Answer
$h'\left( \theta \right) = {\theta ^2}\cos \theta + 2\theta \sin \theta $
Work Step by Step
$$\eqalign{
& h\left( \theta \right) = {\theta ^2}\sin \theta \cr
& {\text{Differentiating}} \cr
& h'\left( \theta \right) = \frac{d}{{d\theta }}\left[ {{\theta ^2}\sin \theta } \right] \cr
& {\text{Use the product rule for derivatives }}\left( {fg} \right)' = fg' + gf' \cr
& {\text{Let }}f = {\theta ^2}{\text{ and }}g = \sin \theta ,{\text{ then}} \cr
& h'\left( \theta \right) = {\theta ^2}\left( {\sin \theta } \right)' + \sin \theta \left( {{\theta ^2}} \right)' \cr
& {\text{Use the Derivatives of Trigonometric Functions }} \cr
& {\text{and }}\frac{d}{{d\theta }}\left[ {{\theta ^n}} \right] = n{\theta ^{n - 1}},{\text{ then}} \cr
& h'\left( \theta \right) = {\theta ^2}\left( {\cos \theta } \right) + \sin \theta \left( {2\theta } \right) \cr
& {\text{Simplify}} \cr
& h'\left( \theta \right) = {\theta ^2}\cos \theta + 2\theta \sin \theta \cr} $$
