# How do you find the derivative of y=1/(1+e^x)?

Sep 22, 2017

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

#### Explanation:

Differentiate using the chain rule
Given $y = f \left(g \left(x\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \cdot \left(g ' \left(x\right)\right) \leftarrow$chain rule
$y = {\left(1 + {e}^{x}\right)}^{-} 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- {\left(1 + {e}^{x}\right)}^{-} 2\right) \cdot {e}^{x}$
$= - {e}^{x} / {\left(1 + {e}^{x}\right)}^{2}$