Answer
$$y'=\cot t-1$$
Work Step by Step
$$y=\ln(2 e^{-t}\sin t)$$
We have $$(\ln t)'=\frac{1}{t}$$ $$(e^{t})'=e^t$$
So the derivative of $y$ is $$y'=\frac{1}{2e^{-t}\sin t}(2e^{-t}\sin t)'=\frac{2\Big(e^{-t}\sin t(-t)'+e^{-t}\cos t\Big)}{2e^{-t}\sin t}$$
$$y'=\frac{-e^{-t}\sin t+e^{-t}\cos t}{e^{-t}\sin t}=\frac{e^{-t}(\cos t-\sin t)}{ e^{-t}\sin t}$$
$$y'=\frac{\cos t-\sin t}{\sin t}=\frac{\cos t}{\sin t}-1$$
$$y'=\cot t-1$$
