# How do you find the derivative of 1/x^8?

Nov 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{8}{x} ^ 9$

#### Explanation:

Recall that $\frac{1}{x} ^ n = {x}^{-} n$

$\therefore \frac{1}{x} ^ 8 = {x}^{-} 8$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 8 {x}^{-} 9 \to$power rule

$= - \frac{8}{x} ^ 9$

OR

Using quotient rule, if $y = \frac{1}{x} ^ 8$, let $u = 1$ and $v = {x}^{8}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 8 {x}^{7}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{8} \cdot 0 - 8 {x}^{7}}{x} ^ 16$

$= \frac{- 8 {x}^{7}}{x} ^ 16$

$= - \frac{8}{x} ^ 9$

$= - 8 {x}^{-} 9$

Nov 2, 2016

$\frac{d}{\mathrm{dx}} \left({x}^{8}\right) = - \frac{8}{x} ^ 9$

#### Explanation:

$y = \frac{1}{x} ^ 8$
$\implies y = {x}^{-} 8$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 8 {x}^{- 8 - 1}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 8 {x}^{-} 9$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{8}{x} ^ 9$