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

Jun 26, 2016

$= - \frac{1}{2 \left(2 x - 3\right)}$

#### Explanation:

$\int \setminus {\left(2 x - 3\right)}^{-} 2 \setminus \mathrm{dx}$

the usual rule is $\int \setminus {x}^{n} \setminus \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1}$

here we have

$\int \setminus {\left(2 x - 3\right)}^{-} 2 \setminus \mathrm{dx}$

with sub $z = 2 x - 3$ to make it more transparent, and so that $\mathrm{dz} = 2 \mathrm{dx}$ the integral becomes

$\frac{1}{2} \int \setminus {z}^{-} 2 \setminus \mathrm{dz}$

$= \frac{1}{2} \int \setminus \frac{{z}^{\left(- 2 + 1\right)}}{- 2 + 1} \setminus \mathrm{dz}$

$= \frac{1}{2} \setminus \frac{{z}^{- 1}}{- 1}$

$= - \frac{1}{2} \frac{1}{z}$

$= - \frac{1}{2 \left(2 x - 3\right)}$