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

Jun 22, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 x}{{x}^{2} + 1} ^ 2$

#### Explanation:

We can also rewrite the function:

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

Then, through the chain rule, we see that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \left(- {\left({x}^{2} + 1\right)}^{-} 2\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right) = \frac{2}{{x}^{2} + 1} ^ 2 \left(2 x\right) = \frac{4 x}{{x}^{2} + 1} ^ 2$