# How do I differentiate this using quotient rule?

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

Feb 4, 2018

$- \frac{x \left({x}^{2} - 2\right)}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

#### Explanation:

$\frac{\Delta y}{\Delta x} \frac{v}{u} = \frac{\left(u \frac{\Delta y}{\Delta x} v\right) - \left(v \frac{\Delta y}{\Delta x} u\right)}{v} ^ 2$

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

using the chain rule:

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

$\frac{\Delta y}{\Delta x} \left(1 - {x}^{2}\right) = 0 - 2 x = - 2 x$

$\frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot - 2 x = \frac{- 2 x}{2 \sqrt{1 - {x}^{2}}}$

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

$\frac{\Delta y}{\Delta x} \left({x}^{2}\right) = 2 x$

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

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

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

$= {x}^{3} / {\left(1 - {x}^{2}\right)}^{\frac{3}{2}}$

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

${x}^{3} / {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} - \frac{2 x}{1 - {x}^{2}} ^ \left(\frac{1}{2}\right) = {x}^{3} / {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} - \frac{\left(2 x\right) \left(1 - {x}^{2}\right)}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

$\left(2 x\right) \left(1 - {x}^{2}\right) = 2 x - 2 {x}^{3}$

${x}^{3} / {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} - \frac{2 x - 2 {x}^{3}}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right) = \frac{2 x - {x}^{3}}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$

or $- \frac{x \left({x}^{2} - 2\right)}{1 - {x}^{2}} ^ \left(\frac{3}{2}\right)$