Answer
$$\frac{{dy}}{{d\theta }} = 2{e^\theta }\cos \theta $$
Work Step by Step
$$\eqalign{
& y = {e^\theta }\left( {\sin \theta + \cos \theta } \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{e^\theta }\left( {\sin \theta + \cos \theta } \right)} \right] \cr
& {\text{use the product rule }} \cr
& \frac{{dy}}{{d\theta }} = {e^\theta }\frac{d}{{d\theta }}\left[ {\sin \theta + \cos \theta } \right] + \left( {\sin \theta + \cos \theta } \right)\frac{d}{{d\theta }}\left[ {{e^\theta }} \right] \cr
& {\text{solve the derivatives using }}\frac{d}{{d\theta }}\left[ {\sin \theta } \right] = \cos \theta,{\text{ }}\frac{d}{{d\theta }}\left[ {\cos \theta } \right] = - \sin \theta {\text{ and }}\frac{d}{{d\theta }}\left[ {{e^\theta }} \right] = {e^\theta } \cr
& \frac{{dy}}{{d\theta }} = {e^\theta }\left( {\cos \theta - \sin \theta } \right) + \left( {\sin \theta + \cos \theta } \right)\left( {{e^\theta }} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{d\theta }} = {e^\theta }\cos \theta - {e^\theta }\sin \theta + {e^\theta }\sin \theta + {e^\theta }\cos \theta \cr
& \frac{{dy}}{{d\theta }} = 2{e^\theta }\cos \theta \cr} $$
