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D/dx log(x^2+y^2)

  1. Sep 4, 2008 #1
    I'm trying to get

    [tex] \frac{\partial}{\partial x} log(x^{2} + y^{2}) [/tex]

    let z = x2+y2

    Do I need to do a change of base to go from log10z to logez before I can do the partial w.r.t. x?

    That would make it

    [tex] \frac{\partial}{\partial x} log(x^{2} + y^{2}) = \frac{1}{x^{2}+y^{2}} \ (2x) \ (log_{10}e)[/tex]

    Does this look right?

    Then

    [tex]\frac{\partial^{2}}{\partial x^{2}}(log(x^{2}+y^{2})) = - \frac{1.72x^{2}}{(x^{2} + y^{2})^{2}} + 0.86(x^{2}+y^{2}) [/tex]

    ?? that doesn't look right.
     
  2. jcsd
  3. Sep 4, 2008 #2

    tiny-tim

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    Hi Somefantastik! :smile:

    Yes, that's fine …

    logab = [tex]\frac{log_eb}{log_a}[/tex]

    and of course logab = 1/logba :wink:
    ha ha!

    how about [tex]\frac{0.86}{(x^{2}+y^{2})} [/tex] ? :smile:
    (and then of course combine both fractions into one)
     
  4. Sep 4, 2008 #3
    Hello tiny-tim, from the sunny south [usa] :)

    That gives -0.86 /(x2 + y2)

    It just seems like a weird number to me.

    That's going to be the same number for the partial with respect to y, isn't it?

    They should cancel out; I'm trying to show that this is a harmonic using the laplacian.
     
  5. Sep 4, 2008 #4

    tiny-tim

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    Hi Somefantastik! :smile:
    Yup …

    [tex]\left(\frac{\partial^{2}}{\partial x^{2}}\ +\ \frac{\partial^{2}}{\partial y^{2}}\right)(log(x^{2}+y^{2})) = - \frac{1.72(x^{2}\,+\,y^2)}{(x^{2} + y^{2})^{2}}\ +\ 0.86/(x^{2}+y^{2})\ +\ + 0.86/(x^{2}+y^{2})\ =\ 0\ [/tex] :wink:
     
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