Are you asking for the proof of this:

$\displaystyle \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)$?.

If so, I think I may be able to help. Let's see:

Let's say f is continuous on an open interval we can call I.

If g(x), h(x) and a are in I, then

$\displaystyle \int_{h(x)}^{g(x)}f(t)dt=\int_{h(x)}^{a}f(t)dt+\in t_{a}^{g(x)}f(t)dt$

$\displaystyle =-\int_{a}^{h(x)}f(t)dt+\int_{a}^{g(x)}f(t)dt$

So, $\displaystyle \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=-f(h(x))h'(x)+f(g(x))g'(x)$

Is that what you meant?. Probably not, but there it is anyway

.