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Jacobian when there's a multivariate function inside it

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data

    differentiate the function F(x,y) = f( g(x)k(y) ; g(x)+h(y) )

    2. Relevant equations

    Standard rules for partial differentiation

    3. The attempt at a solution

    The Jacobian will have two columns because of the variables x and y. But what then? f is a multivariate function inside the Jacobian!
     
  2. jcsd
  3. Feb 3, 2015 #2

    BvU

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    So on top of the standard rules you get the chain rule.
    Show some attempt at solution and help is on the way.

    To demo my ignorance: Differentiating gives two columns, but one row only, right ?
    Is there a significance in the ";" ? You write F ( x , y ) -- a notation which I am also familiar with -- , but then you write f ( u ; v )
     
  4. Feb 3, 2015 #3

    Ray Vickson

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    Do you mean ##F(x,y) = f(u,v)##, where ##u = g(x) k(y)## and ##v = g(x) + h(y)##? If so, just apply the chain rule for derivatives. You need to express the answers in terms of the functions ##f_1, f_2##, where ##f_1(u,v) \equiv \partial f(u,v)/\partial u## and ##f_2(u,v) \equiv \partial f(u,v) / \partial v##.
     
  5. Feb 3, 2015 #4
    Consider the partial derivatives that make up the derivative matrix. It should be a 2x2, you have two functions, and take the derivative of both functions wrt x or wrt y.
     
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