How to prove that :
$$ \frac{\textrm{d}}{\textrm{d}x}\int^{g(x)}_{h(x)}f(t)\textrm{d}t =f(g(x))g'(x)-f(h(x))h'(x). $$
Answer
Let us first assume that $f$ has a primitive, which we shall refer to as $F$. By the fundamental theorem of calculus, we have:
$$\int_{h(x)}^{g(x)}{f(t)\:dt}=F(g(x))-F(h(x))$$
By the chain rule, we have:
$$\frac{d}{dx}\left(f\circ g\right)=f'(g(x))g'(x)$$
As we know that $\frac{d}{dx}F(x)=f(x)$, we have:
$$\frac{d}{dx}\left(F(g(x))-F(h(x))\right)=F'(g(x))g'(x)-F'(h(x))h'(x)\\=f(g(x))g'(x)-f(h(x))h'(x)$$
Which means that:
$$\frac{d}{dt}\int_{h(x)}^{g(x)}f(t)\:dt=f'(g(x))g'(x)-f(h(x))h'(x)$$
Q.E.D.
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