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

Mar 5, 2018

$f ' \left(x\right) = - \frac{x - 9}{x + 3} ^ 3$

#### Explanation:

We use the quotient rule to find $f ' \left(x\right)$:

$\frac{d}{\mathrm{dx}} \frac{u}{v} = \frac{v u ' - u v '}{v} ^ 2$

So

$\frac{d}{\mathrm{dx}} \frac{x - 3}{x + 3} ^ 2 = \frac{{\left(x + 3\right)}^{2} - 2 \left(x + 3\right) \left(x - 3\right)}{x + 3} ^ 4 = \frac{x + 3 - 2 x + 6}{x + 3} ^ 3 = - \frac{x - 9}{x + 3} ^ 3$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = - \frac{x - 9}{x + 3} ^ 3$

#### Explanation:

$f \left(x\right) = \frac{x - 3}{x + 3} ^ 2$
let
$y = f \left(x\right)$
$y = \frac{x - 3}{x + 3} ^ 2$
Let
$u = x - 3$
Then
$\frac{\mathrm{du}}{\mathrm{dx}} = 1$
Let
$v = {\left(x + 3\right)}^{2}$
$\frac{\mathrm{dv}}{\mathrm{dx}} = 2 \left(x + 3\right)$
and
${v}^{2} = {\left(x + 3\right)}^{4}$
We have by quotient rule in diferentiation

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
Substituting
$\frac{d}{\mathrm{dx}} \left(\frac{x - 3}{{\left(x + 3\right)}^{2}}\right) = \frac{{\left(x + 3\right)}^{2} \left(1\right) - \left(x - 3\right) \left(2 \left(x + 3\right)\right)}{x + 3} ^ 4$
Simplifying
$\frac{d}{\mathrm{dx}} \left(y\right) = \frac{\left(x + 3\right) \left(x + 3 - 2 \left(x - 3\right)\right)}{x + 3} ^ 4$

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left(x + 3\right) \left(x + 3 - 2 x + 6\right)}{x + 3} ^ 4$

$= \frac{- x + 9}{x + 3} ^ 3$
$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = - \frac{x - 9}{x + 3} ^ 3$