# How do you find the derivative of f(x)=1/(3-2x)?

Jul 27, 2018

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

#### Explanation:

Here ,

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

Diff.w.r.t. $x$ , $\text{using "color(blue)"chain rule :}$

$f ' \left(x\right)$=$\frac{- 1}{3 - 2 x} ^ 2 \frac{d}{\mathrm{dx}} \left(3 - 2 x\right) \to \left[\because \frac{d}{\mathrm{du}} {\left(u\right)}^{-} 1 = - 1 {\left(u\right)}^{-} 2\right]$

$\therefore f ' \left(x\right) = \frac{- 1}{3 - 2 x} ^ 2 \left(- 2\right)$

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