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Question about second-order partial derivatives

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data

    If V=xf(u) and u=y/x, show that

    x^2.d2V/dx2 + 2xy.d2V/dxdy + y^2.d2V/dy2= 0

    (This a partial differentiation problem so all the d's are curly d's)

    3. The attempt at a solution
    I have tried to work out d2V/dx2 and the other derivatives, then multiply them by x^2 or 2xy or whatever is in front of them and then add them all together, but the answer I get is not 0.

    I think the issue is I don't know how to work out d2V/dx2

    to work out d2V/dx2 I am trying d/dx (dV/du . du/dx). But it does not seem to work

    Any help would be appreciated
    thanks
    cluivee
     
  2. jcsd
  3. Aug 25, 2011 #2

    Hootenanny

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    Welcome to Physics Forums.

    You are correct in saying that (chain rule)

    [tex]\frac{\partial}{\partial x} f(u) = \frac{\partial}{\partial u} f(u)\frac{\partial u}{\partial x}\;.[/tex]

    However, you need to be careful here since [itex]V(u)=xf(u)[/itex]. Therefore,

    [tex]\begin{aligned}\frac{\partial}{\partial x} V(u) & = \frac{\partial}{\partial x} xf(u) \\ &= f(u) + x\frac{\partial}{\partial x} f(u) \\ & = f(u) + x \frac{\partial}{\partial u} f(u)\frac{\partial u}{\partial x}\;.\end{aligned}[/tex]

    Do you follow?
     
  4. Aug 25, 2011 #3
    I think so, but I need a bit of clarification, I've seen you've used the product rule to get

    =f(u)+x∂∂xf(u)
    =f(u)+x∂∂uf(u)∂u∂x.

    but f(u)= u = y/x right? I know that may be a dumb question.

    Then d/du(f(u)) = 1 always? And du/dx = -y/x^2?

    Thanks for your help though Hootenanny
     
  5. Aug 25, 2011 #4

    Hootenanny

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    Why is [itex]f(u)=u[/itex]?
    No problem :smile:
     
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