Discontinuous partial derivatives example

littlemathquark
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Homework Statement
$$f(x,y)=\left\{\begin{array}{ccc} (x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & , & (x,y)\neq (0,0) \\ 0 & , & (x,y)=(0,0) \end{array}\right.$$
Relevant Equations
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$$f(x,y)=\left\{\begin{array}{ccc} (x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & , & (x,y)\neq (0,0) \\ 0 & , & (x,y)=(0,0) \end{array}\right.$$ This function is differentiable at (0,0) point but ##f_x## and ##f_y## partial derivatives not continuous at (0,0) point. I need another examples like this. Thank you.
 
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IIRC,
##f(x,y):=\frac {2xy}{x^2+ y^2}##
 
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Take a classical univariate differentiable but not continuously differentiable example:
<br /> f(x) := \begin{cases} x^2 \sin (1/x), &amp;x\neq 0 \\ 0,&amp;x=0 \end{cases}<br />
Then define
<br /> h(x,y) = f(x) + f(y).<br />
Obviously, one can extend this to arbitrary finite dimension.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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