## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{d\theta }} = {3^{\tan \theta }}{\left( {\ln 3} \right)^2}{\sec ^2}\theta$$
\eqalign{ & y = {3^{\tan \theta }}\ln 3 \cr & {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr & \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{3^{\tan \theta }}\ln 3} \right] \cr & \frac{{dy}}{{d\theta }} = \ln 3\frac{d}{{d\theta }}\left[ {{3^{\tan \theta }}} \right] \cr & {\text{Use the general power rule for differentiation }}\cr & \frac{d}{{d\theta }}\left[ {{a^u}} \right] = {a^u}\left( {\ln a} \right)\frac{{du}}{{d\theta }}{\text{ }} \cr & {\text{For this exercise let }}a = 3{\text{ and }}u = \tan \theta {\text{; then}}{\text{,}} \cr & \frac{{dy}}{{d\theta }} = \ln 3\left( {{3^{\tan \theta }}} \right)\left( {\ln 3} \right)\frac{d}{{d\theta }}\left[ {\tan \theta } \right] \cr & {\text{solve the derivative}} \cr & \frac{{dy}}{{d\theta }} = \ln 3\left( {{3^{\tan \theta }}} \right)\left( {\ln 3} \right)\left( {{{\sec }^2}\theta } \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{d\theta }} = {3^{\tan \theta }}{\left( {\ln 3} \right)^2}{\sec ^2}\theta \cr}