# How do you find the derivative of 5/((x^2) + x + 1)^2?

Mar 12, 2018

$f ' \left(x\right) = \frac{- 10}{{\left({x}^{2} + x + 1\right)}^{2}}$

#### Explanation:

For a question like this, you must use the quotient rule...

If $f \left(x\right) = \frac{\textcolor{red}{u}}{\textcolor{g r e e n}{v}}$ then...

$f ' \left(x\right) = \frac{\textcolor{g r e e n}{v} \textcolor{red}{u '} - \textcolor{g r e e n}{v '} \textcolor{red}{u}}{\textcolor{g r e e n}{{v}^{2}}}$

So for the given function

$f \left(x\right) = \frac{\textcolor{red}{5}}{\textcolor{g r e e n}{{\left({x}^{2} + x + 1\right)}^{2}}}$

$f ' \left(x\right) = \frac{\textcolor{g r e e n}{{\left({x}^{2} + x + 1\right)}^{2}} \textcolor{red}{0} - \textcolor{g r e e n}{2 \left({x}^{2} + x + 1\right)} \textcolor{red}{5}}{\textcolor{g r e e n}{{\left({x}^{2} + x + 1\right)}^{2}}}$

$\implies f ' \left(x\right) = \frac{- 10 \left({x}^{2} + x + 1\right)}{{\left({x}^{2} + x + 1\right)}^{2}}$

$\implies f ' \left(x\right) = \frac{- 10}{{\left({x}^{2} + x + 1\right)}^{2}}$