Answer
$$\frac{{dy}}{{dt}} = \left( {2\ln 5\sin 2t} \right)\left( {{5^{ - \cos 2t}}} \right)$$
Work Step by Step
$$\eqalign{
& y = {5^{ - \cos 2t}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{5^{ - \cos 2t}}} \right] \cr
& {\text{use the rule }}\frac{d}{{dt}}\left[ {{a^u}} \right] = {a^u}\ln a\frac{{du}}{{dx}}. \cr
& {\text{For this exercise}}{\text{, let }}a = 5{\text{ and }}u = - \cos 2t.{\text{ Then}}{\text{,}} \cr
& \frac{{dy}}{{dt}} = {5^{ - \cos 2t}}\left( {\ln 5} \right)\frac{d}{{dt}}\left[ { - \cos 2t} \right] \cr
& \frac{{dy}}{{dt}} = - {5^{ - \cos 2t}}\left( {\ln 5} \right)\frac{d}{{dt}}\left[ {\cos 2t} \right] \cr
& {\text{solve the derivative using }}\frac{d}{{dt}}\left[ {\cos at} \right] = - a\sin at \cr
& \frac{{dy}}{{dt}} = - {5^{ - \cos 2t}}\left( {\ln 5} \right)\left( { - 2\sin 2t} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dt}} = \left( {2\ln 5\sin 2t} \right)\left( {{5^{ - \cos 2t}}} \right) \cr} $$
