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