Answer
$f'\left( x \right) = - \left( {2x\sin {x^2} + \cos {x^2}} \right){e^{ - x}}$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {e^{ - x}}\cos \left( {{x^2}} \right) \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - x}}\cos \left( {{x^2}} \right)} \right] \cr
& {\text{Use the product rule for derivatives}} \cr
& f'\left( x \right) = {e^{ - x}}\frac{d}{{dx}}\left[ {\cos \left( {{x^2}} \right)} \right] + \cos \left( {{x^2}} \right)\frac{d}{{dx}}\left[ {{e^{ - x}}} \right] \cr
& {\text{Apply the chain rule}} \cr
& f'\left( x \right) = {e^{ - x}}\left[ { - \sin \left( {{x^2}} \right)} \right]\frac{d}{{dx}}\left[ {{x^2}} \right] + \cos \left( {{x^2}} \right){e^{ - x}}\frac{d}{{dx}}\left[ { - x} \right] \cr
& {\text{Computing derivatives}} \cr
& f'\left( x \right) = {e^{ - x}}\left[ { - \sin \left( {{x^2}} \right)} \right]\left( {2x} \right) + \cos \left( {{x^2}} \right){e^{ - x}}\left( { - 1} \right) \cr
& {\text{Multiply and simplify}} \cr
& f'\left( x \right) = - 2x{e^{ - x}}\sin \left( {{x^2}} \right) - {e^{ - x}}\cos \left( {{x^2}} \right) \cr
& f'\left( x \right) = - \left( {2x\sin {x^2} + \cos {x^2}} \right){e^{ - x}} \cr} $$
