How can I differentiate the following with respect to x? $$h(x) = \int_a^\infty f(x,a)g(x,a)da$$ that is: $$d\frac{f(x)}{dx} = \frac d{dx}\left(\int_a^\infty f(x,a)g(x,a)da\right)$$
I'm wondering if I can just treat $f(x,a)g(x,a)$ as $h(x,a)$ and use Liebniz's integral rule. If that is a good approach, then I think I get the following: $$\frac d{dx}\left(\int_a^\infty h(x,a)da\right)$$ and applying the rule, $$= \int_a^\infty \frac{\partial h(x,a)}{\partial x}da$$ If this is correct, can I proceed further to put this in terms of $f(x,a)$ and $g(x,a)$ again?
Note that $$\lim\limits_{a \to \infty} f(x) = 0$$ also note $$h(x,a) = f(x,a)*g(x,a)$$
Edits: initially I called the function $f(x)$ but corrected it to $h(x)$