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Integration of absolute value

  1. Nov 18, 2012 #1
    f: R2 to R1 given by f(x,y) = x(|y|^(1/2))
    show differentiable at (0,0)

    so I'm using the definition lim |h| ->0 (f((0,0) + 9(h1,h2)) - f(0,0) - Df(0,0) (h1,h2)) / |h|

    so first for the jacobian for f, when I'm doing the partial with respect to y, do I have to break this into the case y>0 and y<0 and show its differentiable in both cases (and maybe also have to do the same for when h2 >0 or <0) or can I use do it in one step by rewriting and by differentiating (y^2)^1/4 and. I did that and the Df(0,0) just goes away and then limit doesnt go to 0 it seems. Any help is appreciated thanks!
     
  2. jcsd
  3. Nov 18, 2012 #2

    SammyS

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    The title of this thread is misleading. This thread has to do differentiation, rather than integration.

    It looks like you have some typo's in your definition (of whatever, you don't say),
    lim |h| ->0 (f((0,0) + 9(h1,h2)) - f(0,0) - Df(0,0) (h1,h2)) / |h|​

    Clearly, the partial derivative, ∂f(x,y)/∂y, does not exist for y=0, except perhaps when x=0 and y=0 . Of course, that's at least part the matter at hand ...

    I'm no expert on this, but I suggest looking at the directional derivative of f(x,y) at (0,0), for arbitrary direction.
     
  4. Nov 19, 2012 #3
    yes sorry about the typos. I think the directional derivatives is not the issue I'm just wondering about if I need to do this piecewise.
     
  5. Nov 19, 2012 #4

    SammyS

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    √(|y|) is equivalent to (y2)1/4
     
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