# How do you find the third derivative of y=x^2-2/x?

Apr 5, 2018

$\implies f ' ' ' \left(x\right) = \frac{12}{x} ^ \left(4\right)$

#### Explanation:

We have:

$f \left(x\right) = {x}^{2} - \frac{2}{x}$ We can rewrite this as:

$f \left(x\right) = {x}^{2} - 2 {x}^{-} 1$ We use the power rule:

$\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ if $n$ is a constant.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] - \frac{d}{\mathrm{dx}} \left[2 {x}^{-} 1\right]$

$\implies f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] - 2 \cdot \frac{d}{\mathrm{dx}} \left[{x}^{-} 1\right]$

$\implies f ' \left(x\right) = 2 \cdot {x}^{2 - 1} - 2 \cdot - 1 \cdot {x}^{- 1 - 1}$

$\implies f ' \left(x\right) = 2 x - 2 \cdot - {x}^{- 2}$

$\implies f ' \left(x\right) = 2 x + 2 {x}^{- 2}$ Repeat the process.

$\implies f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[2 x\right] + \frac{d}{\mathrm{dx}} \left[2 {x}^{- 2}\right]$

$\implies f ' ' \left(x\right) = 2 \cdot 1 \cdot {x}^{1 - 1} + 2 \cdot - 2 \cdot {x}^{-} 3$

$\implies f ' ' \left(x\right) = 2 - 4 {x}^{-} 3$ Again!

$\implies f ' ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[2\right] - \frac{d}{\mathrm{dx}} \left[4 {x}^{-} 3\right]$

$\implies f ' ' ' \left(x\right) = 0 - 4 \cdot - 3 \cdot {x}^{- 3 - 1}$

$\implies f ' ' ' \left(x\right) = 0 + 12 \cdot {x}^{- 4}$

$\implies f ' ' ' \left(x\right) = 12 \cdot \frac{1}{x} ^ \left(4\right)$

$\implies f ' ' ' \left(x\right) = \frac{12}{x} ^ \left(4\right)$