Taking the derivative of integral with two functions

Question

Solve: $\frac{d^2}{dx^2}\left({\Large\int} _{1}^{e^x} \ f(t) g(t)\,dt\right)$

Here I know have to find out th second derivative which is $\frac{d}{dx}\left(\frac{d}{dx}\left({\Large\int} _{1}^{e^x} \ f(t) g(t)\,dt\right)\right)$

If I just have the function $\frac{d}{dx}\left(\frac{d}{dx}\left({\Large\int} _{1}^{e^x} \ f(t)\,dt\right)\right)$ I can easily find it out using the fundamental theorem of calculus but here I have g(t) too. What can I do for this?

Answer 1

$\frac{d^2}{dx^2} \int_1^{e^x} f(t)g(t) \ dt\\ \frac {d}{dx}\left(\frac{d}{dx} \int_1^{e^x} f(t)g(t) \ dt\right)\\ \frac {d}{dx}\big(f(e^x)g(e^x)e^x\big)$

By the fundamental theorem of calculus.

Now you just need to properly apply the product rule and chain rule.

$f'(e^x)g(e^x)e^{2x}+f(e^x)g'(e^x)e^{2x}+ f(e^x)g(e^x)e^x$

Answer 2

I'd say:

($\int_1^{e^x} f(t)g(t) dt )^{'} = f(e^x)g(e^x)*(e^{x})'$

You can continue from here to get a second derivative of the function defined as the integral simply derivating $f(e^x)g(e^x)$