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