$(2)$ Let $g:\Bbb R\to \Bbb R$ be differentiable. Let $F(x)=\int_0^{g(x)}t^2 dt$. Prove that $F'(x)=g^2(x)g'(x)$ for all $x \in \Bbb R$. If $G(x)=\int_{h(x)}^{g(x)}t^2 dt$, then what is $G'(x)$?
Tried solving this problem directly calculating the derivative. But to complete that proof I found g' to be continuous... Is that really needed? If not kindly give some suggestions to solve this.
First evaluate the definition of $F(x)$:
$$\int_0^{g(x)}t^2\ dt={(g(x))^3\over 3}$$
Now the prove of $F'(x)$ is just applying change rule.
Do the same for the definition of $G(x)$, then apply change rule.
EDIT:
Using the first theorem$\dots$
Let $p:\Bbb R\to \Bbb R$ be differentiable.
Define $p(y)=t\Rightarrow p'(y)dy=dt$. Doing change of variables in $F(x)$
$$=\int_{p^{-1}(a)}^{x} (p(y))^2\ p'(y)\ dy$$
Then applying the first fundamental theorem we get:
$${d\over dx}F(x)={d\over dx}\left(\int_{p^{-1}(a)}^{x} (p(y))^2\ p'(y)\ dy\right)=(p(x))^2\ p'(x)=$$
Now let $G,q:\Bbb R\to \Bbb R$ such that:
$$G(x)=\int_{q(x)}^{p(x)}t^2\ dt$$
Then separate the integral as:
$$\int_{q(x)}^{p(x)}t^2\ dt=\int_a^{p(x)}t^2\ dt\ -\int_a^{q(x)}t^2\ dt$$
Then define $p(y)=t,\quad q(z)=t\quad\Rightarrow\quad p'(y)dy=dt,\quad q'(z)dz=dt$. Doing change of variables in $G(x)$:
$$\int_a^{p(x)}t^2\ dt\ -\int_a^{q(x)}t^2\ dt=\int_{p^{-1}(a)}^{x}(p(y))^2\ p'(y)\ dy\ -\int_{q^{-1}(a)}^{x}(q(z))^2\ q'(z)\ dz$$
then ${d\over dx}G(x):$
$${d\over dx}G(x) ={d\over dx}\left(\int_{p^{-1}(a)}^{x}(p(y))^2\ p'(y)\ dy\ -\int_{q^{-1}(a)}^{x}(q(z))^2\ q'(z)\ dz \right)$$
$$=(p(x))^2\ p'(x)-(q(x))^2\ q'(x) $$
As pointed by @Mathmafia, you need that $p,q$ be differentiable continuos, and by @TedShifrin, this work for any antidifferentiable function inside the integral.