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Derivative of cos of sin

  1. Apr 28, 2005 #1
    Hi I'm new here, great forum!
    On another thread that I can't find I found a message from jamesrc about differentiating cos of sinx that says:

    df(g(x))/dx = df(g)/dg * dg(x)/dx by the chain rule, this part I get and I thank james for the help the thing that confuses me is why df(g)/dg = cos(cosx).
    Aren't we differentiating df(g) wrt (g) which would give -sin(sinx). What am I missing?
     
  2. jcsd
  3. Apr 28, 2005 #2

    Doc Al

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    Staff: Mentor

    You're not missing anything; you are correct: [itex]d/dx [\cos (\sin x)] = - \sin (\sin x) \cos x [/itex]
     
  4. Apr 28, 2005 #3
    Oh thanks Al! :rofl:
    I just found the thread I read and realised that james was differentiating sin of cos not cos of sin, lol.

    :blushing:
     
  5. Apr 28, 2005 #4

    Doc Al

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    Staff: Mentor

    I'm glad you cleared jamesrc's good name! :smile:

    And welcome to PF, by the way.
     
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