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

Sep 10, 2016

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

#### Explanation:

The quotient rule states that:

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

In this example; $f \left(x\right) = x$ and $g \left(x\right) = \left({x}^{2} + 1\right)$

Hence, using the quotient rule:

$\frac{d}{\mathrm{dx}} \left(\frac{x}{{x}^{2} + 1}\right) = \frac{1 \cdot \left({x}^{2} + 1\right) - x \cdot \left(2 x + 0\right)}{{x}^{2} + 1} ^ 2$

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

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