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

Jan 8, 2016

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

#### Explanation:

Use the quotient rule, which states that for a function

$f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

the derivative of the function is

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

Thus, the derivative of $\frac{x}{1 + {x}^{2}}$ is

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

Find each derivative.

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

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

Hence the derivative is equal to

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

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