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Math Help - Differentiable points of function

  1. #1
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    Differentiable points of function

    The problem is stated as follows:

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    The function f(x,y) is defined in R^2 as f(0,0) = 0 and f(x,y) = \frac{(\sin^2{x})(\sin^2{y})}{x^2 + y^2} if (x,y) \ne (0,0).

    In what points is f(x,y) differentiable?

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    It's easily shown that the two first partial derivatives are continuous except at the origin, so f is differentiable at every point that isn't the origin.

    Using the definition of differentiability for the origin,

    \lim_{(h,k) \to (0,0)}{\frac{f(h,k) - f(0,0) -  h f_1(0,0) - k f_2(0,0)}{\sqrt{{h}^2 +  {k}^2}}} = \lim_{(h,k) \to (0,0)}{\frac{f(h,k)}{\sqrt{{h}^2 +  {k}^2}}} = \lim_{(h,k) \to (0,0)}{\frac{\sin^2{h}\sin^2{k}}{(h^2 + k^2)^{3/2}}}

    After trying Maclaurin expansions of the numerator and denominator without getting anywhere (perhaps due to arithmetical mistakes), and fooling around a bit without any more luck, I switched to polar coordinates, i.e. h = r\cos{\theta}, k = r\sin{\theta}. Since r approaches 0 regardless of the angle \theta, the limit is equivalent to showing that \lim_{r \to 0}\frac{\sin^2{(r\cos{\theta})}\sin^2{(r\sin{\thet  a})}}{r^3} = 0.

    By using L'H˘pital's rule three times, the latter limit can be shown, and while the operations are simple, a considerable amount of arithmetic is required to show differentiability in the origin (and hence, in every point in R^2) like this. I strongly suspect that I have missed at least one much easier solution to this. Hints?
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  2. #2
    Super Member girdav's Avatar
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    Use the inequality |\sin t|\leq |t| for t\in\mathbb R.
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  3. #3
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    Quote Originally Posted by girdav View Post
    Use the inequality |\sin t|\leq |t| for t\in\mathbb R.
    Thanks for your reply! I'll give that a try.

    Also, dividing the polar coordinate limit's numerator and denominator by (r\cos{\theta})^2 (r\sin{\theta})^2 works fine. I'm not sure why I thought using l'h˘pital three times was a good idea...
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