UNIT
I
Differential Calculus-I
1.1
INTRODUCTION
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.
dy
d 2y
, second order derivative denoted by
, third
dx
dx 2
The first order derivative of y denoted by
order derivative by
d 3y
dx 3
and so on. Thus by differentiating a function y = f (x), n times, successively,
dny
or D n (y) or yn(x). Thus, the process of finding
dx n
the differential coefficient of a function again and again is called Successive Differentiation.
we get the nth order derivative of y denoted by
1.2 nth DERIVATIVES OF SOME STANDARD FUNCTIONS
Below we obtain formulas for the nth order derivatives of some standard functions.
(1) nth derivative of eax
Let y = eax. Then by differentiating y successively, we obtain
y1 =
y2 =
y3 =
dy
= ae ax ,
dx
d2 y
dx
2
d3y
3
= a2 eax
= a3eax …
dx
...... ..........................
...... ..........................
yn =
d ny
dx
n
= a n e ax
2
A TEXTBOOK OF ENGINEERING MATHEMATICS–I
Thus, we have the formula
D n (e ax ) = a n e ax
...(1)
In particular,
Dn (e x ) = e x
(2) nth derivative of log (ax + b)
...(2)
y = log(ax + b). Then we find, by successive differentiation
Let
y1 =
y2 =
y3 =
y4 =
dy
a
=
dx ax + b
d 2y
dx 2
d 3y
dx 3
d4y
= (−1)
a2
(ax + b) 2
= (−1) (−2) ⋅
a3
(ax + b) 3
= (−1)(−2)(−3)
a4
(ax + b)4
dx 4
... ............................................
... ............................................
yn =
dny
dx n
= (−1) n −1 ⋅
(n − 1)! a n
(ax + b) n
Thus, we have the formula
D n [log(ax + b)] =
(−1)n −1 ⋅ ( n − 1)! an
(ax + b) n
...(3)
In particular,
Dn [logx ] =
(−1)n −1 ⋅ ( n − 1)!
xn
(3) nth derivative of (ax + b)m
y = (ax + b) m
Differentiating successively, we get
Let
y1 =
y2 =
dy
= ma (ax + b) m −1
dx
d2y
dx
2
= m(m − 1) ⋅ a2 (ax + b) m −2
...(4)
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DIFFERENTIAL CALCULUS-I
y3 =
d 3y
3
= m(m − 1) (m − 2)a 3 (ax + b) m− 2
dx
... .....................................................
... .....................................................
yn = m(m − 1) (m − 2)L(m − n + 1)a n (ax + b) m − n
...(5)
This formula is true for all m.
Following are some particular cases.
Case (i): Suppose m = n (a +ve integer)
In equation (5) becomes,
Dn [(ax + b) n ] = n n − n − ⋅ an ax + bn − n
= n! an
...(6)
In particular,
D n ( x n ) = n!
Case (ii): Suppose m is a positive integer and m > n. Then formula (5) becomes
Dn [(ax + b) m ] =
=
...(7)
m(m − 1)… (m − n + 1)(m − n)(m − n − 1)… 2 ⋅ 1 n
⋅ a (ax + b) m − n
(m − n)(m − n − 1)… 2 ⋅ 1
m!
a n (ax + b) m − n
( m − n) !
...(8)
m!
xm −n
(m − n)!
...(9)
In particular,
Dn (x m ) =
Case (iii): Suppose m is a positive integer and n > m. From (6) we note that
Dn [(ax + b) m ] = m! am
In differentiate further, the right-hand side gives zero.
Thus,
Dn [(ax + b) m ] = 0
if
n>m
...(10)
In particular,
for n > m
Dn (x m ) = 0
Case (iv): Suppose m = – 1, in this case formula (5) becomes,
...(11)
 1 
−1− n
n
Dn 
 = (−1) (−2) (−3)K(−n)a (ax + b)
ax
b
+


=
(−1)n ⋅ n! an
(ax + b)n +1
...(12)
4
A TEXTBOOK OF ENGINEERING MATHEMATICS–I
Case (v): Suppose m is a negative integer. Let us get m = – p, so that p is a positive integer. Then
formula (5) becomes,

