Derivative of a function defined by an integral with $e^{-t^{2}}$ I'm facing a bit of a tricky question and I can't figure out how to get to the correct answer. We have to find the derivative of the following functions:
(1) $F(x) = \int^{x}_{3} e^{-t^{2}} dt$
(2) $G(x) = x^{2} . \int^{5x}_{-4} e^{-t^{2}} dt$
(3) $H(x) = \int^{x^3}_{x^2} e^{-t^{2}} dt$
As I have discovered, $e^{-t^{2}}$ has no closed form and is just defined in the error function. I have got the answer to the first question using the fundamental theorem of calculus, getting an answer of $e^{-x^{2}}$ however I can't get the correct answer for the other two. Could someone help me and explain how to get the answers to (2) and (3)? I have seen the answers on Wolfram Alpha and there appears to be some sort of chaining within the integral which I cannot understand.
Thanks! Helen

If $u$ and $v$ are functions in $x$ and $$F(x) = \int^{v}_{u} f(t) dt$$ , then
$$F'(x)=v'f(v)-u'f(u)$$
