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Second order mixed derivative and chain rule

  1. Aug 17, 2011 #1
    I want to find the second order derivative for [tex]f(x,y),x(u,v),y(u,v)[/tex], f depends on x and y, and x and y depends on u and v. I'm trying to find [tex]\frac{{\partial^2 f}}{{\partial v \partial u}}[/tex]


    This is what I did:
    [tex]\frac{{\partial f}}{{\partial u}}=\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}[/tex]

    Then:
    [tex]\frac{{\partial^2 f}}{{\partial v \partial u}}=\frac{{\partial}}{{\partial v}} \left (\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}}\right )+\frac{{\partial}}{{\partial v}} \left (\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}\right )[/tex]

    Finally what I get:

    [tex]\displaystyle\frac{{\partial^2 f}}{{\partial v \partial u}}=\frac{{\partial^2 f}}{{\partial x^2}}\frac{{\partial x}}{{\partial v}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial^2 f}}{{\partial y \partial x}} \frac{{\partial y}}{{\partial v}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial f}}{{\partial x}}\frac{{\partial^2 x}}{{\partial v \partial u}}+\frac{{\partial^2 f}}{{\partial x \partial y}}\frac{{\partial x}}{{\partial v}}\frac{{\partial y}}{{\partial v}}+\frac{{\partial^2 f}}{{\partial y^2}}(\frac{{\partial y}}{{\partial v}})^2+\frac{{\partial f}}{{\partial y}}\frac{{\partial^2 y}}{{\partial v^2}}[/tex]

    Anyone knows if this is right?
     
    Last edited: Aug 17, 2011
  2. jcsd
  3. Aug 17, 2011 #2

    SammyS

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    In the last three terms of the last line, you appear to have changed ∂y/∂u to ∂y/∂v .
     
  4. Aug 18, 2011 #3
    Thank you SammyS, I knew something was wrong :D
     
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