Answer
$$\frac{{dy}}{{dt}} = \cot t - 1$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {2{e^{ - t}}\sin t} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\ln \left( {2{e^{ - t}}\sin t} \right)} \right] \cr
& {\text{use the rule }}\frac{d}{{dt}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dt}};{\text{ for this exercise consider }}u = 2{e^{ - t}}\sin t;{\text{ then:}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{2{e^{ - t}}\sin t}}\frac{d}{{dt}}\left[ {2{e^{ - t}}\sin t} \right] \cr
& \frac{{dy}}{{dt}} = \frac{1}{{{e^{ - t}}\sin t}}\frac{d}{{dt}}\left[ {{e^{ - t}}\sin t} \right] \cr
& {\text{use the product rule }} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{{e^{ - t}}\sin t}}\left( {{e^{ - t}}\frac{d}{{dt}}\left[ {\sin t} \right] + \sin t\frac{d}{{dt}}\left[ {{e^{ - t}}} \right]} \right) \cr
& {\text{solving derivatives }} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{{e^{ - t}}\sin t}}\left( {{e^{ - t}}\cos t + \sin t\left( { - {e^{ - t}}} \right)} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{{e^{ - t}}\sin t}}\left( {{e^{ - t}}\cos t - {e^{ - t}}\sin t} \right) \cr
& \frac{{dy}}{{dt}} = \frac{{{e^{ - t}}\cos t - {e^{ - t}}\sin t}}{{{e^{ - t}}\sin t}} \cr
& \frac{{dy}}{{dt}} = \frac{{\cos t - \sin t}}{{\sin t}} \cr
& \frac{{dy}}{{dt}} = \frac{{\cos t}}{{\sin t}} - \frac{{\sin t}}{{\sin t}} \cr
& \frac{{dy}}{{dt}} = \cot t - 1 \cr} $$
