = ln(cos x + sin y ) by using implicit differentiation.

dx

− − − Page 60/ APR 2009/ MAT183/ Q2a (6 marks)
= + −

( )3. Find dy for x 2e4x+y = sin2 (3x) + ln y 2 by using implicit differentiation.
dx

6 3 3 − 4 2 4 + − 2 4 + Page 78/ OCT 2009/ MAT183/ Q2b (7 marks)

= 2 4 + − 2

4. Using the implicit differentiation, find dy for 2x 2y 2 + e x2y2 = sec(x + 1).

dx

( + 1) ( + 1) − 4 2 − 2 2 2 2 Page 98/ APR 2010/ MAT183/ Q2b (6 marks)

= 4 2 + 2 2 2 2

5. Find dy for 3xe−y + y 2 sin x = xy by using implicit differentiation.
dx

− 3 − − 2 Page 104/ OCT 2010/ MAT183/ Q2b (6 marks)
= −3 − + 2 −

6. dy for 5 cos(x + y ) − 3e y = 4x2 by using implicit differentiation.
Find
dx

8 + 5 ( + ) Page 109/ APR 2011/ MAT183/ Q2b (6 marks)
= −5 ( + ) − 3

( )7. Let y be a function of x. If e2x−3y = ln xy 4 , find dy by using implicit differentiation.
dx

2 2 −3 − Page 114/ SEP 2011/ MAT183/ Q2c (6 marks)
= 4 + 3 2 −3

8. Find dy for sin(x + y ) = x 6 + 3xy by using implicit differentiation.

dx
Page 119/ MAR 2012/ MAT183/ Q2b (6 marks)

6 5 + 3 − ( + )
= ( + ) − 3

83

MAT183/ MAT421 : Calculus I

4.0 Higher Order Differentiation

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

(first derivate) dy = f' (x) =
(second derivate)
(third derivate) dx
(fourth derivate)
(fifth derivate) d 2y = f' ' (x) =
(sixth derivate)
dx 2

d 3y = f' ' ' (x) =

dx 3

d 4y = f' ' ' ' (x) =

dx 4

d 5y = f' ' ' ' ' (x) =

dx 5

d 6y = f' ' ' ' ' ' (x) =

dx 6

Example 1/ APR 2010/ MAT183/ Q2a (3 marks)

Find the second derivative of y = ln(cos 3x)

Solution :

= ( 3 )

1 3 3
= 3 ∙ − 3 ∙ 3 = − 3 = −3 3

2 = −3 2 3 ∙3 = −9 2 3 #
2

84

Chapter 2 : Differentiation

Example 2/ SEP 2011/ MAT183/ Q2a (4 marks)

( )Find d 2y if y = ln x 2 + 1 .
dx 2

Solution :

= ( 2 + 1)

1 2
= 2 + 1 ∙ 2 = 2 + 1

Differentiate (again) wrt x (using Quotient Rule)

2 ′ − ′ = 2 = 2 + 1
2 = 2 ′ = 2 ′ = 2

2 2( 2 + 1) − 2 (2 )
2 =
( 2 + 1)2

2 2 2 + 2 − 4 2
2 = ( 2 + 1)2

2 2 − 2 2
2 = ( 2 + 1)2

2 2(1 − 2)
2 = ( 2 + 1)2 #

85

MAT183/ MAT421 : Calculus I

Example 3/ OCT 2007/ MAT183/ Q1c (5 marks)
If y = 3xe−2x , prove that d 2y + 4 dy + 4y = 0 .

dx 2 dx
Solution :

= 3 −2

Using product rule, = 3 = −2
= 3 = −2 −2
= ′ + ′


= 3 −2 − 6 −2 = −6 = −2
= −6 = −2 −2

Using product rule,

2 = −6 −2 + ′ + ′
2

2 = −6 −2 − 6 −2 + 12 −2
2

2 = −12 −2 + 12 −2
2

Therefore,

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

2 = −12 −2 + 12 −2
2 + 4 + 4

+12 −2 − 24 −2

+12 −2
2
2 + 4 + 4 = 0 # ( )

86

Chapter 2 : Differentiation

Example 4/ MAR 2004/ MAT183/ Q5a(i) (5 marks)

Given y = 1 , show that d 2y + dy + xy 3 = 3x2 − 1.
dx 2 dx ( )x2 ( )x2
x2 +1 5 3

+1 +1

Solution :

= 1 = ( 2 + 1)−12 → 3 = (( 2 + 1)−21 3 = ( 2 + 1)−32

√ 2 + 1 )

Using generalized power rule,

= − 1 ( 2 + 1)−23 ∙ 2
2

= − ( 2 + 1)−23


Using product rule,

2 = ′ + ′ = ( 2 + 1)−23
2
= −
2 = −( 2 + 1)−23 + 3 2( 2 + 1)−52 = −1 = − 3 ( 2 + 1)−52 ∙ 2
2 2

= −3 ( 2 + 1)−25

Therefore,

2 + 3 = −( 2 + 1)−32 + 3 2( 2 + 1)−52 − ( 2 + 1)−32
2 +
+ ( 2 + 1)−23

2 + + 3 = 3 2( 2 + 1)−25 − ( 2 + 1)−23
2

2 + + 3 = 3 2 1 # ( )
2 − √( 2 + 1)3

√( 2 + 1)5

87

MAT183/ MAT421 : Calculus I

5.0 Linear Approximations and Differentials

( ) ≈ ( + ∆ ) ≈ ( ) + ′( ) ∙ ∆

Hot Tips!!
x0 = nearest whole number that gives the answer

as a whole number when f(x0) is solved
x = small changes (less than 1)
f(x0) should be the whole number
f ’(x0) should be the whole number or fraction (never write in decimals)

