We can solve harder problems involving derivatives of integral
functions. For example, what do we do when our upper limit
of integration is not simply our variable $x$, as it must be to
use FTC 1, but is rather a
function of $x$? We will use FTC 2
to solve this FTC 1 problem.

Solution: Let $F(x)$ be the antiderivative of
$\tan^{-1}(x)$. Finding a formula for $F(x)$ is hard, but we
don't actually need the antiderivative , since we will not
integrate. Recall that by FTC 2,
$$\int_1^{x^2} \tan^{-1}(s) \,ds = F\left(x^2\right) - F(1),$$ so
$$\frac{d}{dx} \int_1^{x^2} \tan^{-1}(s)\,ds =
\frac{d}{dx}\left(F\left(x^2\right) - F(1)\right).$$
By the chain rule, we now get
$$F'(x^2)\,2x-0= 2x\, f(x^2) = 2x \tan^{-1}\left(x^2\right).$$
since $F'=f$ by our assumption.

This method generalizes, but please do not try to memorize this; you
do not need to, because it is simply applying FTC
2 and the chain rule, as you see in the box below and
in the following video.

If $f$ is a
continuous function with antiderivative $F$, and $g$ and $h$
are differentiable functions, then $$\frac{d}{dx}
\int_{g(x)}^{h(x)} f(s)\, ds = \frac{d}{dx}
\Big[F\left(h(x)\right) - F\left(g(x)\right)\Big] $$ $$
\hspace{3cm}\quad\quad\quad= F'\left(h(x)\right) h'(x) -
F'\left(g(x)\right) g'(x) $$ $$ \hspace{3cm}\quad\quad =
f\left(h(x)\right) h'(x) - f\left(g(x)\right) g'(x). $$

There is an an
alternate way to solve these problems, using FTC 1 and the chain
rule. We will illustrate using the previous
example.

Solution:
We let $u=x^2$ and let $g(u)=\int_1^u\tan^{-1}(s)\, ds$, and use the fact
that $\frac{d}{dx}g(u)=g'(u)\frac{du}{dx}$ to get $$\frac{d}{dx}\int_1^{x^2}
\tan^{-1}(s)\, ds=\frac{d}{dx} \left(\int_1^{u} \tan^{-1}(s)\,
ds\right)=\frac{d}{dx}\left(g(u)\right)=g'(u)\frac{du}{dx},$$
By FTC 1, $g'(u)=\tan^{-1}(u)$, and
$\frac{du}{dx}=\frac{d}{dx}(x^2)=2x$, giving us
$$\tan^{-1}(u)\frac{du}{dx}=\tan^{-1}(x^2)\cdot 2x.$$