# How do you find the derivative of g(x)=(2x^2+x+1)^-3?

Apr 13, 2018

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

#### Explanation:

We use the chain rule, which states that,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = 2 {x}^{2} + x + 1 , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = 4 x + 1$.

Now, $y = {u}^{-} 3 , \therefore \frac{\mathrm{dy}}{\mathrm{du}} = - 3 {u}^{-} 4$.

And so,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {u}^{-} 4 \left(4 x + 1\right)$

Now, substitute back $u = 2 {x}^{2} + x + 1$ to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {\left(2 {x}^{2} + x + 1\right)}^{-} 3 \left(4 x + 1\right)$

I'd clean this up and rearrange it into:

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