Find f'(x) for f(x) = cos^2(3x^3).

1 answer:

6 0

$f(x) = \cos^2(3x^3)\\f'(x) = \frac{d}{dx}[\cos^2(3x^3)]\\= 2\cos(3x^3) \cdot \frac{d}{dx}[cos(3x^3)]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot \frac{d}{dx}[3x^3]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot 9x^2\\= -18x^2\cos(3x^3)\sin(3x^3)$

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