LECTURE NOTES 9

MATHEMATICS 1 SM015
Ch 9 Differentiation

Mode: LECTURE (1 of 4)

Topic: 9. DIFFERENTIATION

Sub-Topic: 9.1 Derivative of a Function

Learning Outcomes: At the end of this lesson, students should be able to

(a) find the derivatives of a function f x using the first principle f x fx h fx
lim .
h0 h

*Introduce the notation dy .
dx

**Limit to polynomials up to degree 2, surd and reciprocal.

(b) discuss the differentiability of a function at x a.

*f a fx h fx fx h fx

lim lim
xa h xa h

**Relate the differentiation of function to its continuity.

First Principle fx fx h f x , where f x is the gradient of the
If y f x , then dy lim
h0 h
dx
tangent to the curve.

Notation: dy f x d f x .

dx dx

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MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 1

Find dy by using first principle of
dx

(a) f x x2 4x 2

fx x h2 4x h2 x2 4x 2
lim 2 x2 4x
h 2
h0 4x 4h

lim x2 2hx h2 h

h0

lim 2hx h2 4h
h0 h

h 2x h 4
lim
h0 h
2x 4

(b) g x 1x
fx
1 xh 1x
lim
h0 h 1x 1 xh 1x
1 xh 1x
1 xh
lim
h0 h

lim 1 x h 1x

h 0h 1 x h 1x

lim 1
h0 1 x h
1x

1
21 x

2

MATHEMATICS 1 SM015
Ch 9 Differentiation

(c) f x 1
fx 3x 5

11
3 x h 5 3x 5
lim
h0 h
lim 3x 5 (3x 3h 5)
h 0 h 3 x h 5 3x 5

lim 3h
h 0 h 3 x h 5 3x 5

lim 3
h 0 3 x h 5 3x 5

3
3x 5 2

The Derivatives at a point x = a

fa fx fa
lim
xa x a

EXAMPLE 2 x2 3 at x 1.
By using first principle find derivative of f x

f1 x2 3 12 3
lim
x1 x 1
lim x2 1
x 1x 1

x 1x 1
lim
x1 x 1
lim x 1

x1

2

3

MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 3 1 at x 8.
By using first principle find derivative of f x x1

f8 11
lim x 1 8 1
x8 x 8
lim 3 x 1
x83x 1 x 8

lim 3 x 1 3 x1
x1
x83x 1 x 8 3

9 x1 x1
lim
x83x 1 x 8 3

lim x8 1
h83x 1x 83 x

lim 1
h83x 13 x 1

1
33 6

1
54

4

MATHEMATICS 1 SM015
Ch 9 Differentiation

Differentiability of a Function f at x = a

A function f is differentiable at a point x a if the derivative f a fx f a exists,
lim
xa x a

i.e. f a fx fa fx fa
lim lim .
xa x a xa x a

Note:

If a function is differentiable at a point, then the function is also continuous at that point.

However, a function may be continuous at a point but not differentiable. This happens at point

where there are corners or vertical tangents. For example,

Corner Vertical tangent

The slopes have different limits from the left and from the right, hence f is not differentiable at

x a.

EXAMPLE 4 x is differentiable at x 0.
Discuss whether the function f x

fx x
x, x 0,
x, x 0.

f0 lim x 0
x0 x

lim 1

x0

1

f0 lim x (0)
x0 x

lim 1

x0

1

f 0 does not exists since f 0 f 0.

f x is not differentiable at x 0.

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MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 5

Discuss the function f x x2 3, x 1, is continuous but not differentiable at x 1.
x 3, x 1

Continuity
(i) f 1 12 3 4

(ii) lim x 3 4

x1

lim x2 3 4

x1

limf x 4 1.

x1

(iii) lim f x f 1
x1
f x is continuous at x

Differentiability

f1 x 3 12 3
lim
x1
x1 1
1
lim x
x1x

lim 1

x1

1

f1 x2 3 12 3
lim
x1
x1 1
1
lim x2 1x 1
x1 x x1
1
x
lim

x1

lim x

x1

2

f 1 does not exists since f 1 f 1.

f x is not differentiable at x 1.

