# How do you differentiate y=sqrt[(x^2-1)/(x^2+1)]?

Mar 12, 2017

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

#### Explanation:

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

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

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

$\textcolor{w h i t e}{y} = {\left(1 - \frac{2}{{x}^{2} + 1}\right)}^{\frac{1}{2}}$

So using the power rule and chain rule twice we find:

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

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{2} {\left(1 - \frac{2}{{x}^{2} + 1}\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(1 - 2 {\left({x}^{2} + 1\right)}^{- 1}\right)$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{2} {\left(1 - \frac{2}{{x}^{2} + 1}\right)}^{- \frac{1}{2}} \cdot \left(0 + 2 {\left({x}^{2} + 1\right)}^{- 2} \left(2 x\right)\right)$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{1}{2} {\left(1 - \frac{2}{{x}^{2} + 1}\right)}^{- \frac{1}{2}} \cdot \frac{4 x}{{x}^{2} + 1} ^ 2$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{2 x}{{\left({x}^{2} + 1\right)}^{2} \sqrt{1 - \frac{2}{{x}^{2} + 1}}}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{2 x}{{\left({x}^{2} + 1\right)}^{2} \sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}}$