Answer
$f'\left( x \right) = 4x{e^{{{\sin }^2}\left( {{x^2}} \right)}}\sin \left( {{x^2}} \right)\cos \left( {{x^2}} \right)$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {e^{{{\sin }^2}\left( {{x^2}} \right)}} \cr
& {\text{Differentiating}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{{{\sin }^2}\left( {{x^2}} \right)}}} \right] \cr
& {\text{Use the chain rule: }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}} \cr
& f'\left( x \right) = {e^{{{\sin }^2}\left( {{x^2}} \right)}}\frac{d}{{dx}}\left[ {{{\sin }^2}\left( {{x^2}} \right)} \right] \cr
& {\text{Use the general power rule for derivatives}} \cr
& f'\left( x \right) = {e^{{{\sin }^2}\left( {{x^2}} \right)}}\left( {2\sin \left( {{x^2}} \right)} \right)\frac{d}{{dx}}\left[ {\sin \left( {{x^2}} \right)} \right] \cr
& {\text{Use the chain rule: }}\frac{d}{{dx}}\left[ {\sin u} \right] = \cos u\frac{{du}}{{dx}} \cr
& f'\left( x \right) = 2{e^{{{\sin }^2}\left( {{x^2}} \right)}}\sin \left( {{x^2}} \right)\cos \left( {{x^2}} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{Compute the derivative and simplify}} \cr
& f'\left( x \right) = 2{e^{{{\sin }^2}\left( {{x^2}} \right)}}\sin \left( {{x^2}} \right)\cos \left( {{x^2}} \right)\left( {2x} \right) \cr
& f'\left( x \right) = 4x{e^{{{\sin }^2}\left( {{x^2}} \right)}}\sin \left( {{x^2}} \right)\cos \left( {{x^2}} \right) \cr} $$
