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Homework Help: Partial derivative chain rule

  1. May 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A function is called homogeneous of degree n if it satisfied the equation f(tx,ty) =t^(n) f(x,y), for all t, where n is a positive integer and f has continuous 2nd order partial derivatives.

    If f is homogeneous of degree n, show that df/dx (tx,ty) = t^(n-1) df/dx(x,y)

    *Using df/dx for partial derviatives.

    3. The attempt at a solution

    Basically I've taken the partial derivatives of each side of the definition of homogeneous equation above, applying the chain rule (that's what section the problem is in). What I get is:

    d/d(tx) (tx,ty) * t = t^n d/dx(x,y)

    which simplifies to df/d(tx) (tx,ty) = t^(n-1) df/dx(x,y).

    It feels like I'm done, but I can't figure out-conceptually or otherwise- why the partial derivative with respect to (tx) of f(tx,ty) is the same as the partial derivative with respect to x of f(tx,ty).

    Thanks in advance for help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 12, 2008 #2

    HallsofIvy

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    It's not! [itex]\partial f(tx,ty)/\partial (tx)[/itex] is the same as [itex]\partial f(u,ty)/\partial u[/itex] for any variable u. [itex]\partial f(tx,ty)/\partial x= \partial f(xt,ty)/\partial (xt) \partial f(tx)\partial x[/itex].

    In other words [itex]\partial f(tx,ty)/\partial x[/itex] is t times [itex]\partial f(tx,ty)/\partial (tx)[/itex].

     
  4. May 12, 2008 #3
    Yes, that's what I got when I applied the chain rule, although maybe my notation was confusing.

    d(tx,ty)/d(tx) times t = t^n d(x,y)/dx
    (the right side of the equation is the partial derivative with wrt x of the function given in the original definition)

    once you divide by t you get

    d(tx,ty)/d(tx) =t^(n-1) d(x,y)/dx

    What I don't get is how to deal with the partial derivative w.r.t. (tx). The left side of the equation needs to be d(tx,ty) w.r.t to x for the requested proof.
     
  5. May 12, 2008 #4

    Defennder

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    Homework Helper

    Is this even true? Try it out for the case [tex]f(x,y) \ = \ xy^3 + x^2 y^2.[/tex]

    Did you mean to say [tex]t^n \frac{\partial}{\partial x} f(x,y)[/tex] on the RHS instead? I assume t is an arbitrary constant.
     
  6. May 13, 2008 #5
    You're right, it doesn't seem to work for the case you give. That is definitely how the problem is stated, though. It's from Stewart Calculus, 5ed, 14.6, #55. I guess it could be a typo...?
     
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