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Limit of derivative

  1. Apr 30, 2012 #1
    obviously x^2sin(1/x) isn't continuous in its derivative at zero. But it seems if you take
    (x^2 + y^2) sin(1/(x^2+y^2)) and switch to polar coordinates then the derivative is continuous at zero.... but that can't be right... can it?
  2. jcsd
  3. Apr 30, 2012 #2


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    The second function doesn't restrict to the first function on the x-axis, so I'm not sure why you would be surprised their differentiability properties would be different. On the other hand the function

    [tex] (x^2+y^2) \sin( \frac{1}{\sqrt{x^2+y^2}} ) [/tex]
    is the radial extension of your original function to the x-y plane, and when you switch to polar coordinates you get that it is not differentiable at zero (since it's the exact same function once you do that)
  4. Apr 30, 2012 #3
    Please be more specific. What are your calculations, for instance?

    Also, the derivative in polar coordinates aren't that simple. If we have r=f(θ), then f'(θ) isn't necessarily the slope of the line tangent to f at θ.
  5. Apr 30, 2012 #4
    g'(r,alpha) = (for all alpha) = z2r*sin(1/r*sqrt(z)) - r^2/sqrt(r*z) cos(1/r*sqrt(z)) is what I get. where z = cos^2(alpha) + sin^2(alpha). And the limit should exist as r goes to zero by the squeeze theorem., .... right?
  6. Apr 30, 2012 #5
    OH, no.... I made a mistake. I made a bad substitution as I forgot that r = sqrt(x^2+y^2) and then its really trivial. Funny, that I had this idea but did the transformation wrong, still.
    Last edited: Apr 30, 2012
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