What is the derivative of x/(1+x^2)?

Dec 31, 2016

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

Explanation:

Let $y = \frac{x}{1 + {x}^{2}}$.

We will use the following Quotient Rule for the Derivative :-

$y = \frac{u \left(x\right)}{v \left(x\right)} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \left(x\right) u ' \left(x\right) - u \left(x\right) v ' \left(x\right)}{v \left(x\right)} ^ 2$

Hence,

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {x}^{2}}{1 + {x}^{2}}$.