# How do you find the derivative of f(x)= 1/(9x+6)^2?

Feb 22, 2016

$f ' \left(x\right) = - \frac{2}{3 \cdot {\left(3 x + 2\right)}^{3}}$

#### Explanation:

Since this is under the chain rule, then use it.

Substitute $u = 9 x + 6$.

$\frac{\text{d"u}{"d} x}{=} 9$

So plug it in.

$f ' \left(x\right) = \frac{\text{d"}{"d} x}{\frac{1}{9 x + 6} ^ 2}$

$= \frac{\text{d"}{"d} x}{\frac{1}{u} ^ 2}$

= frac{"d"}{"d"u}(u^{-2})*frac{"d"u}{"d"x}

$= \left(- 2\right) \cdot {u}^{- 3} \cdot \left(9\right)$

$= - \frac{18}{9 x + 6} ^ \left\{3\right\}$

$= - \frac{18}{{3}^{3} \cdot {\left(3 x + 2\right)}^{3}}$

$= - \frac{2}{3 \cdot {\left(3 x + 2\right)}^{3}}$