Derivative of $f(x)=\int_{x}^{x^3} e^{t^2} dt$ Find $f'(0)$ and $f'(\sqrt{3})$ when $f(x)=\int_{x}^{x^3} e^{t^2} dt$
I know that $\int_{}^{} e^{t^2} dt$ doesn't return elementary functions, but I am not sure how to solve this. Can anyone give a little hint?
 A: Recall by the Fundamental Theorem of Calculus, $$\frac{d}{dx}\int_a^xB(t)dt=B(x),\space\text{for $a\in \mathbb{R}$}$$
So if we have $$I=\int_{g(x)}^{h(x)}B(t)\space dt=\int_{a}^{h(x)}B(t)\space dt+\int_{g(x)}^{a}B(t)\space dt=\int_{a}^{h(x)}B(t)\space dt-\int_{a}^{g(x)}B(t)\space dt$$
We want to differentiate $I$, namely $$\frac{dI}{dx}=\frac{d}{dx}\left(\int_{a}^{h(x)}B(t)\space dt-\int_{a}^{g(x)}B(t)\space dt
\right)=\underbrace{\frac{d}{dx}\left(\int_{a}^{h(x)}B(t)\space dt\right)}_{\frac{dI_1}{dx}}-\underbrace{\frac{d}{dx}\left(\int_{a}^{g(x)}B(t)\space dt\right)}_{\frac{dI_2}{dx}}$$
So $$\frac{dI}{dx}=\frac{dI_1}{dx}-\frac{dI_2}{dx}$$
Calculate separately: 
$$\text{For $\frac{dI_1}{dx}$, let $u=h(x)$ so that $\frac{du}{dx}=\frac{dh}{dx}$}$$
Hence 
$$\frac{dI_1}{dx}=\frac{d}{du}\left(\int_{h(a)}^u B(t)\space dt\right)\cdot\frac{du}{dx}=B(u)\cdot\frac{du}{dx}=B(h(x))\cdot h'(x)$$
And similarly, $$\text{For $\frac{dI_2}{dx}$, let $u=g(x)$ so that $\frac{du}{dx}=\frac{dg}{dx}$}$$
Hence $$\frac{dI_2}{dx}=\frac{d}{du}\left(\int_{g(a)}^u B(t)\space dt\right)\cdot\frac{du}{dx}=B(u)\cdot\frac{du}{dx}=B(g(x))\cdot g'(x)$$
$$$$
We conclude $$\frac{dI}{dx}=\frac{dI_1}{dx}-\frac{dI_2}{dx}=B(h(x))\cdot h'(x)-B(g(x))\cdot g'(x)$$ or that
$$$$ 

$$\frac{d}{dx}\left(f(x)\right)=\frac{d}{dx}\left(\int_{g(x)}^{h(x)}B(t)\space dt\right)=h'(x)B(h(x))-g'(x)B(g(x))$$

Note in your case, 


*

*$f(x)=\int_x^{x^3}e^{t^2}dt$

*$h(x)=x^3$

*$g(x)=x$

*$B(t)=e^{t^2}$
So really, you're not trying to find an antiderivative of $y_1=e^{x^2}$, but a derivative of $y_2=\int_{g(x)}^{h(x)}e^{t^2}dt$.
A: Hint:
$\int_{u}^{v} f'(t)\mathrm dt=f(v)-f(u)$, we get that $\left(\int_{u}^{v}f'(t)\mathrm dt\right)'=f'(v)v'-f'(u)u'$. Here $u$ and $v$ are functions of $x$. Using this fact which is popularly called Leibniz Integral Rule, you can get the required result by setting $f(t)=e^{t^2}$, $u(x)=x^2$ and $v(x)=x^3$ and you'll be done in no time.
