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Question About the Derivative

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data

    f(x) = ln(sin(x))

    find f'(x)

    3. The attempt at a solution

    I had a question about this problem. The answer on the website said that f'(x) = cot(x) or cos(x)/sin(x) but I disagreed.

    I got d/du ln(u) = du/u so f'(x) = cos(x)/sin(x) * cos(x) << for the chain rule because of the derivative of the "inner" function. So my answer was cos^2(x)/sin(x).

    Are they right or am I? Thanks.
     
  2. jcsd
  3. Feb 28, 2008 #2

    D H

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    The chain rule is for some function of the form f(g(x)), df/dx = df/dg*dg/gx[/itex]. Here, g(x)=sin(x) and f(g)=ln(g). So, df/dg=1/g=1/sin(x) and dg/dx=cos(x). Putting these results together,

    [tex]\frac d{dx} \ln(\sin x) = \frac 1 {\sin x} \cos x = \frac {\cos x}{\sin x} = \cot x[/tex]
     
  4. Feb 29, 2008 #3

    HallsofIvy

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    In other words, you applied the chain rule twice when you shouldn't have!

    The derivative of ln(sin(x)) is 1/sin(x) times the derivative of sin(x)= 1/sin(x) times cos(x)= cos(x)/sin(x)= cot(x). You had already put the cos(x) in the numerator, then multiplied by it again.

    More formally, d(f(u))/dx= df/du * du/dx. Here, u= sin(x) so ln(sin(x))= ln(u) and its derivative is df/du= 1/u= 1/sin(x). Then you multiply by du/dx= cos(x).
     
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