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Math Help - Second order mixed derivative and chain rule

  1. #1
    Member
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    May 2010
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    241

    Second order mixed derivative and chain rule

    I want to find the second order derivative for f(x,y),x(u,v),y(u,v), f depends on x and y, and x and y depends on u and v. I'm trying to find \frac{{\partial^2 f}}{{\partial v \partial u}}


    This is what I did:
    \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}}

    Then:
    \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 )

    Finally what I get:

    \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}}

    Anyone knows if this is right?
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  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398

    Re: Second order mixed derivative and chain rule

    See my reply on that other site on which you posted this.
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