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Two varibale function. Continuity, derivability and differentiability

  1. Aug 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Discuss the continuity, derivability and differentiability of the function

    [itex]f(x,y) = \frac{x^3}{x^2+y^2}[/itex] if (x,y)≠(0,0) and 0 otherwise

    2. Relevant equations
    if f is differentiable then [itex]∇f.v=\frac{∂f}{∂v}[/itex]
    if f has both continuous partial derivative in a neighbourhood of [itex]x_0[/itex] then it's differentiable in [itex]x_0[/itex]


    3. The attempt at a solution
    I have no problem with continuity.
    For derivability i consider the definition of directional derivative in an arbitray direction (α,β)
    [itex]\frac{f(αt,βt)-f(0,0)}{t}=\frac{α^3t^3-0}{t^3}=α^3[/itex]
    all the directional derivatives exist.
    this equation [itex]∇f.v=\frac{∂f}{∂v}[/itex] tells me that the directional derivative should depend linearly on α and β which is not the case, f is not differentialble in [itex]x_0[/itex]

    on the other hand it's easy to calculate the partial derivatives in a neighborood of (0,0) and see that they are continous.
    [itex]f(x,0)=x \ \ \ f(0,y)=0[/itex]
    [itex]\frac{∂}{∂x}f(x,0)=1 \ \ \ \frac{∂}{∂y}f(0,y)=0[/itex]

    i get two different results with two different approaches and i can't figure out what's wrong with one of them, or both :tongue2:

    any help is appreciated
     
  2. jcsd
  3. Aug 22, 2014 #2
    The partial derivatives are not continuous. To see this, compute either of them at ##p \neq (0,0)## and take the limit as ##p \rightarrow 0##.

    If I am not mistaken, what you showed is that the functions ## f_x(x,0), f_y(0,y)## of one variable (the other being fixed to zero) are continuous.
     
  4. Aug 22, 2014 #3
    Oh i see what's the problem. Thanks a lot for the help :biggrin:
     
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