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Partial Derivatives

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Let [tex]u= (x^2 + y^2 + z^2)^\frac {-1} {2}[/tex]


    Find [tex]\frac {\partial^2 u} {\partial x^2} + \frac {\partial^2 u} {\partial y^2} + \frac {\partial^2 u} {\partial z^2}[/tex]

    2. Relevant equations
    3. The attempt at a solution

    [tex]\frac {\partial^2 u} {\partial x^2} = -(x^2 + y^2 +z^2)^\frac {-3} {2} + 3x^2(x^2 + y^2 + z^2)^\frac {-5} {2}[/tex]

    [tex]\frac {\partial^2 u} {\partial y^2} = -(x^2 + y^2 +z^2)^\frac {-3} {2} + 3y^2(x^2 + y^2 + z^2)^\frac {-5} {2}[/tex]

    [tex]\frac {\partial^2 u} {\partial z^2} = -(x^2 + y^2 +z^2)^\frac {-3} {2} + 3z^2(x^2 + y^2 + z^2)^\frac {-5} {2}[/tex]

    So I think all the partials are right, but I feel like I'm getting a crazy answer when I add them together.

    [tex]3x^2 + 3y^2 + 3z^2(x^2 + y^2 + z^2)^\frac {-5} {2} -3(x^2 + y^2 + z^2)^\frac {-3} {2}[/tex]

    Is this right?
     
  2. jcsd
  3. Apr 8, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Why do you think that's crazy? It looks correct to me. But you can express the answer in a much simpler form.
     
  4. Apr 8, 2009 #3
    here's the matlab quick code for the first step - finding d^2/dx^2:

    >> syms u; syms y; syms z;
    >> u = 1/sqrt(x^2+y^2+z^2)

    u =

    1/(x^2+y^2+z^2)^(1/2)


    >> diff(u,x)

    ans =

    -1/(x^2+y^2+z^2)^(3/2)*x


    >> diff(ans,x)

    ans =

    3/(x^2+y^2+z^2)^(5/2)*x^2-1/(x^2+y^2+z^2)^(3/2)
     
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