CHAPTER 7 SUCCESSIVE DIFFERENTIATION

[Pages:17]

CHAPTER 7 SUCCESSIVE DIFFERENTIATION

TOPICS: 1 . Successive differentiation-nth derivative of a function ? theorems. 2. Finding the nth derivative of the given function. 3. Leibnitz's theorem and its applications.





SUCCESSIVE DIFFERENTIATION

Let f be a differentiable function on an interval I. Then the derivative f is a function of x and

if f is differentiable at x, then the derivative of f at x is called second derivative of f at x. It is denoted by f(x) or f(2)(x).similarly, if f" is differentialble at x , then this derivative is called the 3rd derivative of f and it is denoted by f(3)(x). Proceeding in this way the nth derivative of f is the derivative of the function f(n-1)(x) and it the denoted by f(n)(x).

If y = f(x) then

f (n)(x)

is denoted by

dny dx n

or Dn y or y(n) or yn

and f (n) ( x) = lim f (n-1) ( x + h) - f (n-1) ( x)

h0

h

THEOREM

If f(x) = (ax + b)m, m R, ax + b > 0 and n N then f (n) (x) = m(m -1)(m - 2)...(m - n +1)(ax + b)m-n an

Note : If y = (ax + b)m then yn = m(m ? 1)(m ? 2) ...(m ? n + 1)(ax + b)m?n an.

COROLLARY If f(x) = (ax + b)m, m Z, m > 0, n N then (i) m < n f(n)(x) = 0, (ii) m = n f(n)(x) = n! an

(iii) m > n f(n)(x) = m! (ax + b)m-n an . (m - n)!

COROLLARY If f(x) is a polynomial function of degree less than n where n N then f(n)(x) = 0.

THEOREM

If f (x) = 1 ax + b

then f(n)(x) =

(-1)n (ax +

n !a n b)n+1

.(i.e.,

If y = 1 ax + b

yn =

(-1)n (ax +

n !a n b)n+1

)

THEOREM

If

f(x)

=

log

|ax

+

b|

and

n

N

then

f (n)(x)

=

(-1)n-1(n -1)!an (ax + b)n

.

i.e., y = log |ax+b|

yn

=

(-1)n-1(n -1)!an (ax + b)n





THEOREM

If

f(x)

=

sin(ax

+

b)

and

n

N

then

f (n)(x)

= an

sin

ax

+

b

+

n 2

.

THEOREM

If

f(x)

=

cos(ax

+

b)

and

n

N

then

f (n)(x)

=

an

cos

ax

+

b

+

n 2

.

THEOREM

If f(x) = eax+b and n N then f (n) (x) = aneax+b .

THEOREM If f(x) = cax+b, c > 0 and n N then f (n) (x) = ancax+b (log c)n .

THEOREM

If f(x) = eax sin(bx + c) and n N then f (n)(x) = rneax sin(bx + c + n) where a = r cos , b = r sin

and r =

a2

+

b2

,

=

Tan -1

b a

THEOREM

If f(x) = eax cos(bx + c) and n N then f (n) (x) = rneax cos(bx + c + n) where a = r cos , b = r sin

and r =

a2

+

b2

,

=

Tan -1

b a

.

Note: If f, g are two functions in x having their nth derivatives then

(f ? g)(n)(x) = f (n) (x) ? g(n) (x) .

Note: If f is a function in x having nth derivative and k R then (kf )(n) (x) = kf (n) (x) .





EXERCISE ? 7 (a)

1. Find the nth derivative of sin3x.

Sol: we know that sin 3x = 3sin x - 4 sin3 x sin3 x = 3sin x - sin 3x

4

Differentiate n times w.r.t x,

( ) dn

dx n

sin3 x

=

1 4

dn dx n

(3sin

x

- sin 3x )

=

1 4

-3n.sin

3x

+

n 2

+

3sin

x

+

n 2

n

z

2. Find the nth derivative of sin 5x. sin 3x.?

Sol: let y = sin 5x.sin 3x = 1 (2sin 5x.sin 3x)

2

y= 1 (cos 2x - cos8x)

2

y = 1 (cos 2x - cos8x)

2

Differentiate n times w.r.t x,

yn

=

1 2

dn dx n

(cos 2x

- cos8x)

yn

=

1 2

2n

cos

2x

+

n 2

-

8n

.

cos

8x

+

n 2

n

z

3. Find nth derivative of ex .cos x.cos 2x

Sol: cos x.cos 2x = 1 (2cos 2x.cos x) = 1 (cos3x + cos x)

2

2

Let y = ex (cos3x + cos x)

2

Differentiate n times w.r.t x,

( ) yn

=

1 2

dn dx n

ex cos 3x + ex cos x

( ) ( ) ( ) ( ) ( ) yn

=

ex 2

10

n

cos

3x + n tan-1 3

n

+

2

n

+ cos

x

+

n

tan-1 1

n

z

= ex 2

n 10

2

cos

3x + n tan-1 3

+

2n

/

2

cos

x

+

n 4

4.

