# How do you find the derivative of 1/(1+x^2)?

Jul 19, 2016

$- \frac{2 x}{1 + {x}^{2}} ^ 2$

#### Explanation:

Two straightforward ways.

$\textcolor{b l u e}{\text{Method One}}$

Rewrite as ${\left(1 + {x}^{2}\right)}^{- 1}$ and use the power and chain rules:

$h ' \left(x\right) = - {\left(1 + {x}^{2}\right)}^{- 2} \cdot 2 x = - \frac{2 x}{1 + {x}^{2}} ^ 2$

$\textcolor{b l u e}{\text{Method Two}}$

Use the quotient rule:

$\frac{d}{\mathrm{dx}} \left(\frac{f \left(x\right)}{g \left(x\right)}\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

$h ' \left(x\right) = \frac{0 - 2 x}{1 + {x}^{2}} ^ 2 = - \frac{2 x}{1 + {x}^{2}} ^ 2$