 (−1)n ⋅ p( p + 1)… ( p + n − 1) ⋅ a n
1
Dn 
=
P
(ax + b) p + n
 (ax + b) 
= (−1)n ⋅
=
1 ⋅ 2… p − 1 p( p + 1)…( p + n − 1) an
⋅
1 ⋅ 2… p − 1
(ax + b) p+ n
an
(−1)n ⋅ ( p + n − 1)!
⋅
( p − 1)!
(ax + b) p + n
...(13)
In particular
( p + n − 1)! 1
 1 
⋅
Dn  p  = (−1)n ⋅
( p − 1)! x p + n
x 
(4) nth derivative of cos (ax + b)
y = cos(ax + b).
Let
Differentiating this successively, we get
y1 =
y2 =
...(14)
dy
= − a sin(ax + b) = a cos(ax + b + π / 2)
dx
d2y
dx
2
= −a2 sin(ax + b + π / 2)
= a 2 cos(ax + b + 2π / 2)
y3 =
d3y
dx
3
= −a3 sin(ax + b + 2π / 2)
= a3 cos(ax + b + 3π / 2)
..... ................................
..... ................................
yn =
d ny
dx n
Thus, we obtain the formula
= an cos(ax + b + nπ/2)
D n [cos(ax + b)] = a n cos(ax + b + nπ / 2)
In particular,
...(15)
D n (cosx ) = cos(x + nπ / 2)
(5) nth derivative of sin(ax + b)
y = sin(ax + b)
Let
...(16)
5
DIFFERENTIAL CALCULUS-I
Differentiating successively, we get
dy
= y1 = a cos(ax + b) = a sin(ax + b + π / 2)
dx
d2 y
dx
2
= y2 = a 2 cos(ax + b + π / 2) = a2 sin(ax + b + 2π / 2)
d 3y
= y3 = a3 cos(ax + b + 2π/2) = a3 sin(ax + b + 3π/2)
dx 3
....... .....................................................................
....... .....................................................................
dny
n
dx
Thus, we have the formula,
= yn = a n sin(ax + b + nπ / 2)
Dn [sin(ax + b)] = an sin(ax + b + nπ/2)
In particular,
...(17)
Dn (sin x) = sin(x + nπ / 2)
(6) nth derivative of eax sin (bx + c)
...(18)
y = e ax sin(bx + c)
Let
dy
= aeax sin(bx + c) + beax 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
y1 =
a = k cos α,
b = k sin α
α = tan−1 (b / a)
So that k = a2 + b2 ,
Thus,
y1 =
dy
= e ax [k (cos α)sin(bx + c) + k (sin α)cos(bx + c)]
dx
= keax sin(bx + c + α)
Therefore,
y2 =
d2y
dx
2
= k[aeax sin(bx + c + α) + beax cos(bx + c + α)]
= keax [k (cos α)sin(bx + c + α) + k (sin α) cos(bx + c + α)]
= k 2e ax sin(bx + c + 2α)
6
A TEXTBOOK OF ENGINEERING MATHEMATICS–I
Proceeding like this, we obtain
yn =
d ny
dx n
Thus, we have the formula
= k n e ax sin(bx + c + nα)
D n [e ax sin(bx + c)] = (a2 + b2 )n / 2 sin{bx + c + n tan −1 (b/a)}
In particular,
...(19)
Dn [ex sin x ] = 2 n / 2 ⋅ e x sin(x + nπ / 4)
(7) nth derivative of eax cos(bx + c)
...(20)
y = e ax cos(bx + c)
Let
Then
y1 =
dy
= aeax cos(bx + c) − beax sin(bx + c)
dx
= eax [k (cos α)cos(bx + c) − k (sin α)sin(bx + c)]
= keax cos(bx + c + α)
Therefore,
y2 =
d2y
dx 2
= k[aeax cos(bx + c + α) − beax sin(bx + c + α)]
= keax [k (cos α) cos(bx + c + α) − k (sin α)sin(bx + c + α)]
= k 2e ax cos(bx + c + 2α)
Proceeding like this, we obtain
yn =
d ny
dx n
Thus, we have the formula
= k n e ax cos(bx + c + nα)
Dn [eax cos(bx + c)] = (a2 + b2 )n / 2 eax cos{bx + c + n tan−1 (b/a)}
In particular,
...(21)
D n (e x cos x) = 2n / 2 ex cos( x + nπ/4)
(8) nth derivative of amx
...(22)
y = a mx
Let
Taking logarithm on both sides,
log y = mx log a
Differentiating w.r.t. ‘x’, we get
1 dy
⋅
= m log a ⋅ 1
y dx
7
DIFFERENTIAL CALCULUS-I
y1 =
y2 =
dy
= (m log a) ⋅ y
dx
d2y
dx 2
= m log a ⋅ y1
= (m log a)(m log a) ⋅ y
= (m log a) 2 y
y3 =
d 3y
dx 3
= (m log a)2 ⋅ y1
= (m log a) 3 y
..... ...................
..... ...................
yn =
d ny
dx n
= (m log a) n ⋅ y
Dn [amx ] = (m log a)n ⋅ amx
WORKED EXAMPLES
Example 1: Find the nth derivative of the following functions:
(i) sin3x
(ii) cos4x.
Solution:
( i) y = sin3 x
where
sin3 x =
\
D n (sin3 x) =
1
(3 sin x − sin 3x )
4
3 n
1
D (sin x) − Dn (sin 3x)
4
4
nπ  3n 
nπ 
3 
= sin x +  − sin 3x + 
4 
2  4
2 

(ii) y = cos4x
where
cos2 x =
1
(1 + cos 2 x )
2
so that
cos4 x =
1
(1 + cos 2 x) 2
4
1
= [1 + cos2 2 x + 2 cos 2 x ]
4
[Using the formula (15)]
8
A TEXTBOOK OF ENGINEERING MATHEMATICS–I
where
cos2 2 x =
\
cos4 x =
=
cos4 x =
1
(1 + cos 4 x )
2
1 1