If

y

=

(

x

2

- 1) (

x

-

2)

find

y n

Sol:

Given

y=

2

(x -1)(x - 2)

=

x

1 -

2

-

x

1 -1

(

partial

fractions)

Differentiate n times w.r.t x,

yn

=

2

(-1)n n! ( x - 2)n+1

-

(-1)n n! ( x -1)n+1

=

2

( -1)n

n!

(x

1

) - 2 n+1

-

(

x

1

- 1)n

+1





5.

If

y=

2x +1 ,

x2 - 4

find

y n

Sol:

Let

2x +1 x2 - 4

=

A x-2

+

B x+

2

2x +1 = A(x + 2) + B(x - 2) -----(1)

In (1) ,Put x = 2 5 = A(4) A = 5

4

In (1) , x = -2 -3 = B(-4) B = 3

4

Therefore,

y

=

2x +1 x2 - 4

=

5

4(x -

2)

+

3

4(x +

2)

Differentiate n times w.r.t. x,

yn

=

dn dx n

5

4(x - 2)

+

3

4(x + 2)

yn

=

5 4

(-1)n n! ( x - 2)n+1

+

3 4

(-1)n n! ( x + 2)n+1

=

( -1)n

4

n !

(x

5

) - 2 n+1

+

(x

3

) + 2 n+1

1.

Find the nth derivative of (i)

x

(x -1)2 (x +1)

(ii)

1

(x -1)(x + 2)2

(iii)

x3

(x -1)(x +1)

(iv) x

x2 + x +1

Sol: i)

(v) x +1

x2 - 4

(vi) Log (4x2 - 9)

Let

y=

x

(x -1)2 (x +1)

Resolving into partial fractions

(x

x

-1)2 (x

+ 1)

=

A x -1

+

(x

B

- 1)2

+

C x +1

x = A (x -1)(x +1) + B(x +1) + C(x -1)2 ----- (1

In (1 ), put x = 1 1 = B(1+1) = 2B B = 1

2

In (1 ), put x = -1 -1 = C(-1-1)2 = 4C C = - 1

4

Equating the co . efficient of x2 A + B = 0 A = - 1

2

Therefore,

y

=

-

1

2(x -1)

+

1

2(x -1)2

-

1

4(x +1)

Differentiate n times w.r.t. x,

yn

=

dn dx n

-

2

(

1 x-

1)

+

1

2(x -1)2

-

4(

1 x+

1)





yn

=

(-1)n n! 2( x -1)n+1

++1

2

(-2)(-3).....(-2 - n +1) ( x -1)n+2

-

1 4

(-1)n n! ( x +1)n+1

=

(-1)n n! 2.( x -1)n+1

+

(

-1)n 2(x

(n +1)! ) -1 n+2

-

1 4

(-1)n n! ( x +1)n+1

= (-1)n

n!

2(x

1

) -1 n+1

+

n +1

2( x -1)n+2

-

4(x

1

) +1 n+1

(ii)

y

=

(

x

-

1)

1

(x

+

2)2

Resolving

into partial fractions

1

(x -1)(x + 2)2

=

A x -1

+

B x+2

+

(x

C

+ 2)2

1=A (x + 2)2 + B(x -1)(x + 2) + C(x -1) ---(1)

In (1) put x = 1 1 = A (1+ 2)2 = 9A A = 1

9

In (1) put x = -2 1 = C(-2 -1) = -3C C = - 1

3

Equating the co ? efficient of x2 In (1)

A + B = 0 B = -A = - 1 9

y

=

9(

1

x -1)

-

9

(

1 x+

2)

-

3(x

1 +

2)2

Differentiate n times w.r.t. x ,

yn

=

dn dx n

9

(

1

x -1)

-

1

9(x + 2)

-

1

3(x + 2)2

yn

=

(-1)n n! 9( x -1)n+1

-

(-1)n n! 9 ( x + 2)n+1

-

1 3

(

-1)n (x +

(n +1) )2 n+2

= (-1)n

n!

