## University Calculus: Early Transcendentals (3rd Edition)

$$y'=e^{\sin t}\Big(\cos t(\ln t^2+1)+\frac{2}{t}\Big)$$
$$y=e^{\sin t}(\ln t^2+1)$$ 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 $$y'=e^{\sin t}(\sin t)'(\ln t^2+1)+e^{\sin t}\Big(\frac{1}{t^2}(t^2)'+0\Big)$$ $$y'=e^{\sin t}\cos t(\ln t^2+1)+e^{\sin t}\frac{2t}{t^2}$$ $$y'=e^{\sin t}\cos t(\ln t^2+1)+\frac{2e^{\sin t}}{t}$$ $$y'=e^{\sin t}\Big(\cos t(\ln t^2+1)+\frac{2}{t}\Big)$$