Show that $S(x) = \int_{h(x)}^{g(x)}f(t)dt \implies S'(x) = f(g(x))g'(x)-f(h(x))h'(x)$ let $g(x), h(x)$ be derivatible functions in $R$ and let $f(x)$ be a continuous function in $R$.
$S(x) = \int_{h(x)}^{g(x)}f(t)dt$.
How can I prove:
$S'(x) = f(g(x))g'(x)-f(h(x))h'(x)$.

Not very rigorous, but useful to help to remember the rule:
Let $F(x)=\int f(t)\mathrm dt$, so is, a primitive of $f$, then $F'(x)=f(x)$
$S(x)=\int_{h(x)}^{g(x)}f(t)dt=[F(t)]_{h(x)}^{g(x)}=F(g(x))-F(h(x))$
$S'(x)=F'(g(x))g'(x)-F'(h(x))h('(x)=f((g(x))g'(x)-f(h(x))h'(x)$
