FlipWorkbookDUM20132

MULTIPLICATION OF DUM20132 ENGINEERING MATHEMATICS 2
VECTOR BY A UNIT 2: VECTOR
SCALAR
If  is a positive real number then

→

a
→
a

→→

 a is a vector of magnitude  a and in the same direction as

→ →→

a . It also follows that is −  a a vector of magnitude  a and

→

with direction opposite to that ofa .

49

UNIT LEARNING DUM20132 ENGINEERING MATHEMATICS 2
OUTCOMES UNIT 2: VECTOR

After completing the unit, students should be able to:
2.3.1 compute the addition of a vector in two and three dimensions.
2.3.2 find the magnitude and unit vectors of two and three dimentions.

2.3 COMPONENTS OF A VECTOR

Example : Notes

If u = 3i + 2 j and v = −i + 5 j , then Vectors in two and three dimensions
a) Represent graphically u and v . Two dimension and three dimension vectors can be
b) Write 3u + 6v in terms of i and j . expressed in the form
c) Find 3u + 6v .
a = a1i + a2 j
Solution : y
and
a)
a = a1i + a2 j + a2 k
v
where a is a vector and i, j and k are unit vectors in

the direction of the positive x-axis, y-axis and z-axis
respectively.

5j u x Definition
2j
− i O 3i ( )If P is a point with coordinates a1 , a2 , a3 , the

b) 3u + 6v = 33 + 6− 1 position vector of P
 2  5  →
OP = a1i + a2 j + a3k or
=  9  + − 6
 6  30   a1 
→
=  3  OP =  a2 
 36
 a3 
= 3i + 36 j
Magnitude of a vector
c) 3u + 6v = 32 + 342
The magnitude of a vector a is given by
= 9 + 1296
= 1305 a= a12 + a2 2
= 36.12 a12 + a22 + a32
and

a=

1. If s = −4i + 2 j and t = 6i − 3 j , then

a) Represent graphically s and t .
b) Write w = 2s + 3t in terms of i and j .

50

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

2. If a = i − j and b = −9i − 2 j , then
a) Represent graphically a and b .
b) Write y = 5a − b in terms of i and j .

c) Find y

3. Given that p = 2i + 3 j + 6k and q = −i + 5 j − 3k

a) Represent graphically p and q .
1

b) Find 3 p + q .
2

51

Example : DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR
→→
Find AB and magnitude of AB where A is Notes
the point (2,−1) and B is the point (−1,3) .
If a has coordinate A = (a1, a2 ) and b has
→ coordinate B = (b1,b2 )
Find the unit vector in the direction of AB .
Then the magnitude of
Solution :
→
y AB = (b1 − a1 )2 + (b2 − a2 )2
If a is any non-zero vector, then the unit vector
B(−1,3)
in the same direction is given as

uˆ = 1 u
u

O A(2,−1) x

→ → → →→
AB = AO + OB = OB − OA
→→

OA →= 2i − j and OB = −i + 3 j

 AB = − i +3j − (2i − j) or  −31 −  −21 =  −3 
4

→

 AB = −3i + 4 j

Magnitude of AB is = (−3)2 + (4)2 = 5

1. Given that A = (2,3) and B = (− 3,5), find AB .

52

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

2. Find the unit vector that has the same direction as

a) 3i − 4 j b) − 8i + 10 j

3. Given vector position for points A and B are a = −3i + 4 j, b = 3i + j .

Find

a) a

→
b) AB

→

c) Unit vector in direction of AB .

3. Given that C = (8,−3) and D = (5,4) , find in terms of i and j for

→
a) CD

→
b) Vector unit in the direction of CD

53

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

Example : Notes

→ The undirected distance or magnitude between two
Given P(1,3,−1) and Q(3,5,0) , find PQ points P1(x1, y1, z1) and P2 (x2, y2, z2 ) is given by

→ P1P2 = (x2 − x1 )2 + ( y2 − y1 )2 + (z2 − z1 )2

and magnitude of PQ .

