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

Apr 30, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 - 4 x - 15 {x}^{2} - 12 {x}^{3}}{x + {x}^{2} + 2 {x}^{3}} ^ 2$

#### Explanation:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + 2 x + 6 {x}^{2}\right) \left(- 2 - 3 x\right) - \left(x + {x}^{2} + 2 {x}^{3}\right) \left(- 3\right)}{x + {x}^{2} + 2 {x}^{3}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 - 4 x - 12 {x}^{2} - 3 x - 6 {x}^{2} - 18 {x}^{3} + 3 x + 3 {x}^{2} + 6 {x}^{3}}{x + {x}^{2} + 2 {x}^{3}} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 - 4 x - 15 {x}^{2} - 12 {x}^{3}}{x + {x}^{2} + 2 {x}^{3}} ^ 2$

REMEMBER, the quotient rule is

$y = \frac{u}{v}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v ' u - v u '}{v} ^ 2$