# How do you differentiate f(x)= x/(x^3-4x ) using the quotient rule?

Aug 3, 2016

$\frac{- 2 {x}^{3}}{{x}^{3} - 4 x} ^ 2$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

If $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$ then

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{a}{a}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots \left(A\right)$
$\text{--------------------------------------------------------------}$
$g \left(x\right) = x \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = {x}^{3} - 4 x \Rightarrow h ' \left(x\right) = 3 {x}^{2} - 4$
$\text{----------------------------------------------------------}$
Substitute these values into (A)

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

$= \frac{{x}^{3} - 4 x - 3 {x}^{3} + 4 x}{{x}^{3} - 4 x} ^ 2 = \frac{- 2 {x}^{3}}{{x}^{3} - 4 x} ^ 2$