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

Feb 8, 2017

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

#### Explanation:

Since $\left(c f \left(x\right)\right) ' = c f ' \left(x\right)$

We can say

$\left(\frac{3}{y} ^ 3\right) ' = 3 \left(\frac{1}{y} ^ 3\right) '$

Since $\frac{1}{y} ^ 3 = {y}^{-} 3$ then,

$3 \left(\frac{1}{y} ^ 3\right) ' = 3 \left({y}^{-} 3\right) '$

The Power Rule states $\left({x}^{n}\right) ' = n {x}^{n - 1}$, then

$3 \left({y}^{-} 3\right) ' = 3 \left(- 3\right) {x}^{- 3 - 1} = - 9 {x}^{-} 4 = - \frac{9}{x} ^ 4$