# How do you differentiate y=x^3/(1-x^2)?

Feb 20, 2017

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

#### Explanation:

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

find the derivative using the Quotient Rule $y ' = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$
when $f \left(x\right) = {x}^{3}$ and $g \left(x\right) = 1 - {x}^{2}$

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

$y ' = \frac{\left(3 {x}^{2}\right) \left(1 - {x}^{2}\right) - \left({x}^{3}\right) \left(- 2 x\right)}{\left(1 - {x}^{2}\right)} ^ 2 = \frac{3 {x}^{2} - 3 {x}^{4} + 2 {x}^{4}}{1 - {x}^{2}} ^ 2 = \frac{- {x}^{4} + 3 {x}^{2}}{1 - {x}^{2}} ^ 2 = \frac{{x}^{2} \left(3 - {x}^{2}\right)}{1 - {x}^{2}} ^ 2$