Solution : The axes of reference Ox,Oy,Oz are chosen so

→ →→ →→ that they form a right handed set. The symbols i, j, k
PQ = PO + OQ = −OP + OQ
denote unit vectors in the direction Ox,Oy,Oz
− (1,3,−1) + (3,5,0) = (2,2,1)
respectively y
→
Magnitude of PQ = 22 + 22 +12 = B

9 =3

1. If a = (−1 2 3), b = (4  − 2), find b r P(a,b, c)
a
O Ax
c
a) a + b (b) 2a − 2b
C

z

→ OP = a
If OP = ai + bj + ck

→
then OP = a 2 + b2 + c2

2. Find a unit vector in the direction of: 3. Given P(2,3,4) and Q(5,6,7). Find →
PQ
a) (1 0 −1)
b) (1 2 − 3) →
and magnitude of PQ

54

TUTORIAL 2.2 DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR
1. Represent graphically the following vectors. 55
a. v = 5i − 4 j
b. w = −2i − j
c. m = i − 2 j − 7k
d. n = 6i + 6 j + 6k

2. Calculate the direction of the following vectors.
a. A = (−8,10) and B = (−6,−1)
i. AB
ii. 5 AB
iii. BA
iv. 1 BA
2

b. P = 5i − 7 j + k and M = 3i + j − 9k
i. PM
ii. − 3PM
iii. MP
iv. 2 MP
5

3. Given v = −40i + 9 j and w = −16i − j . Find
a. v

b. w

c. vw

d. Unit vectors in direction of v
e. Unit vectors in direction of w
f. Unit vectors in direction of vw
g. Unit vectors in direction of wv

4. Given P = (11,10,18) and Q = (−1,2,−6) . Find
a. P
b. Q
c. PQ

d. Unit vectors in direction of P
e. Unit vectors in direction of Q
f. Unit vectors in direction of PQ
g. Unit vectors in direction of QP

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

kj

5. a. State v and w in coordinate form.
b. Calculate the

i. direction of vw
ii. − 5 vw

iii. unit vector in direction of vw

6. if m = 11i + 13 j − 2k and p = −3i + j + 10k are two vectors in 3 dimensional space. Calculate the
a. direction of mp
b. magnitude of mp
c. unit vector in direction of mp

7. Given that p = 4i + j , m = −2i − 3 j and q = i − 5 j . Find
a. p + m + 2 p
b. the magnitude of vector 3m − 2 p
c. the unit vector in direction of 2q + 3m

8. Given A = (−1,2) and B = (3,−1) . Find 1 AB
2

9. Given that a =  5  , b =  −73 and c=  −1  , find
4 −2

a. a + b
b. − c − 5b
c. 5a − 2c
d. 3a + 5c − 7b

10. Given that OA = 3i – 8j , OB = 6i + 2j and OP = x i+ y j . Find the
value of x and y if OP = 2 OA – 3OB .

11. If a = −i + 2 j + 3k and b = 4i + 3 j − 2k , find

a. a + b
b. 2a − 2b
c. c if c + a = b

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES 2.4.1 perform scalar product in two and three dimensions

2.4 SCALAR PRODUCT (DOT PRODUCT) Notes

Example : If a = a1i + a2 j + a3k and
b = b1i + b2 j + b3k
Given a = 2i − 3 j + 4k
b = i + 3 j − 2k Then, a • b = a1b1 + a2b2 + a3b3
and if  is angle between a and b
Find angle between a and b .
Then,
Solution :
a • b = a b cos
a • b = (21)+ (− 3 3)+ (4)(− 2) = −15
b
a = 22 + (− 3)2 + 42 = 29
b = 12 + 32 + (− 2)2 = 14 

a • b = a b cos a

cos = − 15
29 14

 = cos−1 − 15  = 138.11
 406 

PROPERTIES OF THE SCALAR PRODUCT

(a) Parallel Vectors b
If a and b are parallel

b

a a

a  b = abcos0 or a  b = abcos

a  b = abfor like parallel vectors
a b = −ab for unlike parallel vectors
In the special case when a = b, a.b = a.a = a2 (sometimes a.a is written as a2 )

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

(b) Perpendicular Vectors

If a and b are perpendicular then, a.b = abcos = 0 i.e a.b = 0
2

b
a

In the special case of the Cartesian unit vectors these results give
i.i = j.j = k.k = 1
i.j = j.k = k.i = 0

