How do you find the derivative of #cos^2(2x)#?

Question

8295 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-25T12:59:55

#d/dxcos^2 2x=-2sin4x#

We use the chain rule for functions:

#d/dx[f(x)]^n=n[f(x)]^(n-1)f^'(x)#

So

#d/dxcos^2 2x=2cos^(2-1)2xd/dxcos2x=-4sin2xcos2x#

We can rewrite this in a neater form using

#2sinAcosA=sin2A#

#-4sin2xcos2x=-2*2sin2xcos2x=-2sin4x#