# How do you find the derivative of f(x)=2/(3x^3-x^2-24x-4)?

Nov 9, 2016

$f ' \left(x\right) = - \frac{2 \left(9 {x}^{2} - 2 x - 24\right)}{3 {x}^{3} - {x}^{2} - 24 x - 4} ^ 2$

#### Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $f \left(x\right) = \frac{2}{3 {x}^{3} - {x}^{2} - 24 x - 4}$ Then

$\left\{\begin{matrix}\text{Let "u=2 & => & (du)/dx=0 \\ "And } v = 3 {x}^{3} - {x}^{2} - 24 x - 4 & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = 9 {x}^{2} - 2 x - 24\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore f ' \left(x\right) = \frac{\left(3 {x}^{3} - {x}^{2} - 24 x - 4\right) \left(0\right) - \left(2\right) \left(9 {x}^{2} - 2 x - 24\right)}{3 {x}^{3} - {x}^{2} - 24 x - 4} ^ 2$
$\therefore f ' \left(x\right) = - \frac{2 \left(9 {x}^{2} - 2 x - 24\right)}{3 {x}^{3} - {x}^{2} - 24 x - 4} ^ 2$

Incidental, we could also use the chain rule as follows;

$f \left(x\right) = \frac{2}{3 {x}^{3} - {x}^{2} - 24 x - 4}$
$\therefore f \left(x\right) = 2 {\left(3 {x}^{3} - {x}^{2} - 24 x - 4\right)}^{-} 1$
$\therefore f ' \left(x\right) = 2 \left(- 1\right) {\left(3 {x}^{3} - {x}^{2} - 24 x - 4\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{3} - {x}^{2} - 24 x - 4\right)$
$\therefore f ' \left(x\right) = - \frac{2}{3 {x}^{3} - {x}^{2} - 24 x - 4} ^ 2 \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{3} - {x}^{2} - 24 x - 4\right)$
$\therefore f ' \left(x\right) = - \frac{2}{3 {x}^{3} - {x}^{2} - 24 x - 4} ^ 2 \cdot \left(9 {x}^{2} - 2 x - 24\right)$
$\therefore f ' \left(x\right) = - \frac{2 \left(9 {x}^{2} - 2 x - 24\right)}{3 {x}^{3} - {x}^{2} - 24 x - 4} ^ 2$