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

Oct 23, 2016

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

#### Explanation:

In this exercise we are going to use the derivative of the quotient$\frac{u \left(x\right)}{v \left(x\right)}$ and that of a power function:

color(red)(((u(x))/(v(x)))'=(u'(x)v(x)-v'(x)u(x))/(v(x))^2

$\textcolor{b l u e}{\left({\left(u \left(x\right)\right)}^{n}\right) ' = n \cdot u \left(x\right) \cdot u ' \left(x\right)}$

We will apply the following polynomial identities:
$\textcolor{b r o w n}{{\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}}$
$\textcolor{p u r p \le}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$

Based on the above properties let us compute $f ' \left(x\right)$

f'(x)=color(red)((x'*(x+3)^2-color(blue)(((x+3)^2)')*x)/((x+3)^2)^2

$f ' \left(x\right) = \frac{{\left(x + 3\right)}^{2} - \textcolor{b l u e}{2 {\left(x + 3\right)}^{2 - 1} \cdot \left(x + 3\right) '} \cdot x}{x + 3} ^ 4$
$f ' \left(x\right) = \frac{\textcolor{b r o w n}{{x}^{2} + 6 x + 9} - 2 \left(x + 3\right) \left(1\right) \cdot x}{x + 3} ^ 4$
$f ' \left(x\right) = \frac{{x}^{2} + 6 x + 9 - 2 x \left(x + 3\right)}{x + 3} ^ 4$
$f ' \left(x\right) = \frac{{x}^{2} + 6 x + 9 - 2 {x}^{2} - 6 x}{x + 3} ^ 4$
$f ' \left(x\right) = \frac{- {x}^{2} + 9}{x + 3} ^ 4$
$f ' \left(x\right) = \frac{\textcolor{p u r p \le}{9 - {x}^{2}}}{x + 3} ^ 4$
$f ' \left(x\right) = \frac{\textcolor{p u r p \le}{{3}^{2} - {x}^{2}}}{x + 3} ^ 4$
$f ' \left(x\right) = \frac{\textcolor{p u r p \le}{\left(3 - x\right) \left(3 + x\right)}}{x + 3} ^ 4$

Simplify the fraction by $x + 3$
$f ' \left(x\right) = \frac{\textcolor{p u r p \le}{\left(3 - x\right) \cancel{\left(3 + x\right)}}}{x + 3} ^ \left(\cancel{4} \textcolor{\mathmr{and} a n \ge}{3}\right)$

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