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Homework Help: Partial Derivative, piecewise function

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Let :
    [tex]f(x,y) = \frac{xy(x^2 - y^2)}{(x^2 + y^2)^2}[/tex] if [tex](x,y) \neq (0,0)[/tex]
    [tex]f(x,y) = 0[/tex] if [tex](x,y) = (0,0)[/tex]

    a) Find [tex]f_{xx}(0,0)[/tex]

    b) Find [tex]f_{xy}(0,0)[/tex]

    c) Find [tex]f_{yx}(0,0)[/tex]

    2. Relevant equations
    None

    3. The attempt at a solution
    I'm not sure how to deal with the piecewise defined function, I just took the first and second partial derivatives w/ respect to x (for part a), but then still I can't evaluate the resulting function at (0,0), because at the point it doesn't exist...
     
  2. jcsd
  3. Oct 22, 2008 #2

    Mark44

    Staff: Mentor

    We can't tell whether what you got is correct because you didn't tell us what you got.
     
  4. Oct 23, 2008 #3

    HallsofIvy

    User Avatar
    Science Advisor

    That's not really piecewise. f(0, 0) is defined to be 0 in order to make f continuous and so differentiable. You can do the entire problem just ignoring f(0,0).
     
  5. Dec 31, 2008 #4
    I have exactly the same question. The equation is even more or less the same. Mine is:

    [tex]f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}[/tex] if [tex](x,y) \neq (0,0)[/tex]
    [tex]f(x,y)=0[/tex] if [tex](x,y)=(0,0)[/tex].

    But I'm still puzzled how to threat the point (0,0). According to my textbook, [tex]f_{xy}(0,0)\neq f_{yx}(0,0)[/tex]. So, I know more or less the solution, but I'm not sure that my explanation is correct.

    Ignoring the point (0,0) I have:

    first x, then y:
    [tex]\frac{\partial{f}}{\partial{x}}(x,y)=\frac{-y^5+yx^4+4y^3x^2}{(x^2+y^2)^2}[/tex]

    [tex]\frac{\partial{f}}{\partial{y}\partial{x}}(x,y)=\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}[/tex]

    first y, then x:
    [tex]\frac{\partial{f}}{\partial{y}}(x,y)=\frac{x^5-4x^3y^2-y^4x}{(x^2+y^2)^2}[/tex]

    [tex]\frac{\partial{f}}{\partial{x}\partial{y}}(x,y)=\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}[/tex]

    So, it is obvious that as long [tex]x\neq 0[/tex] and [tex] y\neq 0[/tex] that [tex]\frac{\partial{f}}{\partial{y}\partial{x}}(x,y)=\frac{\partial{f}}{\partial{x}\partial{y}}(x,y)[/tex].

    Now, for (x,y)->(0,0), how do I prove that
    [tex]\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)\neq\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)[/tex]? Clearly, just substituting x and y with 0 results in [tex]\frac{0}{0}[/tex].

    Is it valid to say the following:
    For[tex]\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)[/tex], first let x-> 0 first, resulting in [tex]\frac{\partial{f}}{\partial{y}\partial{x}}(0,y)=\frac{-y^6}{y^6}=-1, \forall y[/tex] and thus[tex]\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)=-1[/tex]?

    For[tex]\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)[/tex], first let y-> 0 first, resulting in [tex]\frac{\partial{f}}{\partial{x}\partial{y}}(x,0)=\frac{x^6}{x^6}=1, \forall x[/tex] and thus[tex]\frac{\partial{f}}{\partial{x}\partial{y}}(0,0)=1[/tex]?

    This is more or less a guess. I have no clue whether it is correct and if it is correct how to justify this. Especially, I have no explanation why for [tex]\frac{\partial{f}}{\partial{y}\partial{x}}(0,0)[/tex], I first let x->0 and not y->0.

    Any help on this would be most welcome and will be highly appreciated.

    (and happy new year to you all)
     
  6. Dec 31, 2008 #5

    HallsofIvy

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    Science Advisor

    The partial derivatives are defined by the limits
    [tex]f_x(x_0,y_0)= \lim_{h\rightarrow 0} \frac{f(x_0+h, y_0)- f(x_0,y_0)}{h}[/tex]
    and
    [tex]f_y(x_0,y_0)= \lim_{h\rightarrow 0} \frac{f(x_0,y_0+h)- f(x_0,y_0)}{h}[/tex]
    Here, [itex](x_0,y_0)= (0,0)[/itex] and with h non-zero, in calculating the limit, we are not at (0,0) and so use the other formula.

    Here,
    [tex]f_x(0,0)= \lim_{h\rightarrow 0} \frac{\frac{(x+h)(0)((x+h)^2- 0^2)}{(x+h)^2- 0^2)- 0}}{h}= \lim_{h\rightarrow 0} 0/h= 0[/itex]
    [tex]f_y(0,0)= \lim_{h\rightarrow 0} \frac{\frac{(0)(y+h)(0^2- (y+h)^2)}{0^2-(y+h)^2- 0}}{h}= \lim_{h\rightarrow 0} 0/h= 0[/itex]
    While [itex]f_x[/itex] and [itex]f_y[/itex] for (x,y) not equal to (0,0) is given by differentiating f in the usual way.

    Once you know [itex]f_x[/itex] and [itex]f_y[/itex] for (x,y) close to (0,0) do exactly the same thing with this new function to find the second derivative.
     
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