# How do you find the derivatives of f(x) = (2x-3) ^ -2?

Apr 27, 2015

One way to approach this problem is to let
$f \left(x\right) = g \left(h \left(x\right)\right)$
where
$g \left(x\right) = {x}^{- 2}$
and
$h \left(x\right) = 2 x - 3$

By the Chain Rule
$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \textcolor{red}{\frac{d g \left(h \left(x\right)\right)}{d h \left(x\right)}} \cdot \textcolor{b l u e}{\frac{d h \left(x\right)}{\mathrm{dx}}}$

$= \textcolor{red}{- 2 {\left(h \left(x\right)\right)}^{- 3}} \cdot \textcolor{b l u e}{2}$

$= \textcolor{red}{- 2 {\left(2 x - 3\right)}^{- 3}} \cdot \textcolor{b l u e}{2}$

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