TITLE: Derivative question
QUESTION [7 upvotes]: I need to find the derivative of:
$$
h(x) = \int_{0}^{x^2} (1-t^2)^{1/3} \, dt
$$
Would the answer to that just be:
$$
(1-x^4)^{1/3}?
$$

REPLY [3 votes]: If $F(x)=\int_{a}^{x}f(t)\;\mathrm{d}t$, then $F^{\prime }(x)=f(x)$. By the
chain rule if $F(x)=\int_{a}^{u(x)}f(t)\;\mathrm{d}t$, then
$$F^{\prime }(x)=F^{\prime }(u)u^{\prime }(x)=f(u(x))u^{\prime }(x).$$ 
In the present case $F(x)=h(x)$, $u(x)=x^{2}$ and $f(t)=(1-t^{2})^{1/3}$.
Hence $u^{\prime }(x)=2x$ and $f(u(x))=f(x^{2})=(1-x^{4})^{1/3}$. 
Thus
$$h^{\prime }(x)=2(1-x^{4})^{1/3}x.$$
A generalization is to find the derivative of an integral where both limits are functions of $x$, as in this question.