# How do you find the derivative of h(x)= 1/(6x^2+x+1)^2?

May 17, 2016

#### Answer:

$- 2 \frac{12 x + 1}{6 {x}^{2} + x + 1} ^ 3$

#### Explanation:

This can be done using the chain rule, we have:

$h \left(x\right) = \frac{1}{6 {x}^{2} + x + 1} ^ 2$

This can be re written as:

$= {\left(6 {x}^{2} + x + 1\right)}^{- 2}$

To apply the chain rule, differentiate the function around the bracket then multiply by the derivative of the function inside the bracket.

$\to h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left\{{\left(6 {x}^{2} + x + 1\right)}^{- 2}\right\} =$

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

$= - 2 {\left(6 {x}^{2} + x + 1\right)}^{- 3} \left(12 x + 1\right)$

Which we can write as:

$- 2 \frac{12 x + 1}{6 {x}^{2} + x + 1} ^ 3$

Hence, our final answer.