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

6 0

3 years ago

I need help with this question someone

Masja [62]

Answer:

Step-by-step explanation:

5 0

3 years ago

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What is the exact area of a circle having diameter 4in

BartSMP [9]

The diameter of a circle is calculated by using the formula D=r2pi so plug in the known values to find r=2/pi. Next plug r into the area of a circle formula(A=r^2pi) do so and get 4/pi

4 0

3 years ago

What property would you use to solve m+6= -4

iris [78.8K]

You would use the subtraction property of equality.

5 0

3 years ago

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