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If $$F(h(x))=\int_{0}^{h(x)}f(t)dt$$ Then $$\frac{d}{dx}F(h(x))=f(h(x))h'(x)$$ How can I show that $$\frac{d}{dx}F(h(x))=F'(h(x))h'(x)$$ I'm under the impression that the following isn't true: $$F'(h(x))=f(h(x))$$

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    If $F$ is a primitive of $f$, then the expression you don't believe to be trye is, actually, the definition of $F$ itself, so it is true. – 2017-02-04

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$F'(h(x))$ means "the function $F'$ evaluated at $h(x)$" so you have to plug in $h(x)$ to $F'$ but because $$F(x)=\int_{0}^{x}f(t)dt$$ you easily see that $F'(x) = f(x)$ and so $$F'(h(x)) = f(h(x))$$