# How do you differentiate f(x) = (2x-3) ^ -2?

Apr 24, 2016

$\frac{- 4}{2 x - 3} ^ 3$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{ chain rule }}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right)$
$\text{----------------------------------------------}$

f(g(x)) $= {\left(2 x - 3\right)}^{- 2} \Rightarrow f ' \left(g \left(x\right)\right) = - 2 {\left(2 x - 3\right)}^{- 3}$

and g(x) = 2x - 3 → g'(x) = 2
$\text{----------------------------------------------}$
Substitute these values into the derivative

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