# Differentiation under the Integral sign : Leibnitz’s Rule

### Leibnitz’s Rule :

If g is continuous on [a, b] and f1(x) and f2(x) are differentiable functions whose values lie in [a , b], then

$\displaystyle \frac{d}{dx} \int_{f_1(x)}^{f_2(x)} g(t) dt = g(f_2(x))f_2′(x) – g(f_1(x))f_1′(x)$

Illustration: $\displaystyle if \; f(x) = \int_{0}^{x} (f(t))^2 dt$ and $\displaystyle \int_{0}^{2} f(t) dt = 1$ , then find the value of f(2)

Sol: Here , $\displaystyle f(x) = \int_{0}^{x} (f(t))^2 dt$

Differentiating both the sides with respect to x we get, f'(x) = (f(x))2 so that

$\displaystyle \int \frac{f'(x)}{(f(x))^2} = \int 1$

$\displaystyle \frac{-1}{f(x)} = x + c$

$\displaystyle f(x) = \frac{-1}{x + c}$

Integrating both the sides from 0 to 2, with respect to x, we get

$\displaystyle \int_{0}^{2} f(x) dx = – \int_{0}^{2} \frac{1}{x + c} dx = 1$

$\displaystyle -ln |x + c|_{0}^{2} = 1$

$\displaystyle -ln \frac{2+c}{c} = 1$

$\displaystyle ln \frac{c}{2 + c} = 1$

$\displaystyle \frac{c}{2+c} = e$

$\displaystyle c = \frac{2 e}{1-e}$

Hence , $\displaystyle f(x) = -\frac{1}{x + \frac{2 e}{1-e} } = \frac{e-1}{x – e x + 2 e}$

and , $\displaystyle f(2) = \frac{e-1}{2-2x + 2e} = \frac{e-1}{2}$

Illustration: $\displaystyle if \; y(x) = \int_{\pi^2/16}^{x^2} \frac{cosx . cos\sqrt{\theta}}{1 + sin^2 \sqrt{\theta}} d\theta$ , then find $\displaystyle \frac{dy}{dx}$ at x = π

Sol: $\displaystyle y(x) = \int_{\pi^2/16}^{x^2} \frac{cosx . cos\sqrt{\theta}}{1 + sin^2 \sqrt{\theta}} d\theta$

$\displaystyle y(x) = cosx \int_{\pi^2/16}^{x^2} \frac{cos\sqrt{\theta}}{1 + sin^2 \sqrt{\theta}} d\theta$

$\displaystyle \frac{dy}{dx} = -sinx \int_{\pi^2/16}^{x^2} \frac{cos\sqrt{\theta}}{1 + sin^2 \sqrt{\theta}} d\theta + \frac{2x cosx . cosx}{1 + sin^2 x}$

Hence, at x = π

$\displaystyle \frac{dy}{dx} = 0 + \frac{2 \pi (-1)(-1)}{1 + 0} = 2 \pi$

♦ $\displaystyle F(t) = \int_{a}^{b} g(x , t) dx$ ;

Then , $\displaystyle \frac{dF}{dt} = \int_{a}^{b} \frac{\partial}{\partial t} g(x ,t) dx$ ;  where $\displaystyle \frac{\partial g}{\partial t}$ represents the differential coefficient of g with respect to t keeping x constant.

Illustration: Evaluate : $\displaystyle \int_{0}^{\pi/2} ln(1 + sin\alpha sin^2 x) cosec^2 x dx$ , (Where α > 0 )

Sol: $\displaystyle I = \int_{0}^{\pi/2} ln(1 + sin\alpha sin^2 x) cosec^2 x dx$

$\displaystyle \frac{dI}{d\alpha} = \int_{0}^{\pi/2} \frac{cos\alpha sin^2 x cosec^2 x}{1 + sin\alpha sin^2 x}dx$

$\displaystyle = \int_{0}^{\pi/2} \frac{cos\alpha}{1 + sin\alpha sin^2 x} dx$

$\displaystyle = \int_{0}^{\pi/2} \frac{cos\alpha sec^2 x}{(1 + sin\alpha)tan^2 x + 1} dx$

Now put tanx = t so that

$\displaystyle \frac{dI}{d\alpha} = \frac{1}{1 + sin\alpha}\int_{0}^{\infty} \frac{cos\alpha dt}{t^2 + \frac{1}{1+sin\alpha}}$

$\displaystyle = \frac{\pi cos\alpha}{2\sqrt{1 + sin\alpha}}$

$\displaystyle = \frac{\pi (cos^2 \frac{\alpha}{2} – sin^2 \frac{\alpha}{2})}{2(sin\frac{\alpha}{2} + cos\frac{\alpha}{2})}$

$\displaystyle = \frac{\pi}{2}(cos\frac{\alpha}{2} – sin\frac{\alpha}{2})$

$\displaystyle I(\alpha) = 2 \frac{\pi}{2}(cos\frac{\alpha}{2} + sin\frac{\alpha}{2}) + c$

$\displaystyle I(0) = \int_{0}^{\pi/2} ln(1 + 0. sin^2 x) cosec^2 x dx = 0$

⇒ c = -π

$\displaystyle I(\alpha) = \pi (cos\frac{\alpha}{2} + sin\frac{\alpha}{2} -1)$

Exercise:

(i) $\displaystyle \int_{\pi/3}^{x} \sqrt{3 – sin^2 t} \; dt + \int_{0}^{y} cost \;dt = 0$ ; then evaluate $\frac{dy}{dx}$

(ii)  $\displaystyle If \; \int_{log 2}^{x} \frac{dx}{\sqrt{e^x – 1}} = \frac{\pi}{6}$ ; then find x .

(iii) Let a + b = 4 ; where a < 2 and let g(x) be a differentiable function of x. If $\frac{dg}{dx} > 0$ > for all x , prove that $\displaystyle \int_{0}^{a} g(x) dx + \int_{0}^{b} g(x) dx$ increases as (b – a) increases.

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