# How do you find the derivative of 1/x^2?

Mar 30, 2015

let $y = \frac{1}{{x}^{2}}$
$\frac{1}{{x}^{2}}$ is the same as ${x}^{- 2}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \cdot {x}^{- 2 - 1}$
$= - 2 {x}^{- 3}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{{x}^{-} 3}$

Mar 30, 2015

To find the derivative of a ratio, apply the formula
$D f \frac{x}{g} \left(x\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{{g}^{2} \left(x\right)}$
In particular, if $f \left(x\right) = 1$, you have that $f ' \left(x\right) = 0$, and the expression becomes
$D \frac{1}{g} \left(x\right) = \frac{- g ' \left(x\right)}{{g}^{2} \left(x\right)}$

Since your $g \left(x\right)$ is ${x}^{2}$, you easily get that $g ' \left(x\right) = 2 x$. So,
$\frac{- g ' \left(x\right)}{{g}^{2} \left(x\right)} = - \frac{2 x}{{x}^{4}} = - \frac{2}{x} ^ 3$