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March 25, 2017

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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 - ,

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

    should say:

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

  • Math, derivatives - ,

    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 - ,

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

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