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Thread: 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? $\displaystyle f(x) = \frac{x}{x^2 ln(x)}$

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

    -Dan

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

    Hello, Boo!

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

    You need Logarithmic Differentiation.

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

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

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

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

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


    Differentiate implicitly:

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

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

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

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

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


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