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Math Help - Logarithmic differentiation

  1. #1
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    Logarithmic differentiation

    Can someone tell me if this is correct?
    y = (lnx)^x
    y' = x(lnx)^x-1 * 1/x
    y' = (lnx) ^ x-1

    It seemed a bit too simple so I wanted to check if I'm doing it right, thanks.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Logarithmic differentiation

    It's not correct.
    Write y = [\ln(x)]^x = e^{x\ln[\ln(x)]}
    Now, compute \frac{dy}{dx}
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  3. #3
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    Re: Logarithmic differentiation

    I'm trying to understand what you did there, but I also have a second method,
    Which is: lny = xln(lnx)
    Would this also be correct?
    So...
    y'/y = ln(lnx) + x * 1/lnx * 1/x
    y' = y(ln(lnx) + 1/lnx)
    y' = (lnx)^x(ln(lnx) + 1/lnx)
    Or is this also wrong?
    Last edited by aimforthehead; October 31st 2012 at 10:47 AM.
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  4. #4
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    Re: Logarithmic differentiation

    y = (\ln{x})^x

    \ln{y} = \ln(\ln{x})^x

    \ln{y} = x \cdot \ln(\ln{x})

    \frac{y'}{y} = x \cdot \frac{1}{x\ln{x}} + \ln(\ln{x}) \cdot 1

    \frac{y'}{y} = \frac{1}{\ln{x}} + \ln(\ln{x})

    y' = y\left[\frac{1}{\ln{x}} + \ln(\ln{x})\right]

    y' = (\ln{x})^x\left[\frac{1}{\ln{x}} + \ln(\ln{x})\right]
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