Answer to Question #102231 in Calculus for DEBASIS BEHERA

Question

d/dx int 1 to rootx tan^2(t^2)dt

Answer 1

Leibniz integral rule:

$\frac{d}{dx}(\int_{a(x)}^{b(x)} f(x,t)dt)= f(x,b(x))\times \frac{d}{dx}b(x)-f(x,a(x))\times\frac{d}{dx}a(x) +\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)dt$

In your case :

$\frac{d}{dx} \int_{1}^{\sqrt{x}} tan^2(t^2)dt= ?$

Then:

$a(x)=1$

$b(x) = \sqrt{x}$

$f(x,t)=tan^2(t^2)$

Take derivatives with respect to x:

$\frac{d}{dx}a(x) = \frac{d}{dx}1 = 0$

$\frac{d}{dx}b(x)=\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}$

$\frac{\partial }{\partial x}f(x,t) = \frac{\partial }{\partial x}tan^2(t^2)=0$

(comment: in this case f(x,t) is independent of the variable x, so partial derivative with respect to t equals zero).

$f(x,b(x))$ means f(x,t) with a fixed value of a variable t, t takes on a value equal to b(x).

Eventually we have:

$\frac{d}{dx} \int_{1}^{\sqrt{x}} tan^2(t^2)dt=\frac{ tan^2(x)}{2\sqrt{x}}- tan^2(1^2)\times 0 +\int_{1}^{\sqrt{x}}0dt= \frac{ tan^2(x)}{2\sqrt{x}}$