## Calculus (3rd Edition)

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(e^x)'=e^x$ Recall that $(\sin x)'=\cos x$. Since we have $$y= e^t \sin t$$ then the first derivative $\frac{dy}{dt}$, using the product rule, is given by \begin{align*} \frac{dy}{dt}&= (e^t)' \sin t+e^t (\sin t)'=e^t \sin t+e^t \cos t\\&=e^t( \sin t+ \cos t).\end{align*} The second derivative $\frac{d^2y}{dt^2}$ is given by \begin{align*}\frac{d^2y}{dt^2}&=(e^t)'( \sin t+ \cos t)+e^t( \sin t+ \cos t)' \\ &=e^t( \sin t+ \cos t)+e^t( \cos t- \sin t)\\ &=2e^t\cos t. \end{align*}