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Differential Calculus
To err is human, to admit superhuman, to forgive divine,
to blame it on others politics, to repeat unprofessional.
ANONYMUS
Calculus is a central branch of Mathematics, developed
from algebra and geometry.
It is built on two major complementary ideas, both of
which rely critically on the concept of limits.
The first is the differential calculus ( Part A ) , which is
concerned with the instantaneous rate of change of quantities
with respect to other quantities.
More precisely, the local behavior of functions which can
be illustrated by the slope of a function's graph.
The second is the integral calculus ( Part B ) , which studies
the accumulation of quantities, such as areas under a curve,
linear distance traveled, or volume displaced.
These two processes act inversely to each other, a fact
delivered conclusively by the Fundamental theorem of
calculus.
nth order derivatives of some standard functions:
1. y = eax
dy
y1 = = a eax
dx
d2y
ax
2 ax
y2 =
=
a.a
e
=
a
e
2
dx
.
.
yn = an eax.
2. y = amx where m is a positive integer.
y1 = D (amx) = m amx log a.
y2 = m log a . (m amx log a)
= (m log a)2 amx
y3 = (m log a)3 amx
.
.
yn = (m log a)n amx.
3. y = (ax + b)m, where m is a positive integer such that m > n .
y1 = m(ax + b)m-1.a
y2 = m(m – 1)(ax + b) m-2. a2
y3 = m(m – 1) (m- 2) (ax + b)m-3 a3
.
.
yn = m(m- 1) (m – 2) … [m – (n – 1)] (ax + b)m-n an.
1
4. y =
ax + b
Let us write y = (ax + b)-1
y1 = -1 (ax + b)-2.a = (-1)1 1! (ax + b)-2.a
y2 = (-1) (-2) (ax + b)-3 a2 = (-1)2 2! (ax + b)-3 . a2
y3 = (-1) (-2) (-3) (ax + b)-4 a3 = (-1)3 3! (ax + b)-4 a3
.
.
yn = (-1)n n! (ax + b) –(n+1) an
(- 1) n n !a n
yn =
(ax + b) n+ 1
5.
y = log (ax + b)
y1 = a(ax + b)-1
y2 = a(-1) (ax + b)-2.a = a2(-1)1 1! (ax + b)-2
y3 = a2(-1) (-2) (ax + b)-3 .a = a3(-1)2 2! (ax + b)-3
.
.
yn = an (-1)n-1 (n – 1) ! (ax + b)-n
(- 1) n- 1 (n - 1)!a n
yn =
(ax + b) n
6. y = sin (ax + b)
y1 = cos (ax + b). a
ép
y1 = a sin ê + ax +
êë2
ù
bú
ú
û
ép
y2 = a cos ê + ax +
êë2
ù
búa
ú
û
ép æp
= a sin ê + çç + ax +
êë2 è 2
2
æ p
= a sin çç2. + ax +
è 2
.
.
2
æp
yn = a sin ççn + ax +
è 2
n
ö
÷
b÷
÷
ø
ö
÷
b÷
÷
ø
öù
b÷
ú
÷
÷
øú
û
æp
Exercise: If y = cos (ax + b), prove that yn = a cos ççn + ax +
è 2
n
7.
y = eax sin (bx + c)
y1 = eax. b cos (bx + c) + aeax sin (bx + c),
= eax [b cos (bx + c) + a sin (bx + c)]
Put
a = r cos , b = r sin 
Then  = tan-1 (b/a) and
a2 + b2 = r2 (cos2  + sin2 ) = r2
ö
÷
b÷
÷
ø
y1 = eax [ r sin  cos (bx + c) + r cos  sin (bx + c)]
Note: sin (A + B) = sin A cos B + cos A sin B
y1 = r eax sin ( + bx + c)
Similarly we get,
y2 = r2 eax sin (2 + bx + c),
y3 = r3 eax sin (3 + bx + c)
.
.
yn = rn eax sin (n + bx + c)
where r = a 2 + b2 and q = tan -1 (b/a).
Exercise:
If y = eax cos (bx + c), yn = rn eax cos (n + bx + c),
where r = a 2 + b2 and q = tan -1 (b/a).
