# How do you find the derivative of y=3/(2x)^3?

Nov 29, 2017

$- \frac{9}{8} {x}^{-} 4 \mathmr{and} - \frac{9}{8 {x}^{4}}$

#### Explanation:

First of all, let's simplify the fraction to $\frac{3}{8 {x}^{3}}$.
Two options:

Using the Quotient Rule

In case you didn't know, if $y = f \frac{x}{g} \left(x\right)$, then
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g {\left(x\right)}^{2}}$
In this case, $f \left(x\right) = 3$ and $g \left(x\right) = 8 {x}^{3}$. Therefore,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[3\right] ' \cdot \left(8 {x}^{3}\right) - \left[8 {x}^{3}\right] ' \cdot 3}{8 {x}^{3}} ^ 2$.
$= \frac{0 \cdot \left(8 {x}^{3}\right) - \left[8 {x}^{3}\right] ' \cdot 3}{8 {x}^{3}} ^ 2$ (because $\left[3\right] ' = 0$)
$= \frac{- \left[8 {x}^{3}\right] ' \cdot 3}{8 {x}^{3}} ^ 2$
$= \frac{- 24 {x}^{2} \cdot 3}{8 {x}^{3}} ^ 2$ (because $\left[8 {x}^{3}\right] ' = 24 {x}^{2}$)
$= \frac{- 72 {x}^{2}}{64 {x}^{6}}$ (simplifying)
$= \frac{- 9 {x}^{2}}{8 {x}^{6}}$ (reducing)
$= - \frac{9}{8 {x}^{4}}$ (reducing exponents)

That is one way of doing it, but it's kind of hard. There is a simpler way of doing it.

Using the Power Rule

$y = \frac{3}{8 x} ^ 3 = \frac{3}{8} {x}^{- 3}$
In case you didn't know, if $y = {x}^{n}$, then $\frac{\mathrm{dy}}{\mathrm{dx}} = n \cdot {x}^{n - 1}$. Therefore,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{8} \cdot \left(- 3\right) \cdot {x}^{- 3 - 1}$
$= - \frac{9}{8} \cdot {x}^{- 4}$
$= - \frac{9}{8 {x}^{4}}$