# How do you use the quotient rule to differentiate (2x+1)/(x^2-1)?

Jan 20, 2016

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

#### Explanation:

Standard form: $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

Given:$\textcolor{w h i t e}{. .} y = \frac{2 x + 1}{{x}^{2} - 1}$

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

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

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

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

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