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

$\frac{dy}{dx} = e^{-x}sin u=e^{-x}sin(e^{-x})$
$\frac{dy}{du} = \frac{d(cos u)}{du} = -sin u$ and, $\frac{du}{dx} = \frac{d(e^{-x})}{dx} = -e^{-x}$ So, $\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}$ $\frac{dy}{dx} = e^{-x}sin u$