Answer
$y' = 4x\sin \left( {{x^2} + 1} \right)\cos \left( {{x^2} + 1} \right)$
Work Step by Step
$$\eqalign{
& y = {\sin ^2}\left( {{x^2} + 1} \right) \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{{\sin }^2}\left( {{x^2} + 1} \right)} \right] \cr
& {\text{Use The Power Rule Combined with the Chain Rule}} \cr
& y' = 2\sin \left( {{x^2} + 1} \right)\frac{d}{{dx}}\left[ {\sin \left( {{x^2} + 1} \right)} \right] \cr
& y' = 2\sin \left( {{x^2} + 1} \right)\cos \left( {{x^2} + 1} \right)\frac{d}{{dx}}\left[ {{x^2} + 1} \right] \cr
& {\text{Compute derivative}} \cr
& y' = 2\sin \left( {{x^2} + 1} \right)\cos \left( {{x^2} + 1} \right)\left( {2x} \right) \cr
& {\text{Simplify}} \cr
& y' = 4x\sin \left( {{x^2} + 1} \right)\cos \left( {{x^2} + 1} \right) \cr} $$
