Answer
$$\frac{{dy}}{{d\theta }} = 2{e^{ - {\theta ^2}}}\sin \left( {{e^{ - {\theta ^2}}}} \right)$$
Work Step by Step
$$\eqalign{
& y = \cos \left( {{e^{ - {\theta ^2}}}} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\cos \left( {{e^{ - {\theta ^2}}}} \right)} \right] \cr
& {\text{use }}\frac{d}{{d\theta }}\left[ {\cos u} \right] = - \sin u\frac{{du}}{{d\theta }};{\text{ for this exercise consider }}u = {e^{ - {\theta ^2}}}.{\text{ then}} \cr
& \frac{{dy}}{{d\theta }} = - \sin \left( {{e^{ - {\theta ^2}}}} \right)\frac{d}{{d\theta }}\left[ {{e^{ - {\theta ^2}}}} \right] \cr
& {\text{use the formula }}\frac{d}{{d\theta }}{e^u} = {e^u}\frac{{du}}{{d\theta }}{\text{ for }}\frac{d}{{d\theta }}\left[ {{e^{ - {\theta ^2}}}} \right] \cr
& \frac{{dy}}{{d\theta }} = - \sin \left( {{e^{ - {\theta ^2}}}} \right)\left( {{e^{ - {\theta ^2}}}} \right)\frac{d}{{d\theta }}\left[ { - {\theta ^2}} \right] \cr
& \frac{{dy}}{{d\theta }} = - \sin \left( {{e^{ - {\theta ^2}}}} \right)\left( {{e^{ - {\theta ^2}}}} \right)\left( { - 2\theta } \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{d\theta }} = 2{e^{ - {\theta ^2}}}\sin \left( {{e^{ - {\theta ^2}}}} \right) \cr} $$
