= Solution
For the <Leibniz integral rule>, the upper endpoint has <derivative> $2\pi x$ and the lower endpoint has <derivative> $1$. The integrand has <partial derivative> $4xt+\sin t$. Therefore
$$
\begin{aligned}
h'(x)={}&\left(2\pi x^4+x\sin(\pi x^2)\right)2\pi x-\left(2x^3+x\sin x\right)\\
&+\int_x^{\pi x^2}(4xt+\sin t)\,dt.
\end{aligned}
$$
The last <integral> is $[2xt^2-\cos t]_x^{\pi x^2}=2\pi^2x^5-2x^3-\cos(\pi x^2)+\cos x$. Combining terms gives
$$
\boxed{h'(x)=6\pi^2x^5-4x^3+2\pi x^2\sin(\pi x^2)-x\sin x+\cos x-\cos(\pi x^2).}
$$
Back to article page
