TITLE: How to calculate $f'(0)$ and $f'$ $(\sqrt{2})$ while $f(x)$= $\int_{x}^{x^3} e^{t^{2}}dt\ $?
QUESTION [5 upvotes]: How to calculate $f'(0)$ and $f'$ $(\sqrt{2})$ while $f(x)$= $\int_{x}^{x^3} e^{t^{2}}dt\  $? 
I thought about using the fundamental theorem of calculus, but im not sure im fully aware of how to use it in this case.
any kind of help/directing, would be appreciated.

REPLY [3 votes]: Some hints. The Fundamental Theorem says that
$$
\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t=e^{u^2}\tag{1}
$$
Let $u=x^3$, then
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\int_0^ue^{t^2}\,\mathrm{d}t
&=\frac{\mathrm{d}u}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\\
&=3x^2\frac{\mathrm{d}}{\mathrm{d}u}\int_0^ue^{t^2}\,\mathrm{d}t\tag{2}
\end{align}
$$
Then note that
$$
\int_x^ue^{t^2}\,\mathrm{d}t=\int_0^ue^{t^2}\,\mathrm{d}t-\int_0^xe^{t^2}\,\mathrm{d}t\tag{3}
$$