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Homework Help: Need some help with a derivative and the chain rule

  1. Mar 4, 2010 #1
    Okay so I'm doing chain rule work to go over the stuff from calc 1 before I take a departmental exam and I've run into this problem:

    1. The problem statement, all variables and given/known data
    Take the derivative of:

    [tex] f(x) = \frac{sin(x^2)}{ln sinx}[/tex]

    2. Relevant equations
    Here's the formula I used (and always do) for the chain rule:

    [tex] \frac{d}{dx}[f(g(x))] = f'(g(x))*g'(x) [/tex]

    3. The attempt at a solution

    [tex] f'(x) = \frac{ln sinx(cos(x^2)2x) - sin(x^2)\frac{1}{sinx}sinx cosx}{[ln sinx]^2}[/tex]

    So, I've done this simple derivative problem 3 times and for some reason I keep getting this answer. E-Mailed my Prof and he said that the sinx right after the 1/sinx is not suposed to be there, but I cannot see why...Can anyone verify the correct answer for me just to see what I'm not grasping here. I'm very frustrated.
  2. jcsd
  3. Mar 4, 2010 #2
    Why do you have a sinx there? sinx2 comes from the original function, cosx/sinx comes from differentiating ln(sinx).
  4. Mar 4, 2010 #3
    I'm thinking [tex]g(x) = sinx[/tex] so by the chain rule the bottom [tex]lnsinx = \frac{1}{sinx}sinxcosx[/tex]

    And you and my prof are saying that [tex]lnsinx = \frac{1}{sinx}cosx[/tex]

    But what happens to the [tex]g(x)[/tex]
  5. Mar 4, 2010 #4
    D [ln(sinx)] = (1/sinx)cosx.
    (where D is the derivative)

    The general formula you are using is D [a(x)/b(x)] = [a'(x)b(x) - b'(x)a(x)] / [b(x)]2
    You have most the terms right, it is the b'(x)a(x) term that is incorrect.
  6. Mar 4, 2010 #5
    But why? If the rule is [tex]f'(x) = f'(g(x))*g'(x)[/tex] than what happens to the [tex]g(x)[/tex] which in this case [tex]= sinx[/tex]
  7. Mar 4, 2010 #6
    That's the quotient rule, but I'm talking about the chain rule that needs to be used to get [tex]f'(x)[/tex]
  8. Mar 4, 2010 #7
    Show step by step where the problem is.
    If you are trying to differentiate ln(sinx), take the g(x)=sinx. then D [ln(g(x))] = 1/g(x) * g'(x). What is g'(x) here? It is cosx. Thus D [ln(g(x))] = 1/g(x) * g'(x) = (1/sinx)cosx
  9. Mar 4, 2010 #8
    I see it now! You're right. I can't thank you enough.. This was driving me crazy. Seriously, thank you so much.
  10. Mar 4, 2010 #9
    No problem! :) Peace
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