Answer
$$\frac{{dy}}{{dt}} = \left( {\frac{2}{t} + \left( {\ln {t^2} + 1} \right)\cos t} \right){e^{\sin t}}$$
Work Step by Step
$$\eqalign{
& y = {e^{\sin t}}\left( {\ln {t^2} + 1} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^{\sin t}}\left( {\ln {t^2} + 1} \right)} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dt}} = {e^{\sin t}}\frac{d}{{dt}}\left[ {\ln {t^2} + 1} \right] + \left( {\ln {t^2} + 1} \right)\frac{d}{{dt}}\left[ {{e^{\sin t}}} \right] \cr
& {\text{use the logarithmic property }}\ln {u^n} = n\ln u \cr
& \frac{{dy}}{{dt}} = {e^{\sin t}}\frac{d}{{dt}}\left[ {2\ln t + 1} \right] + \left( {\ln {t^2} + 1} \right)\frac{d}{{dt}}\left[ {{e^{\sin t}}} \right] \cr
& {\text{solve the derivative and simplify}} \cr
& \frac{{dy}}{{dt}} = {e^{\sin t}}\left( {\frac{2}{t}} \right) + \left( {\ln {t^2} + 1} \right)\left( {{e^{\sin t}}\cos t} \right) \cr
& \frac{{dy}}{{dt}} = \frac{{2{e^{\sin t}}}}{t} + \left( {\ln {t^2} + 1} \right){e^{\sin t}}\cos t \cr
& {\text{Factoring}} \cr
& \frac{{dy}}{{dt}} = \left( {\frac{2}{t} + \left( {\ln {t^2} + 1} \right)\cos t} \right){e^{\sin t}} \cr} $$
