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Differentiate [ln(u)]/[1+ln(2u)]

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Differentiate the function

    [tex]f(u)=\frac{\ln u}{1+\ln(2u)}[/tex]


    2. Relevant equations
    - Quotient rule?
    - Logarithmic differentiation?


    3. The attempt at a solution
    I just learned about logarithmic differentiation so I think the idea here was to use logarithmic differentiation.. however, with ln's on the top and bottom, I was unsure of how to go about that and used the quotient rule instead, which got nasty.

    [tex]f(u)=\frac{\ln u}{1+\ln(2u)}[/tex]

    [tex]f'(u)=\frac{(1+\ln(2u))\frac{d}{du}\ln u-\ln u\frac{d}{du}(1+\ln(2u))}{(1+\ln(2u))^2}[/tex]

    [tex]\frac{(1+\ln(2u))\frac{1}{u}-\ln u(\frac{2}{2u})}{(1+2\ln(2u)+(\ln(2u))^2}[/tex]

    [tex]\frac{\frac{1+\ln(2u)}{u}-\frac{\ln u}{u}}{(1+2\ln(2u)+(\ln(2u))^2}[/tex]

    [tex]\frac{1+\ln(2u)-\ln u}{u(1+2\ln(2u)+(\ln(2u))^2)}[/tex]

    The book gives

    [tex]f'(u)=\frac{1+\ln2}{u[1+\ln(2u)]^2}[/tex]

    Oh my god that was a **** to type up :(
     
  2. jcsd
  3. Jul 29, 2008 #2

    Integral

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    note that ln(2u) = ln2 + lnu
     
  4. Jul 29, 2008 #3
    Thanks! I actually did not know that.. logs aren't my strength :/

    That helps with the numerator, but not with the denumerator. Continuing from the last step:

    [tex]\frac{1+\ln 2 + \ln u-\ln u}{u(1+2\ln 2+2\ln u + (\ln 2 + \ln u)^2)}[/tex]

    [tex]\frac{1+\ln 2}{u(1+2\ln 2+2\ln u + (\ln 2 + \ln u)^2)}[/tex]
     
  5. Jul 29, 2008 #4

    Integral

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    They did not expand the square, look at your first step.
     
  6. Jul 29, 2008 #5
    Ah! Brilliant! Thank you! So I did get it :). Much appreciated.
     
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