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LEIBNITZS THEOREM OF NTH DERIVATIVE

OF THE PRODUCT OF TWO FUNCTION

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If then !n ny x y n

If thenax n axny e y a e

If then log anx x

n ey a y a

11

If then 1 !n nn

ny y a n ax b

ax b

If sin then sin2

n

n

ny ax b y a ax b

1

1

1

1 0 1 1 1

n n n

r r r

n n n

C C C

C C C n n

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Statement: Let u and v are two function of x differentiable ntimes then the nth derivative of their product is

Let y=uv

By actual differentiation, we have

The theorem is seem to be true for n=2,3

Let us assume that the theorem is true for n=m so that

0 1 1 1 2 2 2( ) ...... ...n n n n nn n n n r n r r n nuv c u v c u v c u v c u v c uv

1 1 1y u v uv 2 2 1 1 1 1 2 2 1 1 2

2 2 2

0 2 1 1 1 2 2

2

=

y u v u v u v uv u v u v uv

c u v c u v c uv

3 3 2 1 1 2 3

3 3 3 3

0 3 1 2 1 2 1 2 3 3

3 3y u v u v u v uv

c u v c u v c u v c uv

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Differentiating this w.r.to x we have

The theorem is to seem to be true for n=m+1

Hence, by the method of mathematical induction we

conclude that the theorem is true for any positive integral

value of n.

1 1 1 2 1 2 ......... ...m m m

m m m m r m r r my u v c u v c u v c u v uv

1 1 1 1 1 1 2 2 1 2

2 2 3 1...

m m m

m m m m m m

m m m

m r m r r r m r r

y u v u v c u v c u v c u v

c u v c u v c u v

1 1 1 1 2 1 2

1 1 1

(1 ) ( ) ......

( ) ......

m m mm m m

m m

r r m r r m

u v c u v c c u v

c c u v uv

1 1 1

1 1 1 2 1 2 1 1...... ....m m m

m m m r m r n mu v c u v c u v c u v uv

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It states that if u and v be any two function of x,

Ex. If y=x3eax, find yn

Let, u=eax and v=x3 so that y = uv=x3eax,

Dnu=aneax Dv=3x2

Dn-1

u=an-1

eax

D2

v=6xDn-2u=an-2eax D3v=6

Now by Leibnitzs Theorem we have

1 2 2

1 2( ) ( ) ( ) ( ) .....n n n n n n n n

nD uv D u v c D u Dv c D u D v c uD v

3

3 1 3 2 2 3 3 3 31 2 3

( )

( ) ( ). ( ). ( ).

n ax

n ax n n ax n n ax n n ax

D x e

D e x c D e Dx c D e D x c D e D x

3 1 2 2 3( 1) ( 1)( 2).3 .6 .62 6

n ax n ax n ax n axn n n n na e x na e x a e x a e

3 1 2 2 33 3 ( 1) ( 1)( 2)ax n n n ne a x na x n n a x n n n a

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Ex. If show that(1+x2)y2+(2x-1)y1=0

and (1+x2)yn+2+(2nx+2x-1)yn+1+n(n+1)yn=0

Soln. As is given

Diff. w. r. t. x

1tan x

y e

1tan x

y e

1

1

tan 11

tan

1 2

tan

1.

1

x

x

dy e xdx

y e

x

12 tan

1

2

1

(1 )

(1 )

xx y e

x y y

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Diff. again w. r. to x

Which proves the first result

Now, Diff. (1) n times, by Leibnitzs theorem, we have

2

2 1 1

2

2 1

(1 ) 2

(1 ) (2 1) 0.................(1)

x y xy y

x y x y

2

2 1{(1 ) } {(2 1) } 0n nD x y D x y

2 2 2 2

2 1 1 2 1 1{ (1 ) . (1 ) (1 )} { (2 1) (2 1)} 0n n n

n n n n ny x c y D x c y D x y x y c D x

2

2 1 1

( 1)(1 ) 2 .2 (2 1) .2 0

2n n n n n

n nx y ny x y y x ny

2 2

2 1

2

2 1

(1 ) (2 2 1) ( ) 0

(1 ) (2 2 1) ( 1) 0

n n n

n n n

x y nx x y n n y

x y nx x y n n y

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1 1

2 : If y + y =2m mExample x

2 2 2

2 1

then show that

( 1) (2 1) ( ) 0n n nx y n xy n m y

1 1

2 1

: We have y + y =2

2 1 0

m m

m m

solutions x

y xy

21

22 4 4 = 12

mx x

y x x

2

2 1

12

( 1)

