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

Sep 17, 2016

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

#### Explanation:

We use Chain Rule here. In order to differentiate a function of a function, say $y = f \left(g \left(x\right)\right)$, where we have to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we need to do (a) substitute $u = g \left(x\right)$, which gives us $y = f \left(u\right)$. Then we need to use a formula called Chain Rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$.

Here we have $y = \frac{1}{g} \left(x\right)$ where $g \left(x\right) = 1 + {x}^{2}$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$

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