57

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

1. Determine a •b and the angle between a and b for the following pairs of non-
zero vectors.

a) a = 2i + 4 j − 3k and b = i + 3 j + 2k b) a = i + 3 j + 3k and b = 3i − 4k

c) a = 3 j + 5i − k and b = i + j d) a = 7i − j and q = 4i − 5 j + 9k

 −1  4 
p =  2  anqd=  − 2
e)  3   − 4 f) p = (8 1 − 2) and q = (− 5 0 9)

58

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

TUTORIAL 2.4

1. Given that u = 3 i− 2 j+ 6 k and v = 2 i+ 5 j+ 6 k , find the dot product for u and v
~ ~~ ~ ~ ~~ ~ ~~

2. Taking a = j + 3k and b = 2i - j + 2k and c = 3i – k. Find a.b , a.c and b.c

3. Find the angle between the vector u = i− 2 j+ 2 k and v = −3 i+ 6 j+ 2 k .
~~ ~ ~ ~~ ~
~

4. Find the angle between the vector 3i - j - 2k and i + 2j - 3k.

5. Find the angle between the vector 2i - 3j + k and -3i + j +9k.

6. Find the angle between the vector a = 2i + 3j + 2k and b = i + 2j - k.

59

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES 2.5.1 perform the vector product in two and three dimensions.

2.5 VECTOR PRODUCT (CROSS PRODUCT)

Example : Notes

Given a = i + 3 j − 7k and If a = a1i + a2 j + a3k and
b = b1i + b2 j + b3k
b = −4i + j + 2k a  b = a b sin 
Find a  b
in direction perpendicular to a and b , so that
Solution : a , b and a  b form a right-handed set.

i jk

ab = 1 3 −7

−4 1 2 Also, in matrix form, jk

3 −7 −j 1 −7 +k 1 3 i
=i 2 -4 1
12 a  b = a1 a2 a3
-4
b1 b2 b3
= i(6 − (− 7))− j(2 − 28)+ k(1− (−12))

= 13i + 26 j + 13k

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

→→ →→

1. If A = 2i + 4 j − 3k and B = i + 3 j + 2k , determine A B .

2. Given a = (1,0,−1), b = (1,−1,2). Find ab .

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

3. Find F1  F2 for the following pairs of force.
a) F1 = 4i − 5 j and F2 = 2i + 3 j .

b) F1 = −i + 20 j and F2 = 400i +10 j .

c) F1 = 5i + 4 j + 3k and F2 = i + 30 j − 8k .

d) F1 = i + 40 j and F2 = 5i + j − 5k .

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 2: VECTOR

TUTORIAL 2.5
1. If a = 2i + 3j and b = - 3i – 2j, find a x b.

2. Given vector c = - 8i + 2j and d = 2i + 3j. Find c x d .

3. Find the vector product of ⃗ = 3 − 4 + 2 ⃗ = 2 + 5 − .

4. Given the vectors a = i − 2 j + 3k and b = −2i + 3 j − k . Find

a. a  b
b. b  a

5. Given that a = 2 j − k and b = i − 3 j . Find vector product for vector a and b .
~~ ~~


6. Find the vector product of P = 2i + 4 j + 4k and Q = i + 5 j − 2k

63

UNIT 3

Content

3.1 Introduction of Differentiation

3.1.1 Differentiation from first principle

3.2 Derivative of some standard functions

3.2.1 Derivative of a constant function

3.2.2 Derivative of n
x

3.2.3 Derivative of trigonometric functions

3.2.4 Derivative of exponential function

3.2.5 Derivative of ln x

3.2.6 Logarithmic differentiation

3.3 Technique of differentiation

3.3.1 The constant multiplication rule

3.3.2 Sum rule or difference rule

3.3.3 Differentiate product of functions using Product rule

3.3.4 Differentiate quotient of functions using Quotient rule

3.3.5 Differentiate composite functions using Chain rule

3.4 Higher order derivatives

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES 1 Find the differential of a function from first principles.(C2)
2 Differentiate algebraic functions or polynomials using the basic rule
of differentiation or formula.(C3)
3 Differentiate the trigonometric, logarithmic and exponential functions
by using appropriate methods.(C3)
4 Differentiate product of functions using Product Rule. (C3)
5 Differentiate product of functions using Quotient Rule. (C3)
6 Differentiate composite functions using Chain Rule. (C3)
7 Solve the higher order derivatives, differentiate the previous
derivative’s function.(C3)

