Derivative of the area under $f(x)$ between $a(x)$ and $b(x)$ Consider the following, where $a(x)$, and $b(x)$ are both functions that are continuous in their domain:
$$
g(x) = \int\limits_{a(x)}^{b(x)}f(t)dt
$$
Is it the case that $g'(x)$ is always the following?
$$
g'(x) = f(b(x))\times b'(x) - f(a(x))\times a'(x)
$$
If so, why does this work?

Fix a number $x_0\in\mathbb{R}$. You can then write
$$g(x)=\int_{a(x)}^{b(x)}f(t)dt=\int_{x_0}^{b(x)}f(t)dt-\int_{x_0}^{a(x)}f(t)dt$$
Now, we want to write $\int_{x_0}^{b(x)}f(t)dt$ as a composition of "known" functions, and similarly with $\int_{x_0}^{a(x)}f(t)dt$.
Let $h(x)=\int_{x_0}^x f(t)dt$. By the fundamental theorem of calculus, $h'=f$. Now, notice that
$$g=h\circ b-h\circ a.$$
By the chain rule,
$$g'(x)=h'(b(x))\cdot b'(x)-h'(a(x))\cdot a'(x)=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x)$$
as wanted.