1 + (1 + cos 4 x ) + 2 cos 2 x 
4  2

1 1 1
1
+ + cos 4x + cos 2 x
4 8 8
2
3 1
1
+ cos 2 x + cos 4 x
8 2
8
1 n
n 3 1 n
Dn (cos4 x ) = D   + D (cos 2 x) + D (cos 4 x )
8
 8 2
\
=0+
=
nπ  4n
nπ 
2n


cos  2x +
cos  4 x +
+

2
2  8
2 


n π  4n
nπ 
2n


cos  2 x +
cos  4 x +
+

2
2  8
2 


Example 2: Find the nth derivative of the following:
( i) sin h 2 x sin 4 x
(ii) e − x sin 2 x
2
3
(iii) e x cos x
(iv)
e − x sin x cos 2 x
(v) e ax cos2 x sin x
Solution:
1
( i) sin h 2 x sin 4 x = (e2 x − e −2 x ) ⋅ sin 4 x
2
1
sin h 2 x sin 4 x = [e 2 x sin 4 x − e − 2 x sin 4 x ]
2
1
D n (sin h 2 x sin 4 x ) = [ D n (e2 x sin 4 x) − D n (e −2 x sin 4 x )]
2
1
= [20 n / 2 {e2 x sin(4x + n tan−1 2) − e−2 x sin(4 x − n tan −1 2)}]
2
Using the formula (19)
sin h 2x =
2x
−2 x
e −e
2
9
DIFFERENTIAL CALCULUS-I
(ii) We have sin 2 x =
1
(1 − cos 2x )
2
Therefore,
Dn (e − x sin2 x) =
1 n −x
1
D (e ) − D n (e − x cos2 x )
2
2
1
1
= e− x (−1)n − [5n / 2 e− x cos(2x + n tan−1 (−2))]
2
2
1
= e− x [( −1) n − 5n / 2 cos(2 x − n tan −1 (2))]
2
( iii) We have cos3 x =
1
(cos 3x + 3 cos x )
4
D n (e 2 x cos3 x ) =
3
1 n 2x
D (e cos 3x) + D n (e 2 x cos x )
4
4
1
= [(22 + 32 )n / 2 e2 x cos{3x + n tan−1 (3/ 2)}]
4
3
+ [(22 + 12 ) n / 2 e 2 x cos{x + n tan−1 (1 / 2)}]
4
=
1 2x n / 2
e [13 cos{3x + n tan −1 (3 / 2)}
4
+ 3(5) n / 2 ⋅ cos{x + n tan −1 (1 / 2)}]
( iv) We have sin x cos 2 x =
1
(sin 3x − sin x )
2
Therefore,
Dn (e − x sin x cos 2 x) =
1 n −x
1
D (e sin 3x ) − D n (e − x sin x )
2
2
1
= [((−1) 2 + 32 ) n / 2 e − x sin{3x + n tan−1 (−3)}]
2
1
− [((−1) 2 + 12 ) n / 2 e − x sin{x + n tan −1 (−1)}]
2
1
= e − x [10 n / 2 sin(3x − n tan −1 3) − 2 n / 2 sin(x − nπ / 4)]
2
(v) We note that
1
cos2 x sin x = (1 + cos 2 x ) ⋅ sin x
2
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A TEXTBOOK OF ENGINEERING MATHEMATICS–I
1
1
= sin x + ⋅ 2 sin x cos 2x
2
4
1
1
= sin x + (sin 3x − sin x)
2
4
=
D n (e ax cos2 x sin x ) =
\
=
1
1
sin x + sin 3x
4
4
1 n ax
1
D (e sin x ) + D n (e ax sin 3x)
4
4
1 ax 2
e [( a + 1) n / 2 sin{x + n tan −1 (1/ a)}
4
+ (a 2 + 9) n / 2 sin{3x + n tan−1 (3 / a)}]
Example 3: Find the nth derivative of the following:
x+3
( x − 1)( x + 2)
( i)
( ii)
x2
( x + 2)(2x + 3)
Solution:
( i) y =
x+3
( x − 1)( x + 2)
By partial fractions
x+3
A
B
=
+
( x − 1)(x + 2) x − 1 x + 2
Then
...(1)
x + 3 = A( x + 2) + B( x − 1)
x =1 ⇒ A = 4/3
Taking
x = −2 ⇒ B = −1/ 3
Equation (1), becomes
x+3
4
1
1 1
= ⋅
−
( x − 1)(x + 2) 3 ( x − 1) 3 ( x + 2)
x+3

 4 n 1  1 n 1 
Dn 
 = D  x− − D  x+ 
x
x
(
1
)(
2
)
−
+
1 3
2



 3
=
4 (−1) n ⋅ n ! 1 (−1) n ⋅ n !
−
3 ( x − 1) n +1 3 ( x + 2) n +1


1
4
1
= ( −1)n ⋅ n ! 
−

1
1
n
+
n
+
3
( x + 2) 
 ( x − 1)
[Using the formula (12)]