( 1) (1 )1

m

m

y x x

xy m x x

x

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2 2 2 2

1

quaring both sides, we get

(x -1)y

S

m y

2 2 2

1 2 1 1

2 2

2 1

gain diff. both sides w.r.t.x we get

2(x -1)y y 2 y 2 y

(x -1)y y

A

x m y

x m y

2 2

2 1

2 2

2 1 1 2 1 1

2 2 2

2 1

pplying Leibnitz theorem,we get

(x -1)y y

(x -1)y 2 y 2y 2y y 0

(x -1)y 2( 1)y ( ) 0

n n n

n n n n n n

n n n

A

x m y

c x c c m y

xn n m y

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1Example 3: If y=cos

apply Leibnitz theorem

x

then

1

12

:

now y= cos

1y

1

Sol

x

x

2 2 2

1 12

quaring both sides, we get

1y or y (1 ) 1

1

S

xx

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2 2

1 2 12

2 1

gain diff. both sides w.r.t.x we get

2(1-x )y y 2 y 0

(1-x )y y 0

A

x

x

2

2 1 1 2 1 1

2 2

2 1

pplying Leibnitz theorem,we get

(1-x )y 2 y 2y y y 0(1-x )y 2(2 1) y 0

n n n

n n n n n

n n n

A

c x c x c

n x n y

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Value of the nth derivative of a function for x=0

Step 1. Equate the given function to y

Step 2. Find y1

Step 3. Again find y2

Step 4. Differentiate both sides n times by Leibnitzs theorem.

Step 5. Put x=0 in equations of step(1),(2),(3) and (4)

Step 6. On putting x=0 two cases arises (i) when n=odd (ii) whenn=even integer

Ex. If prove that yn(0)=0 for a odd

yn(0)=2.22.42.62..(n-2)2 ,n2 for n even

1 2(sin )y x

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Ex. If prove that yn(0)=0 for a odd

yn(0)=2.22.42.62..(n-2)2 ,n2 for n even

Soln.

Diff. (1) w. r. t. x

1 2(sin )y x

1 2(sin ) ......................(1)y x

1

12

2 1

1

2 2

1

2sin

1

1 2 sin

(1 ) 4 ..............(2)

xy

x

x y x

x y y

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Diff. (2) again w. r. to x

Diff. (3) n times by Leibnitzs theorem

2 2

1 2 1 1

2

2 1

2

2 1

(1 )2 2 4

(1 ) 2(1 ) 2 0.............(3)

x y y xy y

x y xyx y xy

2

2 1

1 1

2

2 1 1

2

2 2 2 2

2 1 1 2

2

2 1

{ (1 ) . (1 )

{(

(

1 ) } {( ) 1} 0

{ ( ) ( )} 0

( 1)(1 ) 2 .2 ( ) . 0

2

(1 ) ( )

)}

0

1

1 2

n n

n

n n

n n n n n

n

n n

n n n

n n

D x y D xy

y x y c D x

n nx y ny x y y x

y x c y D x c y

ny

x y x n y

D x

n y

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Ex 3. If

Sol.

21 y (0)m

ny x x find

1

2 2

1

112 2 2

y 1 ( 1

1( 1 1 (1 ) .2

2

m

m

dm x x x x

dx

m x x x x

2 1

2

2 1 2 2

2 2

[ 1 ] [1 ]1

[ 1] [ 1] [ 1]

1 1

m

m m

xm x x

x

x x x x x xm m

x x

2

11x y my

2 2 2 2

1

squaring both sides

(1+x ) ..............(2)

on

y m y

2 2 2

1 2 1 1

. (2) w. r. to x

(1+x ) 2 .2 2

Diff

y y y x m yy

2 2

2 1

2 2

2 1

(1+x )

(1+x ) 0..............(3)

y xy m y

y xy m y

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Again diff. (3) n times by Leibnitzs theorem, we have

2 22 1 1 2 1 11 2 .2 .1 0n n nn n n n n nx y c y c y xy c y m y

2 2

2 1 1

2 2 2

2

( 1)1 2 .2 02

1 2 1 0................(4)

n n n n n n

n n n

n nx y nxy y xy ny m y

x y n xy n m y

2

1 1

2 2

2

(0) 1, y (0) , y (0) ,

y (0) ( ) (0)...............(5)n n

y m m

m n y

2 2 2 2

3 1

2 2 2 2 2 2

5 3

2 2 2 2 2 2

When n is odd, Putting n=1,3,5,.......... in (5)

y (0) ( 1 ) (0) ( 1 )

y (0) ( 3 ) (0) ( 1 )( 3 )

.

.

.

y (0) { ( 2) }.........( 3 )( 1 )n

m y m m

m y m m m

m n m m m

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2 2 2 2 2

4 2

2 2 2 2 2 2 26 4

2 2 2 2 2 2 2

When n is even, putting n=2,4,6.......... in (5)

y (0) { 2 }y (0) ( 2 )

(0) ( 4 ) (0) ( 4 )( 2 )

.

.

(0) ( ( 2) }...........( 4 )( 2 )n

m m m m

y m y m m m

y m n m m m

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Assignment-I

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4.

5.