Examples:
1. Find the nth derivative of y = cos h2 3x
2
ée3 x + e- 3 x ù
ú
Solution: Write cos h2 3x = ê
êë 2
úû
1 6x
= (e + e-6x + 2)
4
yn =
1 n 6x
[6 e + (-6) n e-6x ] .
4
Find the nth derivative of : (1) sin h 2x sin 4x
Solution: Dn[sinh 2x sin 4x]
1 n 2x
= (D [e sin 4x] - D n [e-2x sin 4x])
2
1
= 20n/2 {e 2x sin (4x + n tan -1 2) - e -2x sin (4x - n tan
2
(2) y = log (4x2 – 1)
Solution: Let y = log (4x2 – 1) = log [(2x + 1) (2x – 1)]
Therefore y = log (2x + 1) + log (2x – 1).
(- 1) n- 1 (n - 1)!2 n
yn =
+
n
(2 x + 1)
(- 1) n- 1 (n - 1)!2n
(2 x - 1) n
-1
2)}
x2
Find the nth derivative of y =
(x + 2)(2 x + 3)
Solution:
x2
x2
y=
=
(x + 2)(2 x + 3) 2 x 2 + 7 x + 6
1
1
2
2 x + 7 x + 6)- (7 x + 6)
(
2
= 2
2 x2 + 7 x + 6
1é
A
B
= ê1+
2 êë x + 2 2 x +
ù
ú
3ú
û
1 1
(7 x + 6)
= 2 2 ( x + 2)(2 x + 3)
1é
8
9
= ê1+
2 êë x + 2 2 x +
ù
d n éê
x2
8(- 1) n n !
ú
=+
n+ 1
dx n êë(x + 2)(2 x + 3)ú
û
2( x + 2)
9(- 1) n n !2 n
2(2 x + 3) n+ 1
ù
ú
3ú
û
Leibnitz’s Theorem:
If u and v are functions of x possessing derivatives of the nth order,
then
(uv)n =
n
C0 uv n + n C1 u1 v n-1 + n C2 u n-2 v 2 +...+ n Cn-1 u n-1v1 + n Cn u n v.
Proof: The Proof is by the principle of mathematical induction on n.
Step 1: Take n = 1
By direct differentiation, (uv)1 = uv1 + u1v
For
n = 2,
(uv)2 = u2v+ u1v1 + u1v1+ uv2
= u 2 v+ 2C1u1v1 + 2C2 uv 2
Step 2: We assume that the theorem is true for n = m
(uv)m =
m
C0 uv m + m C1 u1 v m-1 + ...
+
m
Cm-1 u m-1v1 + mC m u m v.
Differentiating both sides we get
(uv) m+1 =
m
C0 u v m+1 + m C0 u1 v m +
... + m Cm u m v1 + m Cm u m+1v.
m
C1 u1 v m + m C1 u 2 v m-1 + ...
Note: (i) m Cr-1 + m Cr = (m+1) Cr
(ii) 1 + m C1 = 1+m = (m+1) C1
(iii) m Cm = 1 =
(uv) m+1 =
m
(m+1)
C
m+1
C0 u v m+1 + ( m C0 + m C1 )u1 v m +( mC1 + mC2 )u 2 v m-1 + ...
... +(m Cm-1 + m Cm )u m v1 + m Cm u m+1v.
(uv)m+1 =
m+1
C0 uvm+1 +
m+1
C1 u1 v m + ...
+
m+1
C m u m v1 +
m+1
C m+1u m+1v.
Therefore the theorem is true for m + 1 and hence by the principle of
mathematical induction, the theorem is true for any positive integer n.
Example: If y = sin (m sin-1 x) then prove that
(i) (1 – x2) y2 – xy1 + m2 y = 0
(ii) (1 – x2) yn+2 – (2n + 1) xyn+1 + (m2 – n2) yn = 0.
-1
y1 = cos (m sin x) m
1
1- x 2
1- x 2 y1 = m cos (m sin -1 x)
(1 – x2) y12 = m2 cos2 (m sin-1 x)
= m2 [ 1 – sin2 (m sin –1 x)]
= m2 (1 – y2).
