Application of fundamental theorem of calculus I have this problem:
$$ \frac{d}{dx} \left( \int_{\sqrt{x}}^{x^2-3x} \tan(t) dt \right) $$
I know how found the derivative of the integral from constant $a$ to variable $x$ so:
$$ \frac{d}{dx} \left( \int_a^x f(t) dt  \right) $$
but I don't know how make it between two variables, in this case from $\sqrt{x}$ to $x^2-3x$
Thanks so much.

Hint: Let $F(t)$ be an antiderivative of $\tan t$. We can find an explicit formula for $F(t)$, but it is better not to.
Then our integral is $F(x^2-3x)-F(\sqrt{x})$. Differentiate, using the Chain Rule, and remembering that $F'(u)=\tan u$. 
