# How do you find the derivative y=(-3-2x+2x^3)/(-2-x^2) using the quotient rule?

Jul 10, 2018

The answer is $= - \frac{2 \left({x}^{4} + 7 {x}^{2} + 3 x - 2\right)}{2 + {x}^{2}} ^ 2$

#### Explanation:

The derivative of a quotient is

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Here, we have

$y = \frac{- 3 - 2 x + 2 {x}^{3}}{- 2 - {x}^{2}}$

$u = - 3 - 2 x + 2 {x}^{3}$, $\implies$, $u ' = - 2 + 6 {x}^{2}$

$v = - 2 - {x}^{2}$, $\implies$, $v ' = - 2 x$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- 2 + 6 {x}^{2}\right) \left(- 2 - {x}^{2}\right) - \left(- 2 x\right) \left(- 3 - 2 x + 2 {x}^{3}\right)}{- 2 - {x}^{2}} ^ 2$

$= \frac{4 + 2 {x}^{2} - 12 {x}^{2} - 6 {x}^{4} - 6 x - 4 {x}^{2} + 4 {x}^{4}}{2 + {x}^{2}} ^ 2$

$= \frac{4 - 6 x - 14 {x}^{2} - 2 {x}^{4}}{2 + {x}^{2}} ^ 2$

$= - \frac{2 \left({x}^{4} + 7 {x}^{2} + 3 x - 2\right)}{2 + {x}^{2}} ^ 2$