# How do you find the derivative of 2e^-x?

Oct 31, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {e}^{-} x$

#### Explanation:

Recall that $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

Using chain rule, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$,

Let $u = - x$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{d}{\mathrm{du}} 2 {e}^{u} = 2 {e}^{u} = 2 {e}^{-} x$

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} - x = - 1$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - 1 \cdot 2 {e}^{-} x = - 2 {e}^{-} x$