Differentiating both sides we get
(1 –x2)2y1. y2 + y12 (-2x) = m2 (- 2y. y1)
(1 – x2) y2 – xy1 + m2 .y = 0
Applying Leibnitz’s rule we get
[(1 – x2) yn+2 + nc1 (- 2x) . yn+1 + nc2 (-2) .yn ]
– [x yn+1+ nc1.1. yn ] + m2 yn = 0
(1 – x2) yn+2 – (2n + 1) xyn+1 + (m2 – n2) yn = 0.
Example: If y1/m + y-1/m = 2x, show that
(x2 – 1) yn+2 + (2n + 1)xyn+1 + (n2 – m2)yn = 0.
1/m
Solution: y
+
1
1/ m
y
= 2x
 (y1/m)2 + 1 = 2x (y1/m)
That is, (y1/m)2 – 2x(y1/m) + 1 = 0 which is a quadratic
equation in y1/m.
1/m
y
=
- (- 2 x) ±
y = éêx ±
ë
(- 2 x) 2 - 4.1.1
2
m
x - 1ù
ú
û
2
=x±
x2 - 1
log y = m log éêx ±
ë
x 2 - 1ù
ú
û
Differentiating w.r.t x we get,
ìïï
ü
ïï
1
1
1
y1 = m.
.2 xý
í 1±
2
2
é
ù
ïîï
ïþ
y
2
x
1
x
±
x
1
ï
êë
ú
û
1
1
[ x 2 - 1 ± x]
y1 = m.
.
2
éx ± x 2 - 1ù
y
x
- 1
êë
úû
=±
m
x2 - 1
Squaring and cross multiplying we get (x 2 - 1) y12 = m2 y2 .
(x2 – 1) 2y1 y2 + 2xy12 = m2 (2yy1)
 (x2 – 1) y2 + xy1 - m2y = 0, on dividing by 2y1.
Now differentiating each term n times by Leibnitz theorem , we get
{(x - 1) y
2
n+ 2
+ n . 2x . y n + 1
n(n-1)
+
. 2y n
1.2
}
+ {x . yn+1 + n . 1 . yn} – m2yn = 0
(x2 – 1) yn+2 + 2n xyn+1 + n2 yn – nyn + xyn+1 + nyn – m2 yn = 0
(x2 – 1)yn+2 + (2n + 1) xyn+1 + (n2 – m2)yn = 0
Example: If cos –1 (y / b) = log(x/n)n, then show that
x2yn+2 + (2n + 1) xyn + 1 + 2n2 yn = 0
Solution:
y = b cos [n log (x/n)]
Differentiating w.r.t x we get,
1 1
y1 = - b sin [n log (x/n)].n
.
( x / n) n
 xy1 = - n b sin [n log (x/n)]
Differentiating w.r.t. x again we get ,
xy2 + 1. y1 =
1 1
- n. b cos [n log (x/n)] . n.
.
( x / n) n
Therefore x(xy2 + y1) = -n2b cos [n log (x/n)] = -n2y
Now,
x2y2 + xy1 + n2y = 0
Applying Leibnitz theorem ,
{
x 2 yn+ 2 + n. 2x. y n + 1 +
n(n-1)
. 2 . yn
1.2
}
+ {xyn+1 + n. 1 . yn} + n2yn = 0
x2 yn+2 + (2n + 1) xyn+1 + 2n2 yn = 0
Exercise: If y = e
m cos-1 x
, prove that (1 - x 2 ) y 2 - xy1 = m2 y and
hence show that (1 - x 2 ) y n+2 - (2n + 1) xy n+1 - (n 2 + m2 ) y n = 0.
HINT : y1 = e
m cos-1 x
æ
çççè
ö
my
÷
= ÷
÷
1- x 2
1- x 2 ø
m
1- x 2 y1 = - my
Squaring and differentiating again we get,
(1 - x 2 ) y 2 - xy1 = m 2 y
Differentiating n times using Leibnitz theorem,
(1 – x2)yn+2 – (2n + 1)xyn+1 – (n2 – m2) yn = 0.