6

MATHEMATICS 1 SM015
Ch 9 Differentiation

Mode: LECTURE (2 of 4)

Topic: 9. DIFFERENTIATION

Sub-Topic: 9.2 Rules of Differentiation

Learning Outcomes: At the end of this lesson, students should be able to
(a) apply the rules of differentiation:

(i) Basic Rule;
(ii) Sum Rule;
(iii) Product Rule;
(iv) Quotient Rule; and
(v) Chain Rule.

*Include d f x n n f x n 1 f x .
dx

(b) perform second and third order differentiation.
**Second order differentiation is the differentiation of the first derivatives.

Basic Rules 0
(a) If f x k, where k is any constant, then f x
(b) If f x xn, then f x nxn 1
(c) If f x kxn, then f x knxn 1

EXAMPLE 6
Differentiate the following functions with respect to x.

(a) f x 3 (b) f x 5x

fx 0 fx 5 1 x1 1
5x 0
5

7

MATHEMATICS 1 SM015 (d) f x 12
Ch 9 Differentiation fx
(c) f x x4 x

f x 4x3 1

Sum Rule 12x 2
 If f x u x v x , then
fx 1 12 x 1 1
f x dux 2
dx
2
.
3

6x 2

6

x3

dvx u x v x
dx

EXAMPLE 7

Differentiate 2x2 5 x with respect to x.

d 2x2 5 x d 2x2 d 1
dx dx
dx 5x 2

4x2 1 1 (5)x 1 1
2

2

4x 5x 1
2

2

4x 5
2x

Product Rule u x v x is
If u x and v x are differentiable functions, the derivative of f x

f x vxu x uxv x

The Product Rule can be written as
d uv vu uv
dx

.

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MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 8 3x2 1 7 2x3 with respect to x .
Differentiate p x

u 3x2 1, v 7 2x3
u 6x v 6x2

p x 7 2x3 6x 3x2 1 6x2

6x 7 2x3 3x3 x

6x 5x3 x 7

Quotient Rule u x is
If u x and v x are differentiable functions, the derivative of f x vx

fx vxu x uxv x

vx 2

The Quotient Rule can be written as vu uv
v2
du
dx v

.

9

MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 9
Differentiate the following functions with respect to x.

(a) f x x2 1
x4 1

u x2 1, v x4 1,
u 2x v 4x3

fx x4 1 2x x2 1 4x3

x4 2

1

2x5 2x 4x5 4x3

x4 2

1

2x5 4x3 2x

x4 2

1

(b) f x 1 2x 3x 2
5x 4

u 1 2x 3x 2 , v 5x 4
x 2 6x2 v5

u 1 12x

fx 5x 4 1 12x 5 x 2 6x2

5x 4 2

5x 60x2 4 48x 5x 10 30x2
5x 4 2

30x2 48x 6
5x 4 2

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MATHEMATICS 1 SM015
Ch 9 Differentiation

Chain Rule
If

y f u is a differentiable function of u ,

and

u g x is a differentiable function of x ,

then

y f g x is a differentiable function of x .

The Chain Rule can be written as

dy dy du .
dx du dx

EXAMPLE 10 y u3 3u2 1, u x2 2
Find dy if y u3 3u2 1 and u x2 2. dy 3u2 6u du 2x
du dx
dx

dy dy du
dx du dx

3u2 6u 2x

3 x2 2 6 x2 2 2x
6x3 12x

EXAMPLE 11

Differentiate the following functions with respect to x using the Chain Rule.

(a) y (x 4)3 (b) y 7x 2

(a) dy dy du u x 4, y u3
dx du dx du 1 dy 3u2
dx du
3u2 1
3 x 42

11

MATHEMATICS 1 SM015 u 7x 2, yu
Ch 9 Differentiation du 7
dx 1
(b) dy dy du
dx du dx u2

17 dy 1u 1
2u 2

7 du 2
2 7x 2
1

2u

General Power Rule f x n , involving some function f x raised to a
Any composite function of the form y

rational power n, is called the General Power Rule.
Let

y un, where u f x .

Then

dy nun 1, and du f x .
du dx

By using the Chain Rule dy dy du ,
dx du dx

dy nun 1 f x .
dx

Thus,

d f x n nf x n1f x .
dx

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MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 12

Differentiate the following functions with respect to x using the General Power Rule.

