# How do you differentiate  f(x)= (x^2 + 2x )/( x^2 -4)?

Jun 13, 2016

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

#### Explanation:

Since the given function is in the form of a fraction, we can use the quotient rule.

The quotient rule states that,

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{{v}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Let ${x}^{2} + 2 x$ be $u$.
Let ${x}^{2} - 4$ be $v$.

Plugging in the values into the quotient rule,

$f ' \left(x\right) = \frac{\left({x}^{2} - 4\right) \left({x}^{2} + 2 x\right) ' - \left({x}^{2} + 2 x\right) \left({x}^{2} - 4\right) '}{{x}^{2} - 4} ^ 2$

Use the power rule to differentiate $\left({x}^{2} + 2 x\right)$ and $\left({x}^{2} - 4\right)$.

Note: Use the formula, $\left({x}^{n}\right) ' = n \cdot {x}^{n - 1}$.

$f ' \left(x\right) = \frac{\left({x}^{2} - 4\right) \left(2 x + 2\right) - \left({x}^{2} + 2 x\right) \left(2 x\right)}{{x}^{2} - 4} ^ 2$

Simplifying,

$f ' \left(x\right) = \frac{\left(2 {x}^{3} + 2 {x}^{2} - 8 x - 8\right) - \left(2 {x}^{3} + 4 {x}^{2}\right)}{{\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}}$

Removing the brackets in the numerator,

$f ' \left(x\right) = \frac{2 {x}^{3} + 2 {x}^{2} - 8 x - 8 - 2 {x}^{3} - 4 {x}^{2}}{{\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}}$

$f ' \left(x\right) = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {x}^{3}}}} + 2 {x}^{2} - 8 x - 8 \textcolor{red}{\cancel{\textcolor{b l a c k}{- 2 {x}^{3}}}} - 4 {x}^{2}}{{\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}}$

$f ' \left(x\right) = \frac{- 2 {x}^{2} - 8 x - 8}{{\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}}$

Factor out $- 2$ from the numerator.

$f ' \left(x\right) = \frac{- 2 \left({x}^{2} + 4 x + 4\right)}{{\left(x + 2\right)}^{2} {\left(x - 2\right)}^{2}}$

Factor the leftover expression in the numerator.

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

$f ' \left(x\right) = \frac{- 2 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(x + 2\right)}^{2}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{\left(x + 2\right)}^{2}}}} {\left(x - 2\right)}^{2}}$

$f ' \left(x\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{- \frac{2}{x - 2} ^ 2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$