Differential Calculus-I - New Age International

1.1 INTRODUCTION
UNIT I
Differential Calculus-I
Calculus is one of the more beautiful intellectual achievements of human being. The mathematical
study of change motion, growth or decay is calculus. One of the most important idea of differential
calculus is derivative which measures the rate of change of given function. Concept of derivative is
very useful in engineering, science, economics, medicine and computer science.
2
dy d y
The first order derivative of y denoted by , second order derivative denoted by , third
dx
2
dx
3
d y
order derivative by 3 and so on. Thus by differentiating a function y = f (x),
n times, successively,
dx
n
d y
we get the nth order derivative of y denoted by n or D (y)
dx
n or yn (x). Thus, the process of finding
the differential coefficient of a function again and again is called Successive Differentiation.
1.2 nth DERIVATIVES OF SOME STANDARD FUNCTIONS
Below we obtain formulas for the nth order derivatives of some standard functions.
(1) (1) (1) nth derivative of eax Let y = eax . Then by differentiating y successively, we obtain
dy ax
y = = ae ,
1 dx
2
d y 2
y 2 = = ae
2
dx
3
ax
dy 3 ax
y3 = = ae …
3
dx
................................
................................
yn
n
d y
=
= a
n
dx
n
e
ax

2 A TEXTBOOK OF ENGINEERING MATHEMATICS–I
Thus, we have the formula
In particular,
( )
ax n D e
( )
x n D e
n ax
= a e
...(1)
x
= e
...(2)
(2) (2) nth derivative of log (ax + b)
Let y = log( ax + b).
Then we find, by successive differentiation
y 1
dy a
= =
dx ax + b
2
2
d y a
y2 = = ( −1)
2
2
dx
( ax + b)
3
3
d y
a
y3 = = ( −1)
( −2)
⋅
3
3
dx
( ax + b)
4 4
d y a
y4 = = ( −1)( −2)( −3)
4 4
dx ( ax + b)
... ............................................
... ............................................
yn
Thus, we have the formula
In particular,
D n
n
d y
= = ( −1)
n
dx
n−1
n−1n ( −1) ⋅( n−1)! a
[log( ax + b)]
=
n
( ax + b)
D [logx]
n
(3) (3) nth derivative of (ax + b) m
n−1
( −1) ⋅( n −1)!
=
n
x
Let y m
= ( ax + b)
Differentiating successively, we get
y1
y2
=
dy
= ma ( ax + b)
dx
2
( n −1)!
a
⋅
( ax + b)
m−1
d
y
2
= = m(
m −1)
⋅ a ( ax + b)
2
dx
n
n
m−2
...(3)
...(4)

DIFFERENTIAL CALCULUS-I 3
3
d y
3 m−2
y3
= = m(
m −1)
( m − 2)
a ( ax + b)
3
dx
........................................................
........................................................
yn
n m−n
( m −1)
( m − 2)
L ( m − n + 1)
a ( ax + b
...(5)
= m
)
This formula is true for all m.
Following are some particular cases.
Case Case ( (iiiii): ( ( ): Suppose m = n (a +ve integer)
In equation (5) becomes,
n
n
n n n
D [( ax + b)
] = nn ( 1) ( n 2) 1 a( ax b) −
− − ⋯ ⋅ +
n
= n! a
...(6)
In particular,
( )
n n D x = n!
...(7)
Case Case (
( (ii ii): ii ii ii ): Suppose m is a positive integer and m > n. Then formula (5) becomes
n
D [( ax + b)
m
]
m( m−1) …( m− n+ 1)( m−n)( m−n−1) …2⋅1
= ⋅ a ( ax+ b)
( m−n)( m−n−1) …2⋅1
m! n m − n
= a ( ax + b)
( m − n)
!
In particular,
( )
m n m! m − n
D x = x
( m − n)
!
Case Case ( (
( (iii iii): iii iii iii ): Suppose m is a positive integer and n > m. From (6) we note that
n m−n ...(8)
...(9)
n m
D [( ax + b)
] ! m
= m a
In differentiate further, the right-hand side gives zero.
Thus, n m
D [( ax + b)
] = 0 if n > m ...(10)
In particular,
( )
m n D x = 0 for n > m ...(11)
Case Case ( (
( (iv iv): iv iv iv ): Suppose m = – 1, in this case formula (5) becomes,
n ⎡ 1 ⎤
n
D ⎢ax + b⎥
= ( −1)
( −2)
( −3)
K(
−n)
a ( ax + b)
⎣ ⎦
n n
( 1) n! a
( ax b )
+
− ⋅
=
+
n 1
−1−n
...(12)

