Answer
$$\frac{{dy}}{{dt}} = 3\left( {\cos 3t} \right)\left( {\ln 2} \right){2^{\sin 3t}}$$
Work Step by Step
$$\eqalign{
& y = {2^{\sin 3t}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{2^{\sin 3t}}} \right] \cr
& {\text{use the rule }}\frac{d}{{dt}}\left[ {{a^u}} \right] = {a^u}\ln a\frac{{du}}{{dx}}. \cr
& {\text{For this exercise}}{\text{, let }}a = 2{\text{ and }}u = \sin 3t.{\text{ Then}}{\text{,}} \cr
& \frac{{dy}}{{dt}} = {2^{\sin 3t}}\left( {\ln 2} \right)\frac{d}{{dt}}\left[ {\sin 3t} \right] \cr
& {\text{solve the derivative and simplify}} \cr
& \frac{{dy}}{{dt}} = {2^{\sin 3t}}\left( {\ln 2} \right)\left( {\cos 3t} \right)\left( 3 \right) \cr
& \frac{{dy}}{{dt}} = 3\left( {\cos 3t} \right)\left( {\ln 2} \right){2^{\sin 3t}} \cr} $$
