Posted by **Terry** on Monday, May 5, 2008 at 8:19pm.

f(x) = x² + 2Cos²x, find f ' (x)

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?

- Math, derivatives -
**Terry**, Monday, May 5, 2008 at 8:20pm
sorry, this part:

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

should say:

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

- Math, derivatives -
**Doug**, Monday, May 5, 2008 at 8:57pm
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)

- Math, derivatives -
**Terry**, Monday, May 5, 2008 at 9:24pm
Thanks for your reply! That makes much more sense now. I guess I should go and review my identities.

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