# How do you differentiate f(x)=1/x^2?

Nov 3, 2016

$\frac{d}{\mathrm{dx}} f \left(x\right) = - \frac{2}{x} ^ 3$

#### Explanation:

$f \left(x\right) = \frac{1}{x} ^ 2$
$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 2\right)$
$\frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 2\right) = \frac{d}{\mathrm{dx}} \left({x}^{-} 2\right)$
$\implies \frac{d}{\mathrm{dx}} f \left(x\right) = - 2 \cdot {x}^{-} 3$
$\implies \frac{d}{\mathrm{dx}} f \left(x\right) = - \frac{2}{x} ^ 3$

Nov 3, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{x} ^ 3$

#### Explanation:

Given -

$y = \frac{1}{x} ^ 2$

Method -1
The given function can be written as -

$y = {x}^{-} 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {x}^{- 2 - 1}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {x}^{-} 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{x} ^ 3$

Method - 2
We can apply quotient rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left[\left({x}^{2}\right) \left(0\right)\right] - \left[\left(1\right) \left(2 x\right)\right]}{{\left({x}^{2}\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x}{x} ^ 4$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{x} ^ 3$