f(x) = x² + 2Cos²x, find f ' (x)

Question

a) 2(x+cos x)
b) x - sin x
c) 2x + sin x
d)2(x - sin2x)

I got neither of these answers, since the 2nd part should be chain rule, right?

f(x) = x² + 2Cos²x = x² + 2(Cos x)²
then f '(x) = (2)(2)(cosx)(-sinx)
My answer:
f '(x) = 2x - 4cosxsinx

If I do it this way, then I get that the answer is d:
f(x) = x² + 2Cos²x
f '(x) = x² + 2(-sin x)²
f '(x) = 2x - 2sin²x
f '(x) = 2(x - sin²x)

Is that how I should be doing it?

Answer 1

sorry, this part:
then f '(x) = (2)(2)(cosx)(-sinx)

should say:

then f '(x) = 2x + (2)(2)(cosx)(-sinx)

Answer 2

You're on the right track.

sin 2x= 2sin x cos x (identity)
substitute into your first answer.
f'(x)=2x-2(sin2x)
f'(x)=2(x-sin2x)

Answer 3

Thanks for your reply! That makes much more sense now. I guess I should go and review my identities.