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Differentiation of x^(1/x)

  1. Nov 18, 2008 #1

    madmike159

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    I was looking at results of different numbers in the equation [tex]\sqrt[x]{x}[/tex] and found out that the biggest result came when it was [tex]\sqrt[e]{e}[/tex]. I know this can be re-written as [tex]x^{1/x}[/tex] and that the gradient would be 0 at x = e. How would you differentiate y = [tex]x^{1/x}[/tex], I can't seem to do it using any of the laws I know.
     
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  3. Nov 18, 2008 #2

    D H

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    Find a function [itex]f(x)[/itex] that lets you rewrite [itex]y=x^{1/x}[/itex] in the form [itex]y=\exp(f(x))[/itex].
     
  4. Nov 18, 2008 #3

    madmike159

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    I don't see how that works but I'll give it a go.
     
  5. Nov 18, 2008 #4

    lurflurf

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    use the power rule
    [u^v]'=v*[u^(v-1)]*u'+[u^v]*[log(u)]*v'
     
  6. Nov 18, 2008 #5

    D H

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    Rather than committing the power rule to memory, I find it much easier to remember a couple simple facts which happen to be of use in a lot of other places,
    1. [tex]\frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}\,f'(x)[/tex]

    2. [tex]f(x) = e^{\ln f(x)}\;\Rightarrow\;u(x)^{v(x)} = e^{v(x)\,\ln u(x)}[/tex]
    and thus

    [tex]
    \aligned
    \frac{d}{dx}\left(u(x)^{v(x)}\right)
    &= \frac{d}{dx}\left(e^{v(x)\,\ln u(x)}\right) \\
    &= e^{v(x)\,\ln u(x)}\frac d {dx}\left(v(x)\,\ln u(x)\right) \\
    &= u(x)^{v(x)}\left(v'(x)\,\ln u(x) + \frac{v(x)\,u'(x)}{u(x)}\right)[/tex]
     
  7. Nov 18, 2008 #6

    HallsofIvy

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    Equivalently, if y= x1/x, then ln(y)= (1/x)ln(x). Use "implicit differentiation".
     
  8. Nov 18, 2008 #7

    lurflurf

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    or just use it on
    y^x=x
     
  9. Nov 18, 2008 #8

    lurflurf

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