4 A TEXTBOOK OF ENGINEERING MATHEMATICS–I
Case Case ( (vvvvv): ( ): Suppose m is a negative integer. Let us get m = – p, so that p is a positive integer. Then
formula (5) becomes,
⎡ ⎤
⎢ ⎥
⎣ + ⎦
P
n 1 ( 1) pp ( 1) ( p n 1) a
D
p n
( ax b)
( ax b )
+
− ⋅ + … + − ⋅
=
+
n n
n 12 p 1 p( p 1) ( p n 1) a
( 1)
12 p 1 p n
( ax b )
+
⋅ … − + … + −
= − ⋅ ⋅
⋅ … − +
n n
( 1) ( p n 1)! a
( p 1)! ( ax b )
+
− ⋅ + −
= ⋅
− +
In particular
n ⎛ 1 ⎞ n ( p n 1)! 1
D ⎜ p ⎟ ( 1)
⎝ x ⎠ ( p 1)! p n
x +
+ −
= − ⋅ ⋅
−
(4) (4) nth derivative of cos (ax + b)
Let y = cos( ax + b).
Differentiating this successively, we get
y
y
p n
dy
y1
= =− asin( ax + b) = acos( ax + b+π/
2)
dx
2
dy
=
dx
2 2
3
dy
=
dx
3 3
2
=− a sin( ax+ b+π/
2)
2
= a cos( ax + b + 2π
/ 2)
3
=− a sin( ax+ b+
2 π/
2)
3
= a cos( ax+ b+
3 π/
2)
..... ................................
..... ................................
n
d y n
y n = = a cos( ax+ b+ nπ/2)
n
dx
Thus, we obtain the formula
n
...(13)
...(14)
D [cos( ax b)]
n
+ = a cos( ax + b + nπ
/ 2)
n In particular,
...(15)
D (cosx)
n = cos( x + nπ/
2)
...(16)
(5) (5) nth derivative of sin(ax + b)
Let y =
sin( ax + b)

DIFFERENTIAL CALCULUS-I 5
Differentiating successively, we get
d
dy
dx
2
dx
dx
Thus, we have the formula,
y
2
= y = a cos( ax + b)
= asin(
ax + b + π / 2)
1
2
= y = a cos( ax + b + π / 2)
= a sin( ax + b + 2π
/ 2)
2
3
d y
3
3 3
= y3 = a cos( ax + b + 2 π /2) = a sin( ax + b + 3 π/2)
dx
............................................................................
............................................................................
d
n
y
n
n
= y = a sin( ax + b + nπ
/ 2)
n
D [sin( ax b)]
n
n + = a sin( ax+ b+ nπ
/2)
In particular,
...(17)
D (sinx)
n = sin( x + nπ
/ 2)
...(18)
(6) (6) nth derivative of eax sin (bx + c)
Let y = sin( bx + c)
e ax
dy ax
ax
y 1 = = ae sin( bx + c)
+ be cos( bx + c)
dx
For computation of higher-order derivatives it is convenient to express the constants a and b in
terms of the constants k and a defined by
a = k cosα, b = ksinα
So that
Thus,
Therefore,
2 2 −1
k= a + b , α= tan ( b/ a)
dy ax
y 1 = = e [ k(cos α )sin( bx + c) + k(sin α )cos( bx + c)]
dx
2
ax
= ke sin( bx + c +α)
d y
y ax ax
2 = = k[ ae sin( bx + c +α ) + be cos( bx + c +α)]
2
dx
ax
= ke [ k(cos α )sin( bx+ c+α ) + k(sin α ) cos( bx+ c+α)]
2 ax
= k e sin( bx + c + 2α)
2

