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Math Help - f(x)=x^(lnx) , f'(e)=?

  1. #1
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    f(x)=x^(lnx) , f'(e)=?

    f(x)=x^(lnx)

    This is what I did:

    lny = lnx(lnx)

    (1/y)y' = (lnx)/x + (lnx)/x

    y' = y [ (lnx)/x + (lnx)/x ]

    y' = x^(lnx) [ (2lnx)/x)


    f'(e) = e^(lne) [(2lne/e)

    f'(e) = e^1 (2/e)
    f'(e) = 2

    Is my method of differentiation correct?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: f(x)=x^(lnx) , f'(e)=?

    Quote Originally Posted by NeoSonata View Post
    Is my method of differentiation correct?
    Yes, it is correct.
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  3. #3
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    Re: f(x)=x^(lnx) , f'(e)=?

    Quote Originally Posted by NeoSonata View Post
    f(x)=x^(lnx)

    This is what I did:

    lny = lnx(lnx)
    At this point it would be slightly simpler to write ln(y)= (ln(x))^2 so that
    \frac{1}{y}y'= \frac{2ln(x)}{x} by the chain rule.

    (1/y)y' = (lnx)/x + (lnx)/x

    y' = y [ (lnx)/x + (lnx)/x ]

    y' = x^(lnx) [ (2lnx)/x)


    f'(e) = e^(lne) [(2lne/e)

    f'(e) = e^1 (2/e)
    f'(e) = 2

    Is my method of differentiation correct?
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  4. #4
    Super Member TheChaz's Avatar
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    Re: f(x)=x^(lnx) , f'(e)=?

    Quote Originally Posted by HallsofIvy View Post
    At this point it would be slightly simpler to write ln(y)= (ln(x))^2 so that
    \frac{1}{y}y'= \frac{2ln(x)}{x} by the chain rule.
    I noticed that, too, but thought that it would be simpler to let someone else type it up!
    To move this reply in the direction of legitimacy, let me also comment that

    (ln(x))^2 \; \; is sometimes written as ln^2(x)
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