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Is f Differentiable? - Vector Calculus

  1. Sep 27, 2006 #1
    I have a problem regarding the function f (x,y) = {x*y*(x^2-y^2)/(x^2+y^2) if (x,y)!=(0,0) and f(x,y)=0 if (x,y)=(0,0).

    I am asked if this function is differentiable. Running it through a graphing program it looks differentiable. I know the partial derivatives of it in terms of x and y are 0 and 0 respectively and I know that a function is differentiable at (0,0) if fx(0,0) and fy(0,0) exists which they do and if lim (x,y)->(0,0) [f(x,y)-h(x,y)/||(x,y)-(a,b)||]=0 where h(x,y) describes the tangent plane of the function at point (a,b) or in our case (0,0).

    So after a lot of algebra which I will not repeat here using the above definition I boil it all down to lim (x,y)->(0,0) [x*y*(x^2-y^2)/sqrt(x^2+y^2)] and the question is does this = 0? I have tried to prove it is using delta epsilon proofs but I just get a bunch of gibberish unless I am doing something wrong, I might be I am not sure and I have also tried to prove this is wrong by trying to come into (0,0) at some odd angle like y=mx and showing it doesn't = 0 but all attempts have failed. Help.
  2. jcsd
  3. Sep 28, 2006 #2
    Bump, anyone?
  4. Sep 28, 2006 #3


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    The simplest thing to do with limits of functions of two variables at (0, 0) is convert to polar coordinates. That way the single variable r gives the distance from (0,0). If the limit as r goes to 0 is independent of [itex]\theta[/itex] then the limit as (x,y)-> (0,0) exists and is that value.
  5. Sep 28, 2006 #4
    \left| {\frac{{xy\left( {x^2 - y^2 } \right)}}{{\sqrt {x^2 + y^2 } }} - 0} \right|

    \le \left| {\frac{{x\sqrt {x^2 + y^2 } \left( {x^2 - y^2 } \right)}}{{\sqrt {x^2 + y^2 } }}} \right|

    = \left| x \right|\left| {x^2 - y^2 } \right|

    \le \left| x \right|\left( {\left| x \right|^2 + \left| y \right|^2 } \right)

    \le \sqrt {x^2 + y^2 } \left[ {\left( {x^2 + y^2 } \right) + \left( {x^2 + y^2 } \right)} \right]

    What's between zero and delta? (refer to the definition)
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