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Differentiation of ln(x)

  1. May 29, 2009 #1
    1. The problem statement, all variables and given/known data

    I am unsure how to differentiate ln(x).

    2. Relevant equations

    [tex]\int[/tex] dx/ (x logex)

    3. The attempt at a solution

    I let u = logex

    So it became:
    [tex]\int[/tex] x-1u-1dx

    To integrate I now need to find du/dx... which means differentiate ln(x). How does this work out?
     
  2. jcsd
  3. May 29, 2009 #2

    rock.freak667

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    Homework Helper

    if y=ln(x) then, x=ey

    find dx/dy. Invert to get dy/dx and then figure out what eln[f(x)] works out to be.
     
  4. May 29, 2009 #3
    Thanks so much!
     
  5. May 29, 2009 #4
    Thanks but now I need help on the rest of the question! I'm really stuck. As, when I change it to integral f(x) du, the differentiated ln(x) does not cancel anything out.... Does anyone know how to integrate this equation?

    [tex]\int[/tex] x-1u-1du/e^u


     
  6. May 29, 2009 #5

    Cyosis

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    Homework Helper

    Nothing "cancels" because you're not differentiating log(x) correctly. I don't really know how you can encounter these kind of problems without ever having seen the derivative of log(x), but this is how it works.

    [tex]
    y=\log x \Rightarrow x=e^y, \;\;
    \frac{dx}{dx}=\frac{d e^y}{dx}=e^y \frac{dy}{dx}=1 \Rightarrow \frac{dy}{dx}=\frac{1}{e^y} \Rightarrow \frac{d log x}{dx}=\frac{1}{x}[/tex]
     
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