3.1 INTRODUCTION OF DIFFERENTIATION

y
y=f(x)
Q

Tangent at P

P(x,y)
y

0x x

The above figure shows the points P(x,y) and Q( x + x,y + y ) which are two neighboring points on the
curve y = f(x).
x is known as delta x while y is known as delta y, x and y represents a small increase in x and a
small increase in y respectively.
Gradient of PQ = vertical dis tance

horizontal dis tance

65

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

= (y + y) − y = x
(x + x) − x y

As x become smaller and approaches 0 (x → 0) , point Q is getting nearer and nearer to P until the

chord PQ becomes the tangent to the curve y = f(x) at point P. Since y represent the gradient of the

x

chord PQ, then lim y  represent the gradient of the tangent to the curve y = f(x) at point P.

x→0 x 

It can be concluded that if dy = lim y  then dy represents the gradient of the curve y = f(x) at a certain
dx x→0 x  dx

point.

Therefore

dy = f' (x) = lim f(x + x)

dx x→0 x

The result above is the definition of the derivative of a function f. This process is called differentiation
from first principles.

66

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.1 INTRODUCTION OF DIFFERENTIATION
3.1.1 DIFFERENTIATION FROM FIRST PRINCIPLE

Example :

Find from first principle, the derivative of x2 NOTES

Solution: Derivative from first principle

f (x) = x2 f (x) = lim f (x +  x) − f (x)
 x→0 x

f (x +  x) = (x +  x)2 = x2 + 2( x).x + ( x)2 df = lim  f
dx  x→0  x
 f (x) = lim x2 + 2( x.)x + ( x)2 − x2
 x→0 x

= lim 2( x.)x + ( x)2 = lim 2x +  x
 x→0 x  x→0

= 2x

Using the first principle find the differential of .

67

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.2 DERIVATIVE OF SOME STANDARD FUNCTIONS

DERIVATIVE OF FORMULAE EXAMPLES
3.2.1 A CONSTANT
If f (x) = c , d (c) = 0 a) d (4) = 0
FUNCTION dx dx

DERIVATIVE OF c a constant b) d ( ) = 0
xn dx
3.2.2 If f (x) = xn ,
d (x n ) = nx n−1 a) d x 5 = 5x 5−1 = 5x 4
dx dx
n is a positive integer
b) d 1 = 1 x− 1
2
x2
dx 2

DERIVATIVE OF d (sin x) = cos x
TRIGONOMETRIC dx
3.2.3 FUNCTIONS d (cos x) = − sin x
dx
d (tan x) = sec2 x
dx

3.2.4 DERIVATIVE OF d (e x ) = e x
EXPONENTIAL dx
d (ln x) = 1
FUNCTION dx x

3.2.5 DERIVATIVE OF
ln x

DERIVATIVE OF ( )d ax = ax ln a
3.2.6 a x
dx
( a A CONSTANT)

68

Differentiate the following: DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION
(a) y = 1 (b) y = x6
2 (c) y = x3

(d) y = 5 x (e) y = 1 ( f ) y = x−5
x2

(g) y = ln x (h) y = sin x (i) y = cos x

( j) y = tan x (k) y = ex (l) y = 4x
69
\

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.3 TECHNIQUE OF DIFFERENTIATION
3.3.1 THE CONSTANT MULTIPLICATION RULE

Example : NOTES

Differentiate the following. d (cf (x)) = cf (x)

(a) d 6x 5 = 6(5)x 5−1 = 30x 4 dx
c is a constant
dx f is a differentiable function

(b) d 1 = 5 1  x− 1 = 5 x−1
2 2
5x 2
dx  2  2

(c) d (5 sin x) = 5 cos x

dx

(d) d 4e x = 4e x
dx

(e) d 4 ln x = 4 1 
dx  x 

Differentiate each of the following with respect to x .

