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Homework Help: Simple derivation I just can't get

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data
    g(x)=ln(x*sqrt(x2-11)


    2. Relevant equations
    d/dx lnx=1/x


    3. The attempt at a solution
    I've attempted a few solutions:
    1/(x*sqrt(x2-11)
    (1/x)+1/(sqrt(x2-11)
    etc

    but they have been wrong. I'm not sure how to differentiate a ln of a product...thats why I tried to break it into lnx+ln sqrt(x2-11
     
  2. jcsd
  3. Mar 1, 2010 #2
    You were correct in how you began to differentiate g(x), but after you use D[ln(x)]=1/x, you must apply the chain rule. That is, you must multiply by x*sqrt(x2-11)
     
  4. Mar 1, 2010 #3

    Mentallic

    User Avatar
    Homework Helper

    Actually, while that rule is correct, it's incomplete.

    [tex]\frac{d}{dx}ln\left(f(x)\right)=\frac{f'(x)}{f(x)}[/tex]

    This works even for ln(x) since the derivative of x is 1, so nothing changes.
    All you need to do is multiply everything by the derivative of what's inside the log function.
     
  5. Mar 2, 2010 #4

    Mark44

    Staff: Mentor

    What you started doing here would have worked if you had completed it. Also, for future reference, your problem statement should say what it is that your are trying to do. You didn't make it explicit that you wanted to find the derivative of g(x). The word "derivation" does not imply that you are taking the derivative -- the word for that is differentiation.

    One other thing: you started with an equation -- g(x) = ln(x*sqrt(x2-11) -- each step should have been an equation.

    So, you started with
    g(x) = ln(x*sqrt(x2-11) = ln(x) + ln(sqrt(x2 - 11) = ln(x) + (1/2)ln(x2 - 11)
    ==> g'(x) = d/dx(ln(x)) + d/dx[(1/2)ln(x2 - 11)) = 1/x + (1/2)*1/(x2 - 11)*d/dx(x2 - 11)
    ==> g'(x) = 1/x + 2x/(2(x2 - 11)) = 1/x + x/(x2 - 11)

    In the 2nd line I used the fact that d/dx(kf(x)) = k*d/dx(f(x)) = k*f'(x).
    I also used the chain rule form of the derivative of the natural log function - d/dx(ln(u) = 1/u * du/dx.
     
  6. Mar 2, 2010 #5
    Thanks guys! Finally got the right answer with your help. I'll remember to format my question the right way next time too,
     
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