Answer
$h'v=\frac{3+5v^2}{3\sqrt[3]{1+v^2}^2}$
Work Step by Step
$$h'(v)=(v \sqrt[3]{1+v^2})'=v'\cdot\sqrt[3]{1+v^2}+v\cdot(\sqrt[3]{1+v^2})'=
1\cdot\sqrt[3]{1+v^2}+v\cdot\frac{1}{3}(1+v^2)^{\frac{1}{3}-1}(1+v^2)'=
\sqrt[3]{1+v^2}+\frac{1}{3}v\frac{1}{\sqrt[3]{1+v^2}^2}\cdot2v=
\frac{3(1+v^2)+2v^2}{3\sqrt[3]{1+v^2}^2}=\frac{3+5v^2}{3\sqrt[3]{1+v^2}^2}$$