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.3 TECHNIQUE OF DIFFERENTIATION
3.3.2 SUM RULE OR DIFFERENCE RULE

Example : NOTES

Differentiate each of the following with respect to x . If f (x) = u(x)  v(x) , and if
(a) f (x) = 2x4 − 2x3 + 3x−2 u(x) and v(x) exist, then,

f (x) = 8x3 − 6x2 − 6x−3 d (u  v) = u(x)  v(x)

(b) f (x) = 2sin x + ex dx
f / (x) = 2 cos x + ex
Note: u(x) means d (u)
(c) f (x) = 4x3 + 2tan x
f / (x) = 12x2 + 2 sec2 x dx

v(x) means d (v)

dx

Differentiate each of the following with respect to

71

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.3 TECHNIQUE OF DIFFERENTIATION
3.3.3 DIFFERENTIATE PRODUCT OF FUNCTIONS USING
PRODUCT RULE

Example : NOTES

Differentiate each of the following with respect to x . Product Rule

( )(a) f (x) = x2 + 2 (x + 3) If f (x) = u(x)v(x) , and if u(x) and
v(x) exist
Solution :
d u(x)v(x) = u(x)v(x) + v(x)u(x)
( )f (x) = x2 + 2 d (x + 3)+ (x + 3) d (x2 + 2) dx
dx dx
= (x2 + 2)(1)+ (x + 3)(2x) or
= x2 + 2 + 2x2 + 6x
= 3x2 + 6x + 2 y = uv, dy = u dv + v du
dx dx dx

Differentiate each of the following with respect to

72

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Example :

Differentiate each of the following with respect to(c) f(x)=x2(x+1) .

( )(b) f (x) = ex 2x2 + 3

Solution :

( ) ( )f / (x) = ex d 2x2+3 + 2x2 + 3 d ex
dx dx
( )= e x (4x)+ 2x2 + 3 e x
= 4xe x + 2x 2e x + 3e x

Differentiate each of the following with respect to

73

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Example :

Differentiate each of the following with respect to .

(c) f (x) = ex cos x

Solution :

f / (x) = e x d cos x + cos x d e x

dx dx

= ex (− sin x) + cos x.ex

Differentiate each of the following with respect to

74

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.3 TECHNIQUE OF DIFFERENTIATION
3.3.4 DIFFERENTIATE QUOTIENT OF FUNCTIONS USING
QUOTIENT RULE

Example : NOTES

Differentiate each of the following with respect to . Quotient Rule
(a) f (x) = 3x + 2
If f (x) = u(x) , and if u(x) and
2x2 +1 v(x)

(2x2 +1) d (3x + 2) − (3x + 2) d (2x2 +1) v(x) exist

f (x) = dx dx d  u  = v(x)u(x) − u(x)v(x)
dx  v 
( )2x2 + 1 2 v(x)2

= (2 x 2 + 1)(3) − (3x + 2)(4x) or
( )2x2 + 1 2

( )= − 6x2 − 8x + 3 y=u, dy v du − u dv
2x2 +1 2 v dx dx
=
dx v2

Differentiate each of the following with respect to

75

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Example :

Differentiate each of the following with respect to .

(b) f (x) = e x

cos x

Solution:

( )cos x d e x − e x d cos x
f / (x) = dx dx

(cos x)2

= cos x.e x − e x (− sin x)

cos2 x

= cos xe x + e x sin x
cos2 x

Differentiate each of the following with respect to .

76

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.3 TECHNIQUE OF DIFFERENTIATION
3.3.5 DIFFERENTIATE COMPOSITE FUNCTIONS USING CHAIN
RULE

Example : NOTES

Find dy when y = (3x + 4)5 Chain Rule (Leibniz Notation)
y = f (u) and u = g(x) , are both
dx differentiable functions then
dy = dy . du
Solution: dx du dx

Basic standard form is y = x5  dy = 5x4 or
dx
If y = f g(x)
Replace the single x with (3x + 4)
then, dy = f (g(x))g (x)
 dy = 5(3x + 4)4 d (3x + 4)
dx
dx dx

= 5(3x + 4)4 (3)
= 15(3x + 4)4

Find :

77

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Example :

Differentiate the following with respect to x .
ln(3x2 )

Solution:

Let y = ln(3x2 ) and u = 3x2

Then, y = ln u  dy = 1 and du = 6x
du u dx

 dy = dy  du
dx du dx

= 1 6x = 1 6x = 2
u 3x2 x

or simply, dy = 1 d (3x2 )
dx 3x2 dx

= 1 6x = 2
3x2 x

Differentiate the following with respect to .