(a) f x x5 3 (b) y 2x4 9x 4

1 6

(a) f x 3 x5 2 d x5 1

1 dx

3 x5 1 2 5x4

15x4 x5 2

1

(b) dy 4 2x4 9x 3 d 2x 4 9x 6
dx 4 2x4 9x
6 dx

3 8x3 9

6

Higher Order Derivatives

First derivative y fx dy or df x
fx dx dx
fx or
Second derivative y d dy d2y d2 f x
fn x dx dx dx2 or dx 2
or
Third derivative y d d2y d3y or d3 f x
dx dx2 dx 3 dx 3

nth derivative yn dny dn f x
dx n dx n

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MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 13 2x4 9x3 5x2 7.
Find the first, second and third order derivatives of p x

p x 8x3 27x2 10x

px 24x2 54x 10

p x 48x 54

EXAMPLE 14

Find the first, second and third order derivatives of y 2x .
3x 1

dy 3x 1 2 2x 3 2
dx 3x 1 2 3x 1 2

d 2 y 3x 1 2 0 2 6 3x 1 12
dx2 3x 1 4 3x 1 3

d 3 y 3x 1 3 0 12 9 3x 1 2 108
dx3 3x 1 6 3x 1 4

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MATHEMATICS 1 SM015
Ch 9 Differentiation

Mode: LECTURE (3 of 4)

Topic: 9. DIFFERENTIATION

Sub-Topic: 9.3 Differentiation of Exponential, Logarithmic and Trigonometric Functions

Learning Outcomes: At the end of this lesson, students should be able to
(a) find the derivatives of the functions:

(i) ax , af x , ex , ef x
(ii) ln x, ln f x
(iii) sin x, cosx, tan x, secx, cosecx, cot x;
(iv) sin u, cosu, tan u, secu, cosecu, cot u;
(v) sinn x, cosn x, tann x, secn x, cosecn x, cot n x;
*** u g x is a linear or non-linear function.
*The first principle for exponential, logarithmic and trigonometric functions are not required.
** f x is a linear function.
(b) solve problems involving the combination of differentiation rules.

15

MATHEMATICS 1 SM015
Ch 9 Differentiation

Derivative of ax ax ln a.
If y ax , then dy

dx

Proof Taking ln to both sides
ln y ln ax Differentiating both sides with respect to x

ln y x ln a

1 dy = x 0 ln a 1
y dx
1 dy = ln a
y dx

dy y ln a
dx
dy ax ln a
dx

In general,

d amx mamx ln a
dx

d af x f x af x ln a
dx

If a e, d ef x f x ef x ln e f x ef x .
dx

EXAMPLE 15

Find the derivative of the following functions.

(a) y e2x 1 (b) y e2x e 2x e2x e 2x

(c) y 1 (d) y 2x 3 1
7e3x 5 32x

(a) dy e2x 1 ln e 2 (b) y e4x 1 1 e 4x
dx
dy 4e4x 4e 4x
2e2x 1 dx

(c) y 1 e 3x 5 (d) y 2x 3 3 2x
7
3 dy 2x 3 ln 2 2 3 2x ln 3
dy e 3x 5 ln e dx
dx 7
2x 3 ln 2 2 ln 3
3 e 3x 5 32x
7

16

MATHEMATICS 1 SM015
Ch 9 Differentiation

Derivative of ln x

If y ln x, then dy 1.
dx x

Proof

y ln x

ey x

ey dy 1
dx
dy 1
ey
dx 1
dy

dx x

In general,

d lnf x fx
dx fx

EXAMPLE 16 (b) f x 1 ln 5 3x
Find derivative of the function 2

(a) f x ln 3x (d) y ln 1 2x
1 4x
(c) y ln 2x 1
(b) f x 1 1
(a) f x 13 or f x ln3 lnx 2 5 3x 3
3x fx 3
01 10 6x
1 x
x
1
x

(c) y 1 ln 2x 1 (d) y ln 1 2x ln 1 4x
2
dy 1 2 1 4
dy 1 1 2 dx 1 2x 1 4x
dx 2 2x 1
24
1
2x 1 1 2x 1 4x

