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Finding partial derivatives

  1. Oct 25, 2005 #1
    Greetings, I need help in finding the partials with respect to x and y at (x,y) =/= (0,0) and (x,y) = (0,0)...

    Let f(x,y) = { (xy^2-x^2y+3x^3-y^3) / (x^2+y^2) , (x,y) =/= (0,0)
    { 0 (x,y) = (0,0)

    There was a hint given: first simplify f(x,y). I guess we have to get rid of the x^2+y^2 in the denominator to get rid of dividing by zero. But then I couldn't do it because of the "3" in the 3x^3.

    Thanks for the help!
     
  2. jcsd
  3. Oct 25, 2005 #2
    do you know the quotiant rule

    [[g(x) * f'(x)] - [ f(x) * g'(x)]] / g(x)^2
     
  4. Oct 25, 2005 #3
    yes....i differentiated with respect to x and y...but then I'm not sure how to find the partials of x and y at (0,0).
     
  5. Oct 25, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Use the basic definition.
    [tex]f_x(0,0)= lim_{h\rightarrow0}\frac{f(0+h,0)- f(0,0)}{h}[/tex]
    [tex]= lim_{h\rightarrow 0}\frac{\frac{3(h)^3}{(h)^2}}{h}[/tex].

    [tex]f_x(0,0)= lim_{h\rightarrow0}\frac{f(0,0+h)- f(0,0)}{h}[/tex]
    [tex]= lim_{h\rightarrow 0}\frac{\frac{-(h)^3}}{(h)^2}{h}[/tex].

    That should be easy.
     
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