AP Calculus ABmediummcq1 pt

If g(x)=cos(3x)g(x)=\cos(3x), then g(x)=g''(x)=

A.3sin(3x)-3\sin(3x)
B.9cos(3x)-9\cos(3x)
C.9cos(3x)9\cos(3x)
D.9sin(3x)-9\sin(3x)

Explanation

Core Concept

Differentiating twice, each derivative of cos(3x)\cos(3x) brings a factor of 3-3 and flips between sine and cosine. So f(x)=33cos(3x)=9cos(3x)f''(x)=-3\cdot 3\cos(3x)=-9\cos(3x).

Correct Answer

B9cos(3x)-9\cos(3x)

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