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

Dec 13, 2016

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

#### Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\text{ } \left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $y = \frac{x}{{x}^{2} + 1}$ Then

$\left\{\begin{matrix}\text{Let "u=x & => & (du)/dx=1 \\ "And } v = {x}^{2} + 1 & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = 2 x\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({x}^{2} + 1\right) \left(1\right) - \left(x\right) \left(2 x\right)}{{x}^{2} + 1} ^ 2$
$\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} + 1 - 2 {x}^{2}}{{x}^{2} + 1} ^ 2$
$\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {x}^{2}}{{x}^{2} + 1} ^ 2$