Answer
$$\frac{{dy}}{{dt}} = \left( {\frac{1}{t} - \sin t} \right){e^{\cos t + \ln t}}$$
Work Step by Step
$$\eqalign{
& y = {e^{\left( {\cos t + \ln t} \right)}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^{\left( {\cos t + \ln t} \right)}}} \right] \cr
& {\text{We can use the formula }}\cr
&\frac{d}{{dt}}{e^u} = {e^u}\frac{{du}}{{dt}}\cr
&{\text{ where }}u{\text{ is any differentiable function of }}t \cr
& {\text{For this exercise, you can note that }}\cr
&u = 5\cos t + \ln t,{\text{ then}} \cr
& \frac{{dy}}{{dt}} = {e^{\cos t + \ln t}}\frac{d}{{dt}}\left[ {\cos t + \ln t} \right] \cr
& {\text{Solve the derivative and simplify}} \cr
& \frac{{dy}}{{dt}} = {e^{\cos t + \ln t}}\left( { - \sin t + \frac{1}{t}} \right) \cr
& \frac{{dy}}{{dt}} = \left( {\frac{1}{t} - \sin t} \right){e^{\cos t + \ln t}} \cr} $$
