# How do you differentiate y=e^-x/x?

Oct 21, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {e}^{-} x \left(x + 1\right)}{x} ^ 2$

#### Explanation:

You need to use the quotient rule; $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

so, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \frac{d}{\mathrm{dx}} \left({e}^{-} x\right) - {e}^{-} x \frac{d}{\mathrm{dx}} \left(x\right)}{x} ^ 2$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \left(- {e}^{-} x\right) - {e}^{-} x \left(1\right)}{x} ^ 2$

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {e}^{-} x \left(x + 1\right)}{x} ^ 2$