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Multivariable Derivative Error

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data
    f(x,y) =
    (xy) / (x2 + y4), when (x, y) ≠ (0,0)
    0, when (x,y) = (0,0)

    2. Relevant equations
    Explicitly show that f(x,y) does not satisfy
    lim h -> 0 [ E(v,h) / ||h|| ] = 0 when v = 0
    (h, v, and 0 are all vectors; I'm not sure how to put a hat on them)

    3. The attempt at a solution
    I have no idea how to actually tackle this problem using multivariable calculus methods so I just compared the slopes instead (not sure if this is a valid approach either).
    Further more I don't really get the concept of E(v,h). All my professor said is that it should approach 0 if the function is differentiable. I'd really appreciate an explanation of E(v,h) please (:

    f(h) = f(0,0) + (df/dx(0,0) * x) + (df/dy(0,0) * y) + E(h)
    f(h) = 0 + 0x + 0y + E(h)
    f(h) = E(h)

    Let f(x,y) = f(h,h)
    f(h,h) = h / (1+h2)
    f'(h) = (-h2 + 1) / (1+h2)2

    ||h|| = 2h2
    f'(||h||) = 4h

    lim h-> 0 [ f'(h) / f'(||h||) ] = ∞
     
  2. jcsd
  3. Oct 24, 2015 #2

    SammyS

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    Do you have a definition of E(v,h)?
     
  4. Oct 24, 2015 #3
  5. Oct 24, 2015 #4

    HallsofIvy

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    I get a virus warning when I try to open that.
     
  6. Oct 25, 2015 #5
  7. Oct 25, 2015 #6

    Ray Vickson

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    Some of your equations make no sense, or else are not true for your current function. For example, your "formula" for ##f(h)## has only ##h## on one side and ##x,y## on the other; that does not mean anything. Besides that, even if we were to correct the formula, it might be false for your function, because it makes some assumptions about the nature of the function in a neighborhood of ##(0,0)## that may not hold for your ##f(x,y)##.

    Basically, if we define the directional derivative of f at the point ##(x_0,y_0)## in the direction ##(p,q)## as
    [tex] D_{(p,q)} f(x_0,y_0) = \lim_{t \to 0} \frac{f(x_0 + t p, y_0 + t q) - f(x_0,y_0)}{t}, [/tex]
    then, in order for your result about ##E(h)## to hold, it is necessary (but perhaps not sufficient) that ##D_{(p,q)} f(x_0,y_0)## be linear in ##(p,q)##; that is, it would be necessary to have ##D_{(p,q)} f(x_0,y_0) = a p + b q## for some real numbers ##a## and ##b##. Is that true at ##(x_0,y_0) = (0,0)## in your current example?
     
  8. Oct 25, 2015 #7

    vela

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    Read this page: http://www.math.ku.edu/~lerner/m291F08/Chapter9.pdf
     
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