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Example :

Differentiate the following with respect to x .
sin 4x

Solution:

Let y = sin 4x and u = 4x

Then, y = sin u → dy = cosu and du = 4
du dx

 dy = dy du = cosu.4 = cos 4x.4 = 4cos 4x
dx du dx

or simply

dy = cos 4x d 4x = cos 4x.4 = 4cos 4x
dx dx

Differentiate the following with respect to .

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Example :

Differentiate the following with respect to .
y = e3x ln 5x

Solution:

y = e3x ln 5x

dy = e3x d ln 5x + ln 5x d e3x (Product rule)
dx dx dx (Chain rule)

= e3x 1 d 5x + ln 5x.e3x d 3x
5x dx dx

= e3x 1 .5 + ln 5x.e3x .3
5x

= e3x  1 + 3ln 5x 
x 

Differentiate the following with respect to .

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

3.4 HIGHER ORDER DERIVATIVES

Example : NOTES

Given y = 4x3 − 5x5 , Find dy , d 2 y and d 3 y . The following notations are often
dx dx2 dx3 encountered in engineering problems:

Solution: dy = y = y
dx
dy = y = 12x2 − 25x4 d 2 y = y = y
dx dx3
y, y, y, y(n) is known as the prime
d 2 y = y = 24x + 100 x3
dx2 notation

d3y = y = 24 − 300 x2 y, y, y,  y(n) is known as the Newton’s
dx3
dot notation

In general, if y = f (x) , the nth derivative is
written as d n f or f (n) (x)

dx n
Similarly, d n y = y (n) (x)

dx n

Find and of the following.

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Example :

Find the first and the second derivative of the function y = sin 4x .

Solution:

dy = 4 cos 4x
dx
d 2 y = −16 sin 4x
dx2
Find the first and the second derivative of the following.

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Find d 2 y for the following.
dx2

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

Find d 3 y for the following.
dx3

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

TUTORIAL : DIFFERENTIATION

dy y = x2 tan x ( )Ans : x x sec2 x + 2 tan x

1.Find :

dx

(a)

(b) y = sin 3x Ans : (x + 1)3cos 3x − sin 3x
x +1
(x + 1)2

(c ) y =  x2 − 2  2 20x(x 2 − 2)
2x2 +1
Ans :

(2x 2 + 1)3

(d) y = ln(sin 2x) Ans: 2 cot 2x
(e) y = ln(e4x )
(f) y = 4e x − x Ans: 4
(g) y = e x Ans: 4e x − 1

x Ans: xe x − e x
x2

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 3: DIFFERENTIATION

2. Differentiate the following with respect to x. 3x

x b. f(x) =

a. f(x) = 5x − 2
2x −1
2x +1
5x2 d. f(x) =

c. f(x) = 3x + 2
x +1 x2 −1

3x + 5 f. f(x) =
1− 3x
e. f(x) =

1− x

3. Differentiate the following with respect to x.

a. f(x) = (x −1)3 b. f(x) = (2x + 1)5

c. f(x) = 3(2x −1)4 1

5 d. f(x) =

e. f(x) = (2x + 1)2 x−2
−3

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

dy b. y = 2x3 (x −1)2

4. Find

dx

a. y = x3 (x + 1)2

c. y = (3x −1)(2x +1)3 d. y = (x + 2)(3x −1)4

( )e. y = 3x3 x2 − 5 3 f. y =

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UNIT 4 DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

https://www.mathsisfun.com/calculus/integration-introduction.html

CONTENT
4.1 INTEGRATION AS A REVERSE FUNCTION OF DIFFERENTIATION
4.2 INDEFINITE INTEGRAL OF FUNCTION
4.3 TECHNIQUES OFINTEGRATION
4.4 DEFINITE INTEGRAL

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES 1. Find the integrals of functions by considering integration as a reverse of

differentiation.

4.1 DIFFERENTIATION

Integration is the reverse process of differentiation. The process of obtaining from y (a function



of a) is known as differentiation. Hence, the process of obtaining y from is known as integration.



Integration of y with respect to x, is denoted by ∫ ( ) . The symbols ∫ ( ) denote the
integral of f(x) with respect to the variable x.

https://www.mathsisfun.com/calculus/integration-introduction.html

88

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES 2. Integrate indefinite integrals by rule of integration.

