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Proving a function of 2 variables is diffable

  1. Feb 7, 2008 #1
    1. The problem statement, all variables and given/known data

    i)Prove that f: [tex]R^2\rightarrow R[/tex]

    f(x,y)= [tex]\frac{x^4 + y^4}{(x^2+y^2)^\alpha}[/tex] (x,y) =/= (0,0)
    f(x,y)= 0 (x,y)=(0,0)

    is differentiable on [tex]R^2[/tex] for [tex]\alpha[/tex]<3/2

    2. Relevant equations

    let f be a vector function from n variable sto m variables
    i) f is said to be diffable at a point a (element) R^n iff there is an open set V containing a such that f: [tex]R\rightarrow R^m[/tex] and there is a T (element) L(R^n;R^m) such that the function
    [tex]\epsilon[/tex](h) := f(a+h)-f(a)-T(h) satisfies [tex]\epsilon[/tex](h)/||h|| ---> 0 as h--->0

    ii) f is said to be differentiable on a set E iff E is not empty and f is diffable at every point in E

    3. The attempt at a solution

    I actually have no clue how to do this problem. I had my wisdom teeth pulled the day he covered this in lecture. plz HALP MEH!!!
  2. jcsd
  3. Feb 10, 2008 #2

    i still need help on this problem
  4. Feb 10, 2008 #3


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    Well, suppose that [itex]\alpha<3/2[/itex], and consider the partials of f(x,y) and see if they are continuous. If they are, then f is differentiable.

    [Note: I haven't tried this.]
  5. Feb 10, 2008 #4
    I dunno if this method would work.

    fx = [tex]\frac{(x^2 + y^2)4x^3 - 2ax(x^4+y^4)}{(x^2+y^2)^a+^1}[/tex]

    And this is continuous for all (x,y)==/==0

    but i dont see where 3/2 falls out of there.
    Last edited: Feb 10, 2008
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