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Homework Help: Derivative of 1 / ln x

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of 1 / ln x

    2. Relevant equations


    3. The attempt at a solution

    y = 1/lnx

    First Attempt:
    y' = -1/x/(lnx)^2
    y' = -1 / x(lnx)^2

    Second Attempt:
    ln y = ln (1 / lnx)
    ln y = ln 1 - ln x
    ln y = -lnx
    dy/dx = y(-1/x)
    dy/dx = -1/xlnx

    Third Attempt:
    ln y = -lnx
    y = -x
    y' = -1

    Which one is it? =/
  2. jcsd
  3. Oct 28, 2007 #2


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    In your second attempt this step is wrong
    ln y = ln (1 / lnx)
    ln y = ln 1 - ln x

    It should be lny=ln 1 -ln(lnx)
  4. Oct 28, 2007 #3
    We know the derivative would be the denominator times the derivative of the numerator(which would be zero in this case), minus the numerator times the derivative of the denominator(which is 1/x), over the denominator squared.
  5. Oct 28, 2007 #4


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    Gold Member

    Use the product rule. Sure you can do the quotient rule but the product rule is so easy to remember!
  6. Oct 28, 2007 #5
    What made you doubt your first attempt? Using the product rule:

    [tex]\frac{d}{{dx}}\left( {\frac{1}{{\ln x}}} \right) = \frac{{0 \cdot \ln x - 1\left( {\frac{1}{x}} \right)}}{{\left( {\ln x} \right)^2 }} = \frac{{ - \frac{1}{x}}}{{\left( {\ln x} \right)^2 }} = - \frac{1}{{x\left( {\ln x} \right)^2 }}[/tex]
  7. Oct 29, 2007 #6


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    Science Advisor

    That isn't the product rule!

    What JasonRox meant, I think, was use the chain rule on (ln x)-1.
  8. Oct 29, 2007 #7
    Your first approach was correct, the second one, as already pointed, instead of ln(lnx) you took lnx.
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