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Differentiation Problem -

  1. Jun 18, 2009 #1
    I was working on some of my own equations and today i ended up with this differentiation thinghy, I never expected this in my equation but it just turned up :( so if there's anybody out there who loves to solve math please give this a try :)

    maybe its too simple :) ... i am just having doubts about the order of the operations and their influences...

    me too will try solving it in the mean time...

    [tex]\frac{d}{d\theta}\left\{ \nabla\left(\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}\right)^{\frac{1}{2}}\right\} =?[/tex]

    [tex]\[
    n_{k},n_{j}^{\star}\] [/tex] are independent of [tex]\theta[/tex]
    [tex]\[
    n_{k},n_{j}^{\star}\] [/tex] depend on x,y,z (grad) (Actually these variables are concentrations defined at each point in our domain)
    [tex]\[
    z_{k}^{2}\][/tex] is a constant (actually the valence of the kth chemical species)

    [tex]\theta[/tex] is also independent of the geometry (x,y,z)... i call it the ALIEN variable :) because afterwards i have to kill it by setting it to zero, that is after differentiation...

    Thanks a lot,

    Tim
     
    Last edited: Jun 18, 2009
  2. jcsd
  3. Jun 18, 2009 #2
    Is this Right ???

    [tex]\[
    =-\frac{1}{2}\left(\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}\right)^{-\frac{3}{2}}n_{j}^{\star}\nabla\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}\]


    \[
    +\frac{1}{2}\underset{k}{\sum}\left[n_{k}+\theta n_{j}^{\star}\right]z_{k}^{2}\nabla n_{j}^{\star}\][/tex]

    can someone confirm :)

    Thanks a lot ....
     
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