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Math Help - For a rational/irrational function, determine where f is differentiable

  1. #1
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    For a rational/irrational function, determine where f is differentiable

    This relates to a previous question I asked, so we can somewhat pick up from there.

    For the function f(x,y)=\left\{\begin{array}{cc}x^2+y^2,&x,y\mbox{ both rational }\\0, & \mbox{ otherwise, }\end{array}\right
    determine at which points f is differentiable.

    I already know that f is continuous ONLY at the point (0,0), so that's the only potential location for where f could be differentiable. Showing whether or not it is, however, is the trick.

    Would using this trick work?
    u(h,k)=\frac{f(x+h,y+k)-f(x,y)-f_x(x,y)h-f_y(x,y)k}{\sqrt{h^2+k^2}} when (h,k)\ne (0,0)
    I only need to determine whether or not u(h,k)\rightarrow 0 as (h,k)\rightarrow (0,0) to prove whether or not the function is differentiable at (0,0).
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  2. #2
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    Quote Originally Posted by Runty View Post
    u(h,k)=\frac{f(x+h,y+k)-f(x,y)-f_x(x,y)h-f_y(x,y)k}{\sqrt{h^2+k^2}} when (h,k)\ne (0,0)
    If you define u(h,k) in this way, you need to say what f_x and f_y are. And if you say they are partial derivatives, you need to explain at least why they exist.

    In any case, if the derivative (a linear functional \mathbb{R}^2\to\mathbb{R}) exists, it is zero, so we have to say whether \displaystyle\lim_{(h,k)\to(0,0)}\frac{f(h,k)}{\sq  rt{h^2+k^2}}=0. Well, \displaystyle\frac{f(h,k)}{\sqrt{h^2+k^2}}\le\sqrt  {h^2+k^2}, so yes.
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