Derivatives of $G(x)=\int^{e^x}_1(\log(t))^2dt$ and $H(x)=\int^{x^2}_{-x^2}e^{-t^5}dt$ I have $2$ tasks:
To evaluate $G(x)=\int^{e^x}_1(\log(t))^2dt$ for $x\gt 0$ and $H(x)=\int^{x^2}_{-x^2}e^{-t^5}dt$ for $x \in \Bbb R$
So by the fundamental theorem of calculus:
If $F(x)=\int^x_af$ is differentiable at $c$, then $F'(c)=f(c)$
And by Newton's FTC:
$\int^b_af(x)dx=F(b)-F(a)$
So, what I do is :
$G'(x)=(\log(e^x))^2-(\log(1))^2=x^2$
And 
$H'(x)=e^{-x^{10}}-e^{x^{10}}=\frac{1-e^{2x^{10}}}{e^{x^{10}}}$
But, in the answer sheet, the result is:
$G'(x)=\log^2(e^x)e^x=x^2e^x$
And 
$H'(x)=2x(e^{-x^{10}}+e^{-x^{10}})=4xe^{-x^{10}}$
What am I doing/interpreting wrong? Any help is appreciated!

You are almost correct. You also need to apply the chain rule. If $F(x) = \int_a^x f(t) dt$, then indeed $F'(x) = f(x)$, but suppose $\hat F(x) = \int_a^{x^2} f(t) dt$. Then $\hat F(x) = F(x^2)$ so the chain rule gives $$
{\hat F} {'(x)} = \left(\frac{d}{dx} x^2\right)F'(x^2) = 2x f(x^2) \ne f(x^2)
$$ 
Do you see how to solve the problem now?
