Question

Find dy/dx of y=cos(cos2x)

Answer 1

use the chain rule:
note
derivative of cos x is -sinx
and
that of cos2x is -2 sin2x

Answer 2

@cwrw238 Thank you! That is actually what I wrote down as well. So the answer would then be \[2\sin(\sin2x) \]?

Answer 3

@cwrw238 This is from an online quiz & I typed in that answer and it was marked incorrect but im not sure why?

Answer 4

When you take the derivative of the outer function, the inside should stay the same.\[\Large y=\cos\left[\color{orangered}{\cos2x}\right]\]
\[\Large y'=-\sin\left[\color{orangered}{\cos2x}\right]\color{royalblue}{(\cos2x)'}\]
See how I left the inner function the same when the outer changed to -sine?
Chain rule tells us to make a copy of the inside, take it's derivative and multiply.
So we need derivative of this blue part.

Answer 5

You were on the right track.
Just remember when you apply the chain rule, it should be applied OUTSIDE of the function.
You shouldn't be changing the inner function inside like that :O\[\Large y'=-\sin\left[\cos2x\right](-2\sin2x)\]\[\Large y'=2(\sin2x)\sin\left[\cos2x\right]\]

Answer 6

Hm ok, following what you said I got:\[-\sin(\cos(2x))[-2\sin(2x)]\]

Answer 7

I see what you mean! I totally changed the inner function -_- Therefore, the solution would equal to 2(sin2x)sin(cos2x)?

Answer 8

@zepdrix

Answer 9

Yah looks good! :)
Hmm website must be lagging, I guess my messages are taking a while to show up :3

Answer 10

Its fine! Your great thank you so much! I was on my last try on this question & with your help I got the correct answer :) @zepdrix