# How do you differentiate J(v)= (v^3-2v) (v^-4 + v^-2)?

Apr 6, 2018

$J ' \left(v\right) = \left(3 {v}^{2} - 2\right) \left({v}^{- 4} + {v}^{- 2}\right) - \left({v}^{3} - 2 v\right) \left(4 {v}^{- 5} + 2 {v}^{- 3}\right)$

#### Explanation:

In order to differentiate $J \left(v\right)$, we have to use the $\textcolor{red}{\text{Product Rule}}$, which states that

$\left[f \left(v\right) \cdot g \left(v\right)\right] ' = f ' \left(v\right) g \left(v\right) + f \left(v\right) g ' \left(v\right)$

In our case, $f \left(v\right) = {v}^{3} - 2 v$ and $g \left(v\right) = {v}^{- 4} + {v}^{- 2}$

The derivative of both of these functions can be found using the Power rule:

$f \left(x\right) = {x}^{k} \implies f ' \left(x\right) = k {x}^{k - 1}$

$f \left(v\right) = {v}^{3} - 2 v$

$\therefore f ' \left(v\right) = \left[{v}^{3} - 2 v\right] ' = \left[{v}^{3}\right] ' - 2 \left[v\right] ' = 3 {v}^{2} - 2$

$g \left(v\right) = {v}^{- 4} + {v}^{- 2}$

$\therefore g ' \left(v\right) = - 4 {v}^{- 5} - 2 {v}^{- 3}$

Therefore, we have:

$J ' \left(v\right) = f ' \left(v\right) g \left(v\right) + f \left(v\right) g ' \left(v\right)$

$J ' \left(v\right) = \left(3 {v}^{2} - 2\right) \left({v}^{- 4} + {v}^{- 2}\right) - \left({v}^{3} - 2 v\right) \left(4 {v}^{- 5} + 2 {v}^{- 3}\right)$