Answer
$$\frac{{dy}}{{d\theta }} = {e^{ - 2\theta }}\left( { - 5{\theta ^3}\sin 5\theta - 2{\theta ^3}\cos 5\theta + 3{\theta ^2}\cos 5\theta } \right)$$
Work Step by Step
$$\eqalign{
& y = {\theta ^3}{e^{ - 2\theta }}\cos 5\theta \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{\theta ^3}{e^{ - 2\theta }}\cos 5\theta } \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{d\theta }} = {\theta ^3}{e^{ - 2\theta }}\frac{d}{{d\theta }}\left[ {\cos 5\theta } \right] + \cos 5\theta \frac{d}{{d\theta }}\left[ {{\theta ^3}{e^{ - 2\theta }}} \right] \cr
& {\text{use the product rule for }}\frac{d}{{d\theta }}\left[ {{\theta ^3}{e^{ - 2\theta }}} \right] \cr
& \frac{{dy}}{{d\theta }} = {\theta ^3}{e^{ - 2\theta }}\frac{d}{{d\theta }}\left[ {\cos 5\theta } \right] + \cos 5\theta \left( {{\theta ^3}\frac{d}{{d\theta }}\left[ {{e^{ - 2\theta }}} \right] + {e^{ - 2\theta }}\frac{d}{{d\theta }}\left[ {{\theta ^3}} \right]} \right) \cr
& {\text{solve derivatives using }}\frac{d}{{d\theta }}\left[ {\cos u} \right] = - \sin u\frac{{du}}{{d\theta }}{\text{ and }}\frac{d}{{d\theta }}{e^u} = {e^u}\frac{{du}}{{d\theta }}{\text{ }} \cr
& \frac{{dy}}{{d\theta }} = {\theta ^3}{e^{ - 2\theta }}\left( { - \sin 5\theta } \right)\frac{d}{{d\theta }}\left[ {5\theta } \right] + \cos 5\theta \left( {{\theta ^3}\left( {{e^{ - 2\theta }}} \right)\frac{d}{{d\theta }}\left[ {2\theta } \right] + {e^{ - 2\theta }}\frac{d}{{d\theta }}\left[ {{\theta ^3}} \right]} \right) \cr
& \frac{{dy}}{{d\theta }} = {\theta ^3}{e^{ - 2\theta }}\left( { - \sin 5\theta } \right)\left( 5 \right) + \cos 5\theta \left( {{\theta ^3}\left( {{e^{ - 2\theta }}} \right)\left( { - 2} \right) + {e^{ - 2\theta }}\left( {3{\theta ^2}} \right)} \right) \cr
& {\text{multiply and simplify}} \cr
& \frac{{dy}}{{d\theta }} = - 5{\theta ^3}{e^{ - 2\theta }}\sin 5\theta - 2{\theta ^3}{e^{ - 2\theta }}\cos 5\theta + 3{\theta ^2}{e^{ - 2\theta }}\cos 5\theta \cr
& {\text{factoring out }}{e^{ - 2\theta }} \cr
& \frac{{dy}}{{d\theta }} = {e^{ - 2\theta }}\left( { - 5{\theta ^3}\sin 5\theta - 2{\theta ^3}\cos 5\theta + 3{\theta ^2}\cos 5\theta } \right) \cr} $$
