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

Apr 11, 2018

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

#### Explanation:

Using the Quotient Rule,

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

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

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

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

This is as much simplification as we can do.