Need some help with a derivative and the chain rule

  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. Why do you have a sinx there? sinx2 comes from the original function, cosx/sinx comes from differentiating ln(sinx).
  4. 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. 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. 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. 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. 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. I see it now! You're right. I can't thank you enough.. This was driving me crazy. Seriously, thank you so much.
  10. No problem! :) Peace
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?