# How do you differentiate f(x)=1/(1+x^2)^2?

Oct 17, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 x}{1 + {x}^{2}} ^ 3$

#### Explanation:

We have $y = \frac{1}{1 + {x}^{2}} ^ 2$

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

Differentiate both sides using the chain rule.

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

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \cdot 2 x}{1 + {x}^{2}} ^ 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 4 x}{1 + {x}^{2}} ^ 3$