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MATHEMATICS 1 SM015
Ch 9 Differentiation

Derivative of sin x, cos x, tan x, sec x, cot x and cosec x

(a) d sin x cosx (d) d cot x cosec2 x
dx sin x dx

(b) d cosx sec2 x (e) d secx secx tan x
dx dx

(c) d tan x (f) d cosecx cosecx cot x
dx dx

EXAMPLE 17
Find derivative of the following functions:

(a) y 2sin x 3cosx (b) y 4tan x 5 (c) y 1 2
x
7cosx tan x

dy 2cosx 3 sin x dy 4sec2 x 5 1 x 2 y 1 secx 2cot x
dx dx 7

2cosx 3sin x 4sec2 x 5 dy 1 secx tan x 2 cosec2 x
x2 dx 7
dy 1 secx tan x 2cosec2 x
dx 7

Derivative of sin u, cos u, tan u, sec u, cot u and cosec u

(a) d sin u cosu d u (d) d cot u cosec2 u d u
dx dx dx dx

(b) d cosu sin u d u (e) d secu secu tan u d u
dx dx dx dx

(c) d tan u sec2 u d u (f) d cosecu cosecu cot u d u
dx dx dx dx

EXAMPLE 18

Find derivative of the following functions:

(a) f x cos 5x 4 (b) f x sin ln x

(a) d cos 5x 4 sin 5x 4 d 5x 4 (b) d sin ln x cos ln x d ln x
dx dx dx dx
cos ln x
5sin 5x 4 cos ln x 1
x
x

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MATHEMATICS 1 SM015
Ch 9 Differentiation

Derivative of sinn u, cosn u, tann u, secn u, cotn u and cosecn u

(a) d sinn u n sinn 1 u cosu d u
dx dx

(b) d cosn u n cosn 1 u sin u d u
dx dx

(c) d tann u n tann 1 u sec2 u d u
dx dx

(d) d cot n u n cot n 1 u cosec2 u d u
dx dx

(e) d secn u n secn 1 u secu tan u d u
dx dx

(f) d cosecn u n cosecn 1 u cosecu cot u d u
dx dx

EXAMPLE 19 (b) y sin3 x2 1
Find derivative of the following functions:
(a) f x sec2 x (b) y sin x2 3

(a) f x 2secx d secx 1
dx
dy 3 sin x2 2 d sin x2 1
2secx secx tan x
2sec2 x tan x 1
dx dx

3 sin x2 1 2 cos x2 1 2x

6x sin2 x2 1 cos x2 1

19

MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 20

Find the first derivative of the functions given below:

(a) f x e3x x2 1 (b) f x e2x (c) f x x sin 1
1 ex x

(a) u e3x , v x2 1 (b) u e2x , v1 ex

u 3e3x v 2x u 2e2x v ex

fx 3e3x x2 1 2xe3x 2e2x 1 ex e2xex
e3x 3x2 3 2x 1 ex 2
fx

2e2x 2e3x e3x
1 ex 2

2e2x e3x
1 ex 2

(c) u x, v sin 1 1
u1 x x2

v cos 1
x

cos 1
x

x2

fx sin 1 x cos 1
x

x x2

sin 1 cos 1
x x

x

20

MATHEMATICS 1 SM015
Ch 9 Differentiation

Mode: LECTURE (4 of 4)

Topic: 9. DIFFERENTIATION

Sub-Topic: 9.4 Implicit Differentiations

Learning Outcomes: At the end of this lesson, students should be able to
(a) solve the first and the second derivative implicitly.

Implicit Differentiations
 From the previous lesson, we have learnt how to differentiate the function such as

y 5x4 2x3 1 (Explicit), i.e., dy 20x3 6x2.
dx

 Sometimes, we have to deal with the function such as 2y3 y2 6x3 4x 5 (Implicit),

where y is not easily written as a subject. To find dy for such function, we need to use a
dx

method of differentiation called Implicit Differentiation. To use this method, we need to
differentiate every term with respect to x using the appropriate rules.