6 A TEXTBOOK OF ENGINEERING MATHEMATICS–I
Proceeding like this, we obtain
dx
Thus, we have the formula
D ax n
n
d y
y n = n ax
n = k e sin( bx + c + nα)
2 2 n /2 −1
= ( a + b ) sin{ bx+ c+ ntan ( ba / )}
...(19)
[ e sin( bx + c)]
In particular,
n x
D [ e sin x ] 2 sin( / 4)
2 /
= ⋅ e x + nπ
x n ...(20)
(7) (7) nth derivative of eax cos(bx + c)
Let y = cos( bx + c)
Then
Therefore,
dy
y 1 =
dx
2
e ax
= ae
ax
ax
ax
cos( bx + c)
− be sin( bx + c)
= e [ k(cos α )cos( bx + c) −k(sin α )sin( bx + c)]
= cos( bx + c + α)
ke ax
d y
y 2 = ax
ax
= k[
ae cos( bx + c + α)
− be sin( bx + c + α)]
2
dx
= k e cos( bx + c + 2α)
Proceeding like this, we obtain
n
d y
y n = n ax
= k e cos( bx + c + nα)
n
dx
Thus, we have the formula
n ax
D [ e cos( bx+ c)]
In particular,
ax
= ke [ k(cos α ) cos( bx+ c+α) −k(sin α )sin( bx+ c+α)]
2
ax
D x n n/2 x
2 2 n/2 ax
−1
= ( a + b ) e cos{ bx+ c+ ntan ( ba / )}
...(21)
( e cosx)
= 2 e cos( x+ nπ
/4)
...(22)
(8) (8) nth derivative of amx Let y mx
= a
Taking logarithm on both sides,
logy = mx loga
Differentiating w.r.t. ‘x’, we get
1 dy
⋅ = m
loga ⋅1
y dx

DIFFERENTIAL CALCULUS-I 7
dy
y 1 =
dx
2
= ( m loga)
⋅ y
d y
y 2 = = m loga ⋅ y
2
1
dx
= ( m loga)(
m loga)
⋅ y
3
= ( mloga)
d y
2
y 3 = = ( mlog a) ⋅y
3
dx
= ( mloga)
2
3
y
y
........................
........................
n
d y
y n = n
m a y
n = ( log ) ⋅
dx
n mx
[ ]
D a = ( mlog a) ⋅a
1
n mx
WORKED EXAMPLES
Example Example 1: 1: Find the nth derivative of the following functions:
Solution:
Solution:
(i) 3
y = sin x
(i) sin 3 x (ii) cos 4 x.
where 3
x
\ (sin )
3 D x
n
(ii) y = cos 4 x
where 2
x
so that 4
cos x
1
sin = ( 3sin
x − sin3x)
4
3 n 1 n
= D (sinx)
− D (sin3x)
4 4
=
3 ⎛ nπ
⎞ 3 ⎛ nπ
⎞
sin⎜
x + ⎟ − sin⎜3x
+ ⎟
4 ⎝ 2 ⎠ 4 ⎝ 2 ⎠
1
cos = ( 1 + cos2x)
2
1
= ( 1+
cos2x)
4
1 2
=
[ 1+
cos 2x
+ 2cos2x]
4
2
n
[Using the formula (15)]

8 A TEXTBOOK OF ENGINEERING MATHEMATICS–I
where cos 2x
2
\ 4
x
\
1
= ( 1 + cos4x)
2
1 ⎡ 1
⎤
cos = ⎢1
+ ( 1+
cos4x)
+ 2 cos2x
4
⎥
⎣ 2
⎦
=
1
4
+
1
8
1 1
+ cos4x
+ cos2x
8 2
4 3 1 1
cos x = + cos2x
+ cos4x
8 2 8
n 4 n ⎛ 3⎞
1 n 1 n
D (cos x ) = D ⎜ ⎟ + D (cos2x)
+ D (cos4x)
⎝ 8 ⎠ 2
8
n n
2 ⎛ nπ⎞ 4 ⎛ nπ⎞
= 0 + cos 2x cos 4x
2
⎜ +
2
⎟+ +
8
⎜
2
⎟
⎝ ⎠ ⎝ ⎠
n n
2 ⎛ nπ⎞ 4 ⎛ nπ⎞
= cos 2x cos 4x
2
⎜ +
2
⎟+ +
8
⎜
2
⎟
⎝ ⎠ ⎝ ⎠
Example Example 2: 2: Find the nth derivative of the following:
(i) sin h 2x
sin4x
(ii)
−x
2
e sin x
(iii) e x
(v)
Solution:
Solution:
x
(i)
1 2x
−2x
sinh
2x
sin4x
= ( e − e ) ⋅ sin4x
2
sin h2x x 3 2 cos (iv) e x x
x
e x
−
sin cos 2
ax
cos sin
2
1 2x
−2x
sin h2x sin4x
= [ e sin4x
− e sin4x]
2
n 1 n 2x n −2x
D (sinh 2x
sin4x)
= [ D ( e sin4 x) −D
( e sin4 x)]
2
2x −
−2x
e e
=
2
1 n/2 2x −1−2x −1
[20 { e sin(4x ntan2) e sin(4x ntan2)}]
=
2
+ − −
Using the formula (19)

