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Math Help - Partial derivatives question

  1. #1
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    Partial derivatives question

    consider the function:

    (2x(x^2-y^2))/(x^2+y^2) ---> (x,y) /= 0

    f(x,y) ={

    0 ---> (x,y)= 0


    Show that fx (x partial) and fy(y partial) exist at (0,0) and determine their value...


    I get an x-partial derivative of (2x^4+8x^2y^2-2y^4)/((x^2 + y^2)^2)

    I get a y-partial derivative of (-8yx^3)/((x^2+y^2)^2)

    I have done them a million times and just keep getting that - how on earth are they supposed to exist at (0,0)??

    I was thinking that the numerator would have a factor of (x^2+y^2)^2 but no such luck?

    any ideas?
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  2. #2
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    This is almost unreadable. Is the function

    \displaystyle \begin{cases}\frac{2x(x^2-y^2)}{x^2+y^2}\,\textrm{if}\,(x,y) \neq (0,0)\\ 0 \,\textrm{if}\,(x,y) = (0,0)\end{cases}?
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  3. #3
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    that is exactly it
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  4. #4
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    Do NOT try to find the derivative at (0, 0) by finding the derivative at other points and taking the limit as (x, y) goes to (0, 0). Derivatives are NOT necessarily continuous.

    The partial derivative of f(x,y), with respect to x, at (0, 0) is
    \displaytype\lim_{h\to 0}\frac{f(h,0)- f(0,0)}{h}

    The partial derivative of f(x,y), with respect to y, at (0, 0) is
    \displaytype\lim_{h\to 0}\frac{f(0,h)- f(0,0)}{h}

    Here, f(0,0)= 0 and f(h, 0)= \frac{2h(h^2- 0^2)}{h^2+ 0^2}= \frac{2h(h^2)}{h^2}= 2h

    f(0, h)= \frac{2(0)(0^2- h^2)}{0^2+ h^2}= 0
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