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Wondering if anyone can point out a function $f$ from an open disc in $\mathbb R^2$ to $\mathbb R$, that:

  • is defined and continuous at a point $p$;
  • has(they exist) both of its partial derivatives at $p$;
  • at least one of the partials is discontinuous at $p$;
  • is NOT differentiable at $p$;

Context: (not necessary for providing an answer): While learning about the differentiablity of multivariable functions in my calculus course, started thinking about this particular case after finding out that the differentiablity of a continuous multivariable function (at a point) depends not solely on the existence of the partial derivatives. Naturally the question: “what other condition is necessary for differentiablity then?” arises. Learned that if both partials are continuous at p, that is sufficient. But apparently even if they aren't continuous at p, f can still be differentiable. Found one of those functions after much work but then wondered what would a function look like, that had everything set up for differentiablity at a point, but isn't, because of the discontinuity of one of the partials.

Would appreciate the function, or even a good intuition on how to find one.

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Define $$f(r\cos(t),r\sin(t))=r^{1/2}\sin(2t)\quad(r>0,t\in\Bbb R).$$

Then $f$ is continuous, and not differentiable at the origin (for example if $\phi(x)=f(x,x)=|x|^{1/2}$ then $\phi'(0)$ does not exist).

But $$f(x,0)=f(0,y)=0,$$hence $$\frac{\partial f}{\partial y}(0,0)=\frac{\partial f}{\partial y}(0,0)=0.$$

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