9 ( x

1

) -1 n+1

-

9(x

1

) + 2 n+1

-

n +1

3(x + 2)n+2

(iii)

y

=

(

x

x3

-1) (

x

+

1)

Ans:

( -1)n

2

n!

( x

1

) -1 n+1

+

(x

1

) +1 n+1

( iv)

x

x2 + x +1

Ans:

yn

=

( -1)n

r n +1

n!

cos

(n

+ 1)

-

1 3

sin

(

n

+

1)

(v)

y

=

x +1 x2 - 4

Ans:

( -1)n

4

n!

( x

3

) - 2 n+1

+

(x

1

) + 2 n+1





(vi) y = log (4x2 - 9)

Given y = log (4x2 - 9) = log[2x - 3][2x + 3]

= log (2x - 3) + log (2x + 3)

Differentiating n times,

yn

=

dn dxn

( log ( 2x

- 3)

+

log ( 2x

+ 3))

yn

=

(

) -1 n-1 2n (2x -

(n 3)n

-

1)!

+

( ) -1 n-1 2n (n -1)! (2x + 3)n

=

(

) -1 n-1

2n

(

n

-

1)!

(

2x

1 -

3)n

+

(

2x

1 +

3)n

2.

If y =

a + bx c + dx

then show that

2y1y3

=

3y

2 2

Sol: Given y = a + bx

c + dx

Differentiate w.r.t.x ,

dy dx

=

(c

+

dx) b - (a + (c + dx)2

bx ) .d

y1=

bc + bdx - ad - bdx = (c + dx)2

bc - ad

(c + dx)2

Again diff. w.t.t x,

y2

=

(bc - ad)(-2) d (c + dx)3

=

-2d (bc - ad) (c + dx)3

Diff.wrt.x, we get

y3

=

-2d (bc - ad)(-3).d = (c + dx)4

6d2 (bc - ad) (c + dx)4

L.H.S.= 2y1y3

=

2 ( bc (c +

- ad

dx )2

)

.

6d2 (bc - ad (c + dx)4

)

=

12d2 (bc - ad)2 (c + dx)6

=

3

-2d (bc - ad (c + dx )3

)

2

=

3y

2 2

=R.H.S.

3. If y = sin (sinx ), then show that y2 + (tan x) y1 + y cos2 x = 0

Sol: Given y = sin (sinx) Diff. wrt x,

y1 = cos(sin x) cos x

Diff. wrt x, y2 = cos x - sin (sin x ) cos x - cos (sin x )sin x

= - cos2 x.sin (sin x) - sin x.cos (sin x) LHS= y2 + (tan x) y1 + y cos2 x

= - cos2 x sin (sin x ) - sin x.cos (sin x ) + sin x cos x (sin x ) + sin x.cos(sin x) = 0 = RHS.

cos x





4. If y = axn+1 + bx-n , then show that x2y2 = n (n +1) y .

Sol: y = axn+1 + bx-n

Diff. wrt. X,

y1 = a (n +1) xn - bnx-(n+1) Diff. wrt x, y2 = a.n (n +1) xn-1 + bn.(n +1) x-(n+2) x2.y2 = n (n +1).x2 a.xn-1 + b.x-n-2

( ) = n (n +1) axn+1 + bx-n = n (n +1) y

5. If y = aenx + be-nx , then show that y2 = n2y Sol: y = aenx + be-nx

y1 = a.n.enx - bn.e-nx

( ) y2 = an2enx + bn2 .e-nx = n2 aenx + be-nx

y2 = n2y

6.

If

- kx

y = e 2 (a cos nx + bsin nx) then show that

y2

+

ky1

+

n2

+

k2 4

y

=

0

.

- kx

Sol. y = e 2 (a cos nx + bsin nx)

Differentiating w.r.to x.

y1

=

- kx

e2

[-an sin

nx

+

bn

cos

nx]

+

-

k 2

e

-lx 2

[a

cos

nx

+

b sin

nx]

y1

=

-kx

+e 2 n (-a sin nx + bcos nx) -

k 2

y

y1

+

k 2

y

=

- kx

+n.e 2

(-a sin nx + b cos nx) - (1)

Differentiating w.r.to x.

y2

+

k 2

y1

=

- kx

+n.e 2

-

k 2

(-a sin nx

+

b cos nx)

- kx

+n.e 2

[-an.conx

- bn sin nx]

=

-

k 2

y1

+

k 2

y

-

n2

y

=

-

k 2

y1

-

k2 4

y - n2y



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