2023-02-20

$\frac{d}{dx}\left(\frac{1}{x}\right)=..;x\ne 0$$\frac{-1}{{x}^{2}}$$\frac{1}{{x}^{2}}$not possible$1$

Joanna Sosa

The ideal decision is A $\frac{-1}{{x}^{2}}$
$Explanationforthecorrectoption:option\left(A\right):y=\frac{1}{x}=\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}{x}^{-1}=-1\left({x}^{-1-1}\right)=-1{x}^{-2}=\frac{-1}{{x}^{2}}$
therefore, option A is correct.

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