DIFFERENTIAL CALCULUS-I 9
1
(ii) We have sin ( 1 cos2
)
2
2
x = − x
Therefore,
n −x 2 1 n −x 1 n −x
x = −
D ( e
sin
1
(iii) We have cos (cos3
3cos
)
4
3
x = x + x
n
D ( e
2x
3
)
D ( e ) D ( e cos2 x)
2 2
1 −x n 1 n/2 −x −1
= e ( −1) − [5 e cos(2x+ ntan
( −2))]
2 2
1 −x n n/2
−1
= e [( −1) −5cos(2x−ntan (2))]
2
cos x)
= D ( e
4
1 n 2x
2x
3 n
cos3x)
+ D ( e
4
cosx)
1 2 2 n/2 2x −1
[(2 3 ) e cos{3x ntan
(3/ 2)}]
= + +
4
3 2 2 / 2 2
−1
e x n
+ [( 2 + 1 ) cos{ x + n tan
4
1 2x n / 2
−1
= e [ 13 cos{ 3x
+ n tan
4
n / 2
−1
+ 3(
5)
⋅ cos{ x + n tan
1
(iv) We have sin x cos2x
= (sin3x
− sin x)
2
Therefore,
n −x
1
D ( e sin x cos2x)
= D ( e
2
(v) We note that
= [(( −1)
2
1 n
sin3x)
− D ( e
2
n −x
−x
+ 3
)
( 3/
2)}
( 1/
2)}]
sin x)
1 2 2 / 2 −
−1
e x n
− [(( −1)
2
= e
2
sin{ 3x
+ n tan
1 2 2 / 2 −
−1
+ 1 ) e sin{ x + n tan
x n
1 −x
n / 2
−1
n / 2
[ 10
sin( 3x
− ntan
2 1
cos xsinx =
( 1 + cos2x)
⋅ sin x
2
3)
− 2
( 1/
2)}]
( −3)}]
( −1)}]
sin( x − nπ
/ 4)]

10 A TEXTBOOK OF ENGINEERING MATHEMATICS–I
\ 2
D ( e cos xsin
x)
ax n
=
1 1
= sin x + ⋅ 2sin
x cos2x
2 4
1 1
= sin x + (sin3x
− sin x)
2 4
1 1
= sin x + sin3x
4 4
1
4
n
ax
D ( e
1 n
sinx)
+ D ( e
4
ax
sin3x)
1 ax 2 n /2 −1
= e [( a + 1) sin{ x+ ntan (1/ a)}
4
n / 2
−1
+ ( a + 9)
sin{ 3x
+ n tan ( 3/
a)}]
Example Example 3: 3: 3: Find the nth derivative of the following:
(i)
Solution:
Solution:
x + 3
( x −1)(
x + 2)
(i)
x + 3
y =
( x −1)(
x + 2)
By partial fractions
2
(ii)
x + 3
( x −1)(
x + 2)
A B
= +
x − 1 x + 2
Then x + 3 = Ax ( + 2) + Bx ( −1)
Taking x = 1 ⇒ A = 4/3
Equation (1), becomes
x =−2 ⇒ B =−1/3
x + 3 4 1 1 1
= ⋅ −
( x −1)(
x + 2)
3 ( x− 1) 3( x+
2)
2
x
( x + 2)(
2x
+ 3)
⎡ x + 3 ⎤
D ⎢
⎥
⎣(
x −1)(
x + 2)
⎦
n 4 n⎛ 1 ⎞ 1 n⎛
1 ⎞
= D D
3
⎜ −
x− 1
⎟
3
⎜
x+
2
⎟
⎝ ⎠ ⎝ ⎠
n n
4( −1) ⋅n! 1( −1) ⋅n!
= −
3 ( x − 1) 3(
x + 2)
n+ 1 n+
1
1 n ⎡ 4 1 ⎤
= ( −1) ⋅n! ⎢ −
3 n+ 1 n+
1⎥
⎣( x− 1) ( x+
2) ⎦
...(1)
[Using the formula (12)]