kramtus51

2021-12-17

Please, find the derivative of $\frac{1}{{x}^{2}}$.

### Answer & Explanation

Esther Phillips

Let $y=\frac{1}{{x}^{2}}$ which is the same as ${x}^{-2}$
$\frac{dy}{dx}=-2×{x}^{-2-1}=-2{x}^{-3}$
$\frac{dy}{dx}=-\frac{2}{{x}^{-3}}$

Alex Sheppard

$\frac{d}{dx}\left(\frac{1}{{x}^{2}}\right)$
$=\frac{d}{dx}\left({x}^{-2}\right)$
Apply the Power Rule: $\frac{d}{dx}\left({x}^{a}\right)=a×{x}^{a-1}$
$-2{x}^{-2-1}$
$=\frac{-2}{{x}^{3}}$

RizerMix

Use the formula:
$D\frac{f\left(x\right)}{g\left(x\right)}=\frac{{f}^{\prime }\left(x\right)g\left(x\right)-f\left(x\right){g}^{\prime }\left(x\right)}{{g}^{2}\left(x\right)}$
If f(x)=1, you have f'(x)=0 and you get:
$D\frac{1}{g\left(x\right)}=\frac{-{g}^{\prime }\left(x\right)}{{g}^{2}\left(x\right)}$
Since $g\left(x\right)={x}^{2}$, you get ${g}^{\prime }\left(x\right)=2x$. Thus,
$\frac{-{g}^{\prime }\left(x\right)}{{g}^{2}\left(x\right)}=-\frac{2x}{{x}^{2}}=-\frac{2}{{x}^{3}}$

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