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Change of variable, second order differentiation

  1. Nov 5, 2009 #1
    Okay, so my problem lies within taking the second derivative of a change of variable equation.

    w = f(x,y); x = u + v, y = u - v

    so far I have the first derivative:

    dw/dx = (dw/dv)(dv/dx) + (dw/du)(du/dx) = (d/dv + d/du)w

    Now I'm having problems in finding my second derivative:

    d2w/dx2 = ?

    I don't necessarily want the answer, but more of how to get to the answer. I figure if I get how to do it with dw/dx, dw/dy would follow suit.
    Any help would be greatly appreciated!
  2. jcsd
  3. Nov 6, 2009 #2


    Staff: Mentor

    All your derivatives are partial derivatives, and your formula for dw/dx is completely wrong.

    [tex]\frac{\partial w}{\partial x}[/tex] is just itself; you can't do much with it without knowing more about f(x, y). What I think you want is the partial of w with respect to u, and possibly the partial of w with respect to v. From these you can get the second partials.

    Here's the formula for the chain rule in Leibniz notation for your problem:

    [tex]\frac{\partial w}{\partial u}~=~\frac{\partial w}{\partial x} \frac{\partial x}{\partial u}+ \frac{\partial w}{\partial y} \frac{\partial y}{\partial u}[/tex]

    This formula can be simplified, since
    [tex]\frac{\partial x}{\partial u}~=~\frac{\partial x}{\partial v}~=~1[/tex]


    [tex]\frac{\partial y}{\partial u}~=~1[/tex]
    [tex]\frac{\partial y}{\partial v}~=~-1[/tex]

    [tex]\frac{\partial w}{\partial v}[/tex] is calculated in a similar manner as was used for the partial of w with respect to u, by replacing u everywhere it appears in the formula above with v.
  4. Nov 6, 2009 #3
    First I would like to thank you very much, also I knew they were partials, but I'm a noob to this forum, and didn't know how to put down partials!

    The way I started to set it up, was with how the book explained to do it, but your way would probably be my normal approach. I did manage to get the right answer thanks to your help!

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