# How do you differentiate f(x)=(x-5)/(x-3)^2 using the quotient rule?

Dec 18, 2015

$f ' \left(x\right) = \frac{- x + 7}{x - 3} ^ 3 = - \frac{x - 7}{x - 3} ^ 3$

#### Explanation:

How to differentiate using the quotient rule?

Given $f \left(x\right) = \frac{x - 5}{x - 3} ^ 2$

$f ' \left(x\right) = \frac{{\left(x - 3\right)}^{2} \frac{d}{\mathrm{dx}} \left(x - 5\right) - \left(x - 5\right) \frac{d}{\mathrm{dx}} {\left(x - 3\right)}^{2}}{{\left(x - 3\right)}^{2}} ^ 2$

$= \frac{{\left(x - 3\right)}^{2} \left(1\right) - \left(x - 5\right) \left(2 \left(x - 3\right) \left(1\right)\right)}{x - 3} ^ 4$

$= \frac{\textcolor{red}{\left(x - 3\right)} \left(x - 3\right) - 2 \textcolor{red}{\left(x - 3\right)} \left(x - 5\right)}{\textcolor{red}{\left(x - 3\right)} {\left(x - 3\right)}^{3}}$

Factor out the greatest common factor $\left(x - 3\right)$

$f ' \left(x\right) = \frac{\left(x - 3\right) - 2 \left(x - 5\right)}{x - 3} ^ 3$

$= \frac{x - 3 - 2 x + 10}{x - 3} ^ 3$

$f ' \left(x\right) = \frac{- x + 7}{x - 3} ^ 3 = - \frac{x - 7}{x - 3} ^ 3$