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
 A: Hint: Use the Fundamental Theorem of calculus
For instance
$(1)$ $F'(x) = e^{-x^2}$.
$(2)$ Use the product Rule for derivatives and the chain rule. 
$$G'(x) = 2x \int_{-4}^{5x} e^{-t} dt + x^2 \frac{d}{dx}\bigg(\int_{-4}^{5x} e^{-t} dt\bigg)$$ 
(3) Remember that 
$$\int_{a}^{b} f(t) dt = \int_{a}^{c} f(t) dt + \int_{c}^b f(t) dt$$
A: 2) $$G(x) = x^{2} . \int^{5x}_{-4} e^{-t^{2}} dt$$
First use the product rule of differentiation. In this calculation the derivative of $$\int^{5x}_{-4} e^{-t^{2}} dt$$ will appear. The derivative of a function $h(ax)$ with $a$ a constant is $ah'(ax)$. Take $a = 5$ and use this to get the derivative of the integral.
3) Hint: $$F(x) = \int^{{x}^3}_{{x}^{2}} e^{-t^{2}} dt = \int^{{x}^3}_{0} e^{-t^{2}} dt+\int^{0}_{{x}^{2}} e^{-t^{2}} dt = \int^{{x}^3}_{0} e^{-t^{2}} dt-\int^{{x}^{2}}_{0} e^{-t^{2}} dt$$
A: (1) $F(x)=\int_3^x{e^{-t^2}dt}$
Then,applying the Fundamental Theorem of Calculus, $F'(x)=e^{-x^2}$
(2) $G(x)=x^2\cdot\int_{-4}^{5x}e^{-t^2}=uv$
$u'=2x$
Applying the chain rule,  $v'=5e^{-{(5x)}^2}$
Applying the product rule, $G'(x)=uv'+vu'=5x^2e^{-(5x)^2}+2x\int_{-4}^{5x}{e^{-{t}^2}}$
(3) $F(x)=\int_{0}^{x^3}{e^{-t^2}}-\int_0^{x^2}e^{-t^2}$
Applying the chain rule and the addition rule
$F'(x)=3x^2e^{-x^6}-2xe^{-x^4}$
I hope I have not made any mistake, but this is what I know.
A: If $u$ and $v$ are functions in $x$ and $$F(x) = \int^{v}_{u} f(t) dt$$ , then
$$F'(x)=v'f(v)-u'f(u)$$
