# How do you integrate y=1/(-2+2x^2) using the quotient rule?

Oct 30, 2016

I assume mean differentiate, not integrate, as the quotient rule is an differentiation tool!

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

#### Explanation:

You need to use the quotient rule; $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

So if $y = \frac{1}{- 2 + 2 {x}^{2}}$ we have;

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- 2 + 2 {x}^{2}\right) \left(\frac{d}{\mathrm{dx}} 1\right) - \left(1\right) \left(\frac{d}{\mathrm{dx}} \left(- 2 + 2 {x}^{2}\right)\right)}{- 2 + 2 {x}^{2}} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{0 - \left(4 x\right)}{- 2 + 2 {x}^{2}} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 x}{- 2 + 2 {x}^{2}} ^ 2$