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Confusing chain differentiation rule

  1. Aug 9, 2008 #1

    frb

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    If I have a function f from RxR to R, and a function g from RxR to RxR. What are the partial derivatives of the composition f(g)? I end up multiplying the derivative of f with g, but g is a vector? The partial derivative should have its image in R.
     
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  3. Aug 9, 2008 #2

    arildno

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    Well, the total derivative of g is a matrix, whereas the total derivative of f is a vector. Together, they yield a vector.

    Let's take a concrete example:
    [tex]h(x,y)=f(u(x,y),v(x,y)), g(x,y)=(u(x,y),v(x,y))[/tex]
    We have therefore, for example:
    [tex]\frac{\partial{h}}{\partial{x}}=\frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}}+\frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{x}}[/tex]
    This is then one of the two partial derivatives of h, the other being differentiation with respect to y.
     
  4. Aug 9, 2008 #3

    frb

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    what does
    [tex]\frac{\partial{f}}{\partial{u}}[/tex]
    mean? Is the following formula correct?
    [tex]\frac{\partial{h}}{\partial{x}}(a,b)=\frac{\partial{f}}{\partial{x}}(u(a,b),v(a,b))\frac{\partial{u}}{\partial{x}}(a,b)+\frac{\partial{f}}{\partial{y}}(u(a,b),v(a,b))\frac{\partial{v}}{\partial{x}}(a,b)[/tex]
     
    Last edited by a moderator: Aug 9, 2008
  5. Aug 9, 2008 #4

    HallsofIvy

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    It means the partial derivative of f with respect to u, of course. What else could it mean?

    It doesn't make sense. If f(u(a,b),v(a,b)) makes any sense then f is a function of u and v, not x and y. You must mean [tex]\frac{\partial f}{\partial u}[/tex] not [tex]\frac{\partial f}{\partial x}[/tex].
     
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