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Taking second derivative of a derivative

  1. Jul 5, 2010 #1
    Hello--

    I'm in the process of implementing a PML for FDTD modeling.

    I would like to take the derivative of the partial derivative shown below, but I am uncertain with respect to how I might proceed.

    [tex]
    \[
    \frac{\partial }{{\partial x}} \to \frac{1}{{1 + \frac{{i\sigma \left( x \right)}}{\omega }}}\frac{\partial }{{\partial x}}
    \]
    [/tex]

    Essentially what I would like to do is take the derivative of a partial derivative, and also deal with the [tex]\[{i\sigma \left( x \right)}\] [/tex] term, which is a function of position [tex]x[/tex].

    This would result in the calculation of [tex] \[\frac{{\partial ^2 }}{{\partial x^2 }}\][/tex]
     
    Last edited: Jul 5, 2010
  2. jcsd
  3. Jul 5, 2010 #2
    Perhaps this would be the way to take the second derivative :

    [tex]
    \[
    \frac{{\partial ^2 }}{{\partial x^2 }} \to -\left( {1 + \frac{{i\sigma \left( x \right)}}{\omega }} \right)^{ - 2} \left( {\frac{{\partial \sigma \left( x \right)}}{{\partial x}}\frac{i}{\omega }} \right)\frac{\partial }{{\partial x}} + \frac{{\partial ^2 }}{{\partial x^2 }}\left( {1 + \frac{{i\sigma \left( x \right)}}{\omega }} \right)^{ - 1}
    \]

    [/tex]
     
    Last edited: Jul 5, 2010
  4. Jul 5, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Use the product rule:
    [tex]\frac{\partial }{\partial x}\left((1+ \frac{i\sigma}{\omega})^{-1}[/tex][tex]\frac{\partial y}{\partial x}\right)[/tex][tex]= \frac{\partial (1+ \frac{i\sigma}{\omega})^{-1}}{\partial y}{\partial x}[/tex][tex]+ (1+ \frac{i\sigma}{\omega})^{-1}\frac{\partial^2 y}{\partial x^2}[/tex]
     
    Last edited: Jul 5, 2010
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