# How do you find the derivative of y=(1+e^x)^(-2)?

Jul 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {e}^{x} {\left(1 + {e}^{x}\right)}^{-} 3$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given " y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \text{ chain rule}$

$y = {\left(1 + {e}^{x}\right)}^{-} 2$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {\left(1 + {e}^{x}\right)}^{- 3} \times \frac{d}{\mathrm{dx}} \left(1 + {e}^{x}\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - 2 {e}^{x} {\left(1 + {e}^{x}\right)}^{- 3}$