AP Calculus ABmediummcq1 pt

If f(x)=cotxf(x)=\cot x, then f(x)=f'(x)=

A.csc2x\csc^2 x
B.sec2x-\sec^2 x
C.sec2x\sec^2 x
D.csc2x-\csc^2 x

Explanation

Core Concept

The derivative of cotx\cot x is csc2x-\csc^2 x. It pairs with ddx[tanx]=sec2x\dfrac{d}{dx}[\tan x]=\sec^2 x but with cofunctions and a minus sign.

Correct Answer

Dcsc2x-\csc^2 x

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