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Avinash_Bhat

LEIBNITZ'S RULE

_f(x)

^g(x) H( t , x ) dt = y

Then :

_f(x)

^g(x) [ Partial differentiation of { H( t , x ) } with respect to x ] dt + H( g(x) , x ).g'(x) - H( f(x) , x ).f '(x) = dy/dx

So, if:

_f(x)

^g(x) H( t ) dt = y

Then :

H( g(x) ).g'(x) - H( f(x) ).f '(x) = dy/dx

SAMPLE PROBLEM 1 :

If y = ₀

^x f(t).sin( k{x - t} ) dt , Then : d²y/dx² + k²y =

(a) 0

(b) f(x)

(c) k.f(x)

(d) k².f(x)

k => constant ; f => function

Solution :

Direct application of the formula will give you:

dy/dx = ₀

^x f(t). cos ( k{x - t} ).k dt

Apply the formula once again to get:

d²y/dx² = - k²₀

^x f(t). sin( k{x - t} ) dt + k. f(x)

= - k².y + k. f(x)

So:

d²y/dx² + k².y = k.f(x)

SAMPLE PROBLEM 2 :

If : f(x) = ₀

^x log ( t + sqrt(t² + 1) ) dt , Then the function f(x) is :

(a) an even function

(b) an odd function

(c) a periodic function

(d) none of these

Solution :

Direct application of the formula will give you:

f '(x) = log (x + sqrt(x² + 1) )

f(x) =

log (x + sqrt(x² + 1) )dx

= x.log (x + sqrt(x² + 1) ) + (x² + 1)^3/2 / 3 ------------- (1)

f(-x) = -x.log (-x + sqrt(x² + 1) ) + ((-x)² + 1)^3/2 / 3
= x.log(-x + sqrt(x² + 1) )^-1 + (x² + 1)^3/2 / 3

= x.log (x + sqrt(x² + 1) ) + (x² + 1)^3/2 / 3 ------------- (2)

From (1) and (2), f(x) = f(-x) ; i.e. , f(x) is an even function.

SAMPLE PROBLEM 3 :

lim { x^-3. ₀^{x^2} sin (sqrt(t)) dt } is equal to :
^x=>0

(a) 1/3

(b) 2/3

(c) -1/3

(d) -2/3

Solution :

lim { x ^-3. ₀

^{x^2} sin (sqrt(t)) dt } => 0/0 form
x=>0

Using the formula over the numerator part for

differentiating and directly differentiating the denominator

part, (as per L'Hospital's Rule) to give :

lim { (sin x)(2x)(x^-2) / 3 } = 2/3

x=>0

Test yourself :

Q. If f(x) = ₁

^{x^3} (1 + t⁴)^-1 dt. Then find f '' (x) .

Q. Find : lim { h^-1. _x

^x+h ( z + sqrt(z² + 1) )^-1 dz }

^{h =>0}

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