I know that $f(x,y)=\sqrt{|xy|}$ when $(x,y)\neq{(0,0)}$ is differentiable since partial derivatives exist and they're continous. When I do by definition $f_x (0,0)$ and $f_y (0,0)$, both are equal to $0$. Due to "Sufficient Condition for the Differentiability of Functions", can I conclude that it is differentiable? My problem is that I think it isn't diff. because of an analogy to single-variable $f(x)=|x|$, but not sure. Thanks.
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1You may want to second-guess existence of partial derivatives for $(x,y)\ne(0,0)$. – Apr 13 '20 at 20:01
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I thought about using other directions. However the theorem says that partial derivatives are needed, or maybe i'm confusing myself. – Fabrizio Gambelín Apr 13 '20 at 20:05
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Nevermind! I was misreading your post, thank you very much! – Fabrizio Gambelín Apr 13 '20 at 20:06
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https://math.stackexchange.com/questions/477913/fx-y-sqrtxy?rq=1 – Surb Apr 13 '20 at 21:29
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No, it is not differentiable (since, for instance, its restriction to $\{(x,x)\mid x\in\mathbb R\}$ is not differentiable). Note that, if $x,y>0$, $\frac{\partial f}{\partial x}(x,y)=\frac12\sqrt{\frac yx}$. And we don't have $\lim_{(x,y)\to(0,0)}\frac12\sqrt{\frac yx}=0=\frac{\partial f}{\partial x}(0,0)$. So, $\frac{\partial f}{\partial x}$ is not continuous at $(0,0)$.
José Carlos Santos
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1Sorry, but if the partial derivative isn't continuous at a point, it doesn't mean that the function is not differentiable, does it? – Fabrizio Gambelín Apr 14 '20 at 05:28
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No. That's why my first sentence contains a proof that $f$ is not differentiable. The rest of my answer shows that you were wrong when you stated that the partial derivatives are continuous. – José Carlos Santos Apr 14 '20 at 06:34
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