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Math Help - x/x^2lnx

  1. #1
    Boo
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    x/x^2lnx

    All!
    Please, how do I get f'(x) and f''(x) if I have f(x)=x/(x^lnx)?

    Many thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: x/x^2lnx

    Quote Originally Posted by Boo View Post
    All!
    Please, how do I get f'(x) and f''(x) if I have f(x)=x/(x^lnx)?

    Many thanks!
    Which do you have? f(x) = \frac{x}{x^2 ln(x)}

    or
    f(x) = \frac{x}{x^{ln(x)}}

    -Dan

    Or maybe:
    f(x) = \frac{x}{x^{2ln(x)}}
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  3. #3
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    Re: x/x^2lnx

    Hello, Boo!

    \text{Given: }\:f(x) \:=\:\dfrac{x}{x^{\ln x}}. \quad \text{Find }f'(x)\text{ and }f''(x).

    You need Logarithmic Differentiation.

    We have: . y \;=\;\dfrac{x}{x^{\ln x}}

    Take logs: . \ln(y) \;=\;\ln\left(\frac{x}{x^{\ln x}}\right)

    . . . . . . . . \ln(y) \;=\;\ln(x) - \ln\left(x^{\ln x}\right)

    . . . . . . . . \ln(y) \;=\;\ln(x) - \ln x\ln x

    . . . . . . . . \ln(y) \;=\;\ln(x) - [\ln(x)]^2


    Differentiate implicitly:

    . . . . . . . . \frac{y'}{y} \;=\;\frac{1}{x} - 2\ln(x)\!\cdot\!\frac{1}{x}

    . . . . . . . . \frac{y'}{y} \;=\;\frac{1}{x}\left[1 - 2\ln(x)\right]

    . . . . . . . . y' \;=\;y\cdot \frac{1-2\ln(x)}{x}

    . . . . . . . . y' \;=\;\frac{x}{x^{\ln x}}\cdot\frac{1-2\ln(x)}{x}

    Hence: . . y' \;=\;\frac{1-2\ln(x)}{x^{\ln x}}


    Now repeat the process to find the second derivative, y''.
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