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.

In general, by chain rule, we have$$\dfrac{d}{dx} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = \dfrac{d g(x)}{dx} \dfrac{d }{d g(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right) + \dfrac{d f(x)}{dx} \dfrac{d }{d f(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right)$$
Now make use the fundamental theorem of calculus to conclude that
$$\dfrac{d }{d g(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = h(g(x))$$
$$\dfrac{d }{d f(x)} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = -h(f(x))$$
Hence,
$$\dfrac{d}{dx} \left(\int_{f(x)}^{g(x)} h(t) dt\right) = \dfrac{d g(x)}{dx} h(g(x)) - \dfrac{d f(x)}{dx} h(f(x))$$
A slightly more generalized version is the Leibniz integral rule.
