Question

Alright, this may seem extremely dumb, and elementary, but I would like some clarification as to the derivative of...
cos^2(2x)
1. Can that be re-written as: (cos(2x))^2 ?

Any and all help is greatly appreciated! :)

Answer 1

Yes

Answer 2

You can then separate the factors and use the derivative of a product.

\(f(x) = \cos^2 2x = (\cos 2x)^2 = (\cos 2x)(\cos 2x) \)

Answer 3

Do you know how to calculate the derivative of a product?

Answer 4

Okay, then would the derivative be taken like such:

d/dx (cos(2x))^2 = 2(cos(2x)) * (-sin(2x)) * 2

Answer 5

\(f(x) = g(x)h(x)\)

\(f'(x) = g(x)h'(x) + h(x) g'(x) \)

Answer 6

Oh, does it need expanded like that to take the derivative?

Answer 7

No. Anyway you can do it is fine.

Answer 8

So is: d/dx (cos(2x))^2 = 2(cos(2x)) * (-sin(2x)) * 2 correct?

Answer 9

\(2(\cos(2x)) * (-\sin(2x)) * 2\)

You got the line above.
Now you can use a trig identity to make it look simpler.

Answer 10

Or... -4(cos(2x)^2)(sin(2x)

Answer 11

Right, at least do this:

\(-4\sin 2x \cos2x \)

Answer 12

Is that the final derivative?

Answer 13

Now you can do one more thing, if you want it to be even simpler looking.

Use the identity

\(\sin 2x = 2 \sin x \cos x\)

Answer 14

Ohhh, so.... -2sin(4x) ?? :D

Answer 15

\(-4\sin 2x \cos2x\)

\(=-2 (2\sin 2x \cos2x) \)

\(= -2 \sin 4x\)

Answer 16

Is that accurate?

Answer 17

Yes, you beat me to it. :-)

Answer 18

Yay! You are an amazing helper... Thank you for actually working with me, not just regurgitating textbook terminology at this high-school brain.

Answer 19

*thank you very much for the medal as well :)

Answer 20

You are very welcome. I can tell you know your stuff. Good job!