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A simple derivative that I'm messing up on

  1. Aug 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Find f'(x) for f(x) = sin x + 2 cos^3 x

    2. Relevant equations

    Other than basic derivative properties and formulas, no.

    3. The attempt at a solution

    f'(x) = sin x + 2cos^3 x
    = cos x - 6sin^2 x

    The book says the answer is f'(x) = cos x - 6cos^2 x sin x however I don't understand where the sin x comes from... any help?
  2. jcsd
  3. Aug 12, 2011 #2


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    Do you know the chain rule?
  4. Aug 12, 2011 #3
    Yeah, d/dx[f(g(x))] = f'(g(x))g'(x). My bad, now I got the correct answer of f'(x) = cos x - 6cos^2 x sin x.. I think I used the power rule on 2cos^3 x by mistake before.
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