# How do you use the quotien rule to differentiate 1/ (3x^2+2)^3?

Nov 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 18 x}{3 {x}^{2} + 2} ^ 4$

#### 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.

We will need to differentiate the denominator (using the chain rule), so let's do that first :

$\frac{d}{\mathrm{dx}} {\left(3 {x}^{2} + 2\right)}^{3} = 3 {\left(3 {x}^{2} + 2\right)}^{2} \left(6 x\right)$
$\therefore \frac{d}{\mathrm{dx}} {\left(3 {x}^{2} + 2\right)}^{3} = 18 x {\left(3 {x}^{2} + 2\right)}^{2}$

So with $y = \frac{1}{3 {x}^{2} + 2} ^ 3$ the quotient rule gives;

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({\left(3 {x}^{2} + 2\right)}^{3}\right) \left(\frac{d}{\mathrm{dx}} \left(1\right)\right) - \left(1\right) \left(\frac{d}{\mathrm{dx}} {\left(3 {x}^{2} + 2\right)}^{3}\right)}{{\left(3 {x}^{2} + 2\right)}^{3}} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{0 - 18 x {\left(3 {x}^{2} + 2\right)}^{2}}{3 {x}^{2} + 2} ^ 6$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 18 x}{3 {x}^{2} + 2} ^ 4$