Differential Calculus

To err is to admit to forgive

to blame it on others to repeat

ANONYMUS

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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

2ax 2 ax

2 2y = a.a e = a e

d y

dx=

ax

1y = a e dy

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

yn = (m log a)n amx.

.

.

y3 = (m log a)3 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.

.

.

14. 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

( 1) !

( )

n n

n n

n ay

ax b +

-=

+

.

.

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( 1) ( 1)!

( )

n n

n n

n ay

ax b

-- -=

+

.

.

6. y = sin (ax + b)

y1 = cos (ax + b). a

1y = sin 2

a ax bpé ù

+ +ê úê úë û

2y = cos 2

a ax b apé ù

+ +ê úê úë û

2 = a sin 2.

2ax b

pæ ö÷ç + + ÷ç ÷çè ø

2= a sin

2 2ax b

p pé ùæ ö÷ç+ + +ê ú÷ç ÷çè øê úë û

n

ny = a sin 2n ax bpæ ö

÷ç + + ÷ç ÷çè ø

.

.

n

nExercise: If y = cos (ax + b), prove that y = a cos 2

n ax bpæ ö

÷ç + + ÷ç ÷çè ø

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 θ, b = r sin θ

Then θ = tan-1 (b/a) and

a2 + b2 = r2 (cos2 θ + sin2 θ) = r2

y1 = eax [ r sin θ cos (bx + c) + r cos θ sin (bx + c)]

y1 = r eax sin (θ + bx + c)

Note: sin (A + B) = sin A cos B + cos A sin B

Similarly we get,

y2 = r2 eax sin (2θ + bx + c),

y3 = r3 eax sin (3θ + bx + c)

.

.

yn = rn eax sin (nθ + bx + c)

2 2 -1where r = and = tan (b/a).a b q+

Exercise: If y = eax cos (bx + c), yn = rn eax cos (nθ + bx + c),

2 2 -1where r = and = tan (b/a).a b q+

Examples:

1. Find the nth derivative of y = cos h2 3x

Solution: Write cos h2 3x =

23 3

2

x xe e-é ù+ê úê úë û

6x -6x1(e + e + 2)

4=

n 6x n -6x1[6 e + (-6) e ] .

4ny =

Find the nth derivative of : (1) sin h 2x sin 4x

Solution: Dn[sinh 2x sin 4x]

n 2x n -2x1(D [e sin 4x] - D [e sin 4x])

2=

n/2 2x -1 -2x -1120 {e sin (4x + n tan 2) - e sin (4x - n tan 2)}

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( 1) ( 1)!2

(2 1)

n n

n n

ny

x

-- -= +

+

1( 1) ( 1)!2

(2 1)

n n

n

n

x

-- -

-

( )( )

2

Find the nth derivative of y =2 2 3

x

x x+ +

Solution:( )( )

2 2

22 2 3 2 7 6

x xy

x x x x= =

+ + + +

1 8 91

2 2 2 3x x

é ù= - +ê ú

ê ú+ +ë û

( )( )

2

2 2 3

n

n

d x

dx x x

é ùê úê ú+ +ë û 1

8( 1) !

2( 2)

n

n

n

x +

-= - +

+ 1

9( 1) !2

2(2 3)

n n

n

n

x +

-

+

( ) ( )2

2

1 12 7 6 7 6

2 22 7 6

x x x

x x

+ + - +=

+ +

1 1 (7 6)

2 2 ( 2)(2 3)

x

x x

+= -

+ +

11

2 2 2 3

A B

x x

é ù= - +ê ú

ê ú+ +ë û

Leibnitz’s Theorem:

If u and v are functions of x possessing derivatives of the nth order,

then

Proof: The Proof is by the principle of mathematical induction on n.

Step 1: Take n = 1

By direct differentiation, (uv)1 = uv1 + u1v

n

n n n n n

0 n 1 1 n-1 2 n-2 2 n-1 n-1 1 n n

(uv) =

C uv + C u v + C u v +...+ C u v + C u v.

For n = 2, (uv)2 = u2v+ u1v1 + u1v1+ uv2

Step 2: We assume that the theorem is true for n = m

Differentiating both sides we get

2 2

2 1 1 1 2 2= u v+ C u v + C uv

m

m m m m

0 m 1 1 m-1 m-1 m-1 1 m m

(uv) =

C uv + C u v + ... + C u v + C u v.

m m m m

m+1 0 m+1 0 1 m 1 1 m 1 2 m-1

m m

m m 1 m m+1

(uv) = C u v + C u v + C u v + C u v + ...

... + C u v + C u v.

m m (m+1)

r-1 r rNote: (i) C + C = C

m (m+1)

1 1(ii) 1 + C = 1+m = C

m (m+1)

m m+1 (iii) C = 1 = C

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.

m m m m m

m+1 0 m+1 0 1 1 m 1 2 2 m-1

m m m

m-1 m m 1 m m+1

(uv) = C u v + ( C + C )u v +( C + C )u v + ...

... +( C + C )u v + C u v.

m+1

m+1 m+1 m+1 m+1

0 m+1 1 1 m m m 1 m+1 m+1

(uv) =

C uv + C u v + ... + C u v + C u v.

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

1 2

1y = cos (m sin x) m

1 x-

2 -1

11 y = m cos (m sin x)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

1/

1Solution: y + = 2x

my

⇒ (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.

21/m ( 2 ) ( 2 ) 4.1.1

y =2

x x- - ± - - 2= x ± 1x -

2y = x ± 1m

xé ù-ê úë û

2log y = m log x ± 1xé ù-ê úë û

122

Differentiating w.r.t x we get,

1 1 1 y = m. 1 .2

2 11x

y xx x

ì üï ïï ï±í ýé ù ï ï-± - ï ïî þê úë û

2

122

1 1 [ 1 x]y = m. .

x 11

x

y x x

- ±

é ù -± -ê úë û

2= ±

1

m

x -

2 2 2 2

1Squaring and cross multiplying we get (x - 1) = m y . y

(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

{ }2

2 n 1 n

n(n-1)( 1) . 2x . y . 2y

1.2nx y n+ +- + +

+ {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 1y = - 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 ,

{ }2

2 n 1 n

n(n-1). 2x. y . 2 . y

1.2nx y n+ ++ +

+ {xyn+1 + n. 1 . yn} + n2yn = 0

x2 yn+2 + (2n + 1) xyn+1 + 2n2 yn = 0

-1 cos 2 2

2 1

2 2 2

n+2 n+1 n

Exercise: If y = , prove that (1 - x ) y - xy = m y and

hence show that (1 - x ) y - (2n + 1) xy - (n + m ) y 0

= .

m xe

-1 cos

1 2

HINT : y = 1

m x me

x

æ ö÷ç- ÷ç ÷÷çè ø-

2= -

1

my

x-

2

11 = - myx y-

Squaring and differentiating again we get,

2 2

2 1(1 - x ) y - xy = m y

Differentiating n times using Leibnitz theorem,

(1 – x2)yn+2 – (2n + 1)xyn+1 – (n2 – m2) yn = 0.