 For example,

(a) y2 4x 6x2

Differentiate each term in both sides with respect to x,

d y2 4x d 6x2
dx dx
12x
2y dy 4
dx

Thus, let dy as subject,
dx

dy 12x 4
dx 2y

6x 2
y

21

MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 21 (b) ln xy ex y
Find dy in terms of x and y if

dx
(a) y3 6x x2

(a) 3y2 dy 6 2x 1 ex y
dx 2ex y
dy 2x 6 (b) ln xy 2 2ex y
dx 3y2 ln xy

ln x ln y

d ln x ln y d 2ex y
dx dx
2ex y 1 dy
1 1 dy
x y dx dx
1 1 dy 2ex y 2ex y dy
x y dx
1 dy 2ex y dy dx
y dx dx 2ex y 1
dy 1 2yex y
dx y x
2xex y 1
dy
dx x

y 2xex y 1
x 1 2yex y

22

MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 22

Given that equation of a curve is 3x2 4y2 15xy 6. Find the value of dy and d2y at the point
dx dx 2

1,3 .

d 3x2 4y2 d 15xy 6
dx dx
15x dy y 15
6x 8y dy
dx dx

8y dy 15x dy 15y 6x
dx dx
dy 15y 6x
8y 15x
dx

dy 15 3 6 1
dx x 1,y 3 8 3 15 1
13
3

d2y 8y 15x 15 dy 6 15y 6x 8 dy 15
dx dx

dx2 8y 15x 2

d2y 8 3 15 1 15 13 6 15 3 6 1 8 13 15
dx 2 3 3
8 3 15 1 2
x 1,y 3
236
81

23

MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 23

Given that xy sin 3x , show that x d2y 2 dy 9xy 0.
dx 2 dx

d xy d sin 3x
dx dx
3 cos3x
x dy y 1
dx 9sin 3x
0
x d2y dy dy 0 shown
dx 2 dx dx

x d2y 2 dy 9sin 3x
dx 2 dx

x d2y 2 dy 9xy
dx 2 dx

Parametric Differentiations x f t and y g t .
Given

The first parametric derivative is

dy dy dt
dx dt dx

The second parametric derivative is d dy
dt dx
d2y
dx 2 dx
dt

24

MATHEMATICS 1 SM015
Ch 9 Differentiation

EXAMPLE 24 (b) x cos3t and y 2sin2 3t
Find dy in terms of the parameter t if

dx
(a) x 2t 3 and y 4t 2 1

(a) dx 6t 2, dy 8t (b) dx 3sin 3t, dy 4sin 3t 3cos3t
dt dt dt dt

dy dy dt dy 12sin 3t cos3t 1
dx dx 3sin 3t
dt dx
4 cos3t
8t 1
6t 2

4

3t

EXAMPLE 25

Find the value of dy if x t 1, y 2t 4 when t 2.
dx t t

dx 1 1 , dy 2 4
dt t2 dt t2

t2 1 2t 2 4

t2 t2

dy dy dt

dx dt dx
2t 2 4 t 2
t2 t2 1
2t 2 4
t2 1

dy 24 4
dx t 2 41

4
5

25

MATHEMATICS 1 SM015
Ch 9 Differentiation
EXAMPLE 26

Given that x t xt and 2ty y2 3. By using implicit differentiation, find dx and dy in
dt dt

terms of x and y. Hence, find the values of dy when x 2.
dx

1 dx 1 t 1 dx x 1
dt dt

dx 1 t x1
dt
x1
dx 1t

dt x1
1x
xt xt
x1 x1 t t
x 12
x t
1t
y 2 2t dy 2y dy 0
dt dt 2ty y2
t
dy 2t 2y 2y
dt

dy 2y

dt 2t 3 y2 3
2y 3 y2

2y2 2y

y2 3

dy dy dt
dx dt dx

2y2 1
y2 3 x 12

2y2
3 y2 x 1 2

When x 2, When t 2, 3
2 t 2t 2 2 y y2 0
y2 4y 3 0
t2
y 1y 3

y 1, y 3

dy 2 12 1
dx x 2, y 1 3 12 2 1 2

26

MATHEMATICS 1 SM015
Ch 9 Differentiation

dy 2 32 3
dx x 2, y 3 3 32 2 1 2

27