AP Calculus ABmediummcq1 pt

If f(x)=1x2f(x)=\dfrac{1}{x^2}, then f(x)=f'(x)=

A.2x-\dfrac{2}{x}
B.2x3\dfrac{2}{x^3}
C.1x3-\dfrac{1}{x^3}
D.2x3-\dfrac{2}{x^3}

Explanation

Core Concept

Rewrite 1x2=x2\dfrac{1}{x^2}=x^{-2}, so f(x)=2x3f'(x)=-2x^{-3}. In fraction form this is 2x3-\dfrac{2}{x^3}.

Correct Answer

D2x3-\dfrac{2}{x^3}

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