Procedure to estimate/approximate using differentials :
 Identify the suitable values for 0 and
 Identify ( ) and find the value of ( 0)
 Find ′( ) and the value of ′( 0)
 Use ( 0 + ) ≈ ( 0) + ′( 0) to estimate/approximate the given question

Example 1/ APR 2011/ MAT183/ Q3b (5 marks)
Use differential to estimate the value of √36.02 and give your answer correct to four decimal
places.
Solution :

√36.02 = √ = √ 0 + ∆ = √36 + 0.02
 find , ∆

= 36.02, 0 = 36, ∆ = +0.02

 find ( ) ( )
( ) = √

( 0) = (36) = √36 = 6

 find ′( ) ′( )
1
( ) = √ =
2

′( ) = 1 −21 = 1
2 2√

′( 0 ) = ′(36) = 1 = 1 1
2√36 2(6) = 12

 Estimate/ Approximate
√36.02 ≈ ( 0) + ′( 0) ∙ ∆

1
≈ 6 + (12) (0.02)
≈ 6.0017 # (4 . )

88

Chapter 2 : Differentiation

Example 2/ APR 2009/ MAT183/ Q3a (5 marks)

Given f (x) = 3 . If x changes from 1 to 1.05, estimate the value of f (1.05) using

2+ x2
differentials. Give your answer correct to THREE (3) decimal places.
[Hint : ( 0 + ) ≈ ( 0) + ′( 0) ]

Solution :

1.05 = = 0 + ∆ = 1 + 0.05

 find , ∆
= 1.05, 0 = 1, ∆ = +0.05

 find ( ) ( )
3

( ) = 2 + 2
3

( 0) = (1) = 2 + (1)2 = 1

 find ′( ) ′( )

( ) = 2 3 = 3(2 + 2)−1
+ 2

′( ) = −3(2 + 2)−2 ∙ 2 = (2 −6
+ 2)2

′( 0 ) = ′(1) = (2 −6(1) = 6 2
+ (1)2)2 −9 = − 3

 Estimate/ Approximate

(1.05) ≈ ( 0) + ′( 0) ∙ ∆
2

≈ 1 + (− 3) (0.05)
≈ 0.967 # (3 . )

89

MAT183/ MAT421 : Calculus I

Example 3/ OCT 2010/ MAT183/ Q3a (5 marks)
Use differential to estimate sin  + 0.01 correct to four (4) decimal places.

5 

Solution :


(5 + 0.01) = = 0 + ∆ = 5 + 0.01

 find , ∆

= + 0.01, 0 = , ∆ = +0.01

5 5

 find ( ) ( )
( ) =


( 0) = (5) = (5)

 find ′( ) ′( )

( ) =

′( ) =

′( 0 ) = ′ =
(5) (5)

 Estimate/ Approximate


(5 + 0.01) ≈ ( 0) + ′( 0) ∙ ∆

+ (0.01)
≈ (5) (5)

≈ 0.5959# (3 . )

90

Chapter 2 : Differentiation

Example 4/ APR 2008/ MAT183/ Q3a (5 marks)
Use differential to approximate √1 + (1.1) correct to two (2) decimal places.

Solution :

(1.1) = = 0 + ∆ = 1 + 0.1

 find , ∆ x = the given decimal number.
= 1.1, 0 = 1, ∆ = +0.1 0 = the nearest whole number
∆ = small changes and between (−0.9999 to 0 .9999)

 find ( ) ( ) The best 0 is 1 because it gives ( 0) a
( ) = √1 + whole number, 1
( 0) = (1) = √1 + (1) = √1 + 0 = 1

 find ′( ) ′( )

1

( ) = √1 + = (1 + )2

′( ) = 1 (1 + )−12 1 1
2 ∙ = 2 √1 +

′( 0 ) = ′(1) = 1 (1) = 1
2(1)√1 + 2

 Estimate/ Approximate

√1 + (1.1) ≈ ( 0) + ′( 0) ∙ ∆
1

≈ 1 + (2) (0.1)
≈ 1.05 # (2 . )

91

MAT183/ MAT421 : Calculus I

Example 5/ SEP 2011/ MAT183/ Q3b (5 marks)
Given ( ) = √20 − 2, use differential to estimate the value of √20 − (1.99)2. Give your
answer correct to three (3) decimal places.

Solution :

(1.99) = = 0 + ∆ = 2 − 0.01 = 2 + (−0.01)

 find , ∆ x = the given decimal number.
= 1.99, 0 = 2, ∆ = −0.01 0 = the nearest whole number
∆ = small changes and between (−0.9999 to 0 .9999)

 find ( ) ( ) The best 0 is 2 because it gives ( 0) a
( ) = √20 − 2 whole number, 4
( 0) = (2) = √20 − (2)2 = √16 = 4

 find ′( ) ′( )

( ) = √20 − 2 = (20 − 2 1

)2

′( ) = 1 (20 − 2)−12 ∙ −2 = −
2 √20 − 2

′( 0 ) = ′(2) = −2 22 = 1
√20 − −2

 Estimate/ Approximate

√20 − (1.99)2 ≈ ( 0) + ′( 0) ∙ ∆
1

≈ 4 + (− 2) (−0.01)
≈ 4.005 # (3 . )

92

Chapter 2 : Differentiation

Example 6/ MAR 2012/ MAT183/ Q3a (5 marks)
Use differential to estimate the value of 3√7.96 and give your answer correct to four (4) decimal
places.
Solution :

Try This !

93

MAT183/ MAT421 : Calculus I

Example 7/ OCT 2012/ MAT183/ Q3b (5 marks)
Given ( ) = √4 2 − 7. Estimate the value of (2.01) using differential. Give your answer
correct to three (3) decimal places.

Solution :

Try This !

END OF CHAPTER 2
94