Answer
$$y'=e^{\sin t}\Big(\cos t(\ln t^2+1)+\frac{2}{t}\Big)$$
Work Step by Step
$$y=e^{\sin t}(\ln t^2+1)$$
We have $$(\ln t)'=\frac{1}{t}$$ $$(e^{t})'=e^t$$
We also need to apply the Product Rule and the Chain Rule here. The derivative of $y$ is $$y'=e^{\sin t}(\sin t)'(\ln t^2+1)+e^{\sin t}\Big(\frac{1}{t^2}(t^2)'+0\Big)$$
$$y'=e^{\sin t}\cos t(\ln t^2+1)+e^{\sin t}\frac{2t}{t^2}$$
$$y'=e^{\sin t}\cos t(\ln t^2+1)+\frac{2e^{\sin t}}{t}$$
$$y'=e^{\sin t}\Big(\cos t(\ln t^2+1)+\frac{2}{t}\Big)$$