4.2

= ( ) ∫ ( ) Example
+
+1 ∫ 3  = 3 +

+ 1 + ∫ 5  = 6 +
+ 6
provided ≠ | + | +
1 1
1 | + | + ∫ =   +

1 ( + ) +1 1
+ ( + 1) + ∫ + 3 = | + 3| +
1
+ 11
∫ 4 − 5 = 4 |4 − 5| +
(Refer 4.3 Substitution Type II)
∫(4 + 5)7 = (4 + 5)8
( + ) 4(8)

(Refer 4.3 Substitution Type I) (4 + 5)8
= 32

+ ∫ = +

1 + ∫ −3 = − 1 −3 +
  3

− + ∫   = − +

1 1
− + ∫ 3   = − 3 3 +

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DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

∫   = +
  1
2 + 1
 2   ∫ 3   = 3 3 +

′( ) 1 ∫ 2   = +
∫ ( ) +
∫ 2 4   = 1 4 +
(Refer 4.3 Substitution Type III) 4

| ( )| + ∫   = +

Let ( ) = 1 +
⇒ ′( ) =

∴ ∫ 1 + = |1 + | +

( ) ∫ ( ) + ∫ 6 = 6 +

[ ( ) ± ( )] ∫ ( )  ± ∫ ( )  ∫ 3 + = 4 − +
4
1
2 + 2 1 −1 ∫ 36 1 2 = 1 −1 +
+ 6 6
1
√ 2 − 2 −1 ∫ √9 1 2 = −1 +
− 3
1
2 − 2 1 + 4 1 2 dx = 1 |2+ | +
2 | − | −x 4
2−

90

EXERCISE 4.2 DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION
1. Find the following indefinite integral.
(c) ∫ 3  
(a) ∫ 5  (b) ∫ 1  
3

(d) ∫ 7  (e) ∫ 5 4  (f) ∫ 2 2 
3

(g) ∫ −6 4  (h) ∫ 12 3  (i) ∫ 1
3

(k) ∫ 1 (l) ∫( 2 + 2 + 10) (m) ∫ +1
√ 3

91

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

2. Find the following indefinite integral.

(a) ∫ 2 (b) ∫ 5 (c) ∫ 1 

+5

(d) ∫ 4 (e) ∫ 1 (f) ∫ 4
1+ 2 3

(g) ∫ 1  (h) ∫ 2 (i) ∫ 1
7 +3 3+4
3 +2

(k) ∫ 3 (l) ∫ 5 (m) ∫ 4
5 −2 4−3 1−

92

3. Find the following indefinite integral. DUM20132 ENGINEERING MATHEMATICS 2
(a) ∫( + 2)8 UNIT 4: INTEGRATION

(b) ∫(2 + 3)4

(c) ∫(3 − 7)5 (d) ∫(1 − 5 )6

(e) ∫(3 − 6 )−4 (f) ∫(1 − )−12

93

4. Find the following indefinite integral. DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION
(a) ∫ 7 (b) ∫ 5
(c) ∫ 4 2

(d) ∫ 1 5 (e) ∫ −10 (f) ∫ 3
2 3

5. Find the following indefinite integral.

(a) ∫ 5   (b) ∫ 3 3   (c) ∫ 6  

(d) ∫ 5 2   (e) ∫ 1 (f) ∫ 2 2  
2

94

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

(g) ∫ (2 − 1 2 ) (h) ∫( 3 + 3 )  (i) ∫(2 3   − 5 ) 
2 2

6. Find the following indefinite integral.

(a) ∫ (b) ∫ −2
+3 2 +3

7. Find the following indefinite integral.

(a) ∫ 1 (b) ∫ 2
16+ 2 2+1

95

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

(c) ∫ 1 (d) ∫ 3
√81− 2 √64− 2

(e)∫ 1 2 (f) ∫ 1 2
49− 100−

96

DUM20132 ENGINEERING MATHEMATICS 2
UNIT 4: INTEGRATION

TUTORIAL 4.2 (b) ∫ 6
1. Find the following indefinite integral. 4
(a) ∫ − 25

(c) ∫(3 2 − 4 3) (d) ∫ (3 − 4 + 6 3)
2

(e) ∫(1 + 4 − 6 2) (f) ∫ ( + 1 2)

97