This question arose after my calculus test in which the told us: Show that $$f(x,y)=\begin{cases}\frac{2x^3}{x^2 +y^2}&\text{ if }(x,y)\neq(0,0)\\0&\text{ if }(x,y)=(0,0).\end{cases}$$ is not differentiable at $(0,0)$. I showed that the partials do not exist in $(0,0)$ thus the function is not differentiable.
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1$\begingroup$ Yes, this is correct. If in doubt, the first thing to do should have been to look at the definition of differentiability that was given in your course. $\endgroup$– AlbertJun 8, 2018 at 12:38
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2 Answers
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Yes, you are right. Because if the function $f$ is differentiable at a point $p$ of its domain, then the directional derivatives exist for every direction $v$. And the partial derivatives are a particular case of directional derivatives.
The only problem is that in your specific example, the partial derivatives do exist. In fact $\frac{\partial f}{\partial x}(0,0)=2$ and $\frac{\partial f}{\partial y}(0,0)=0$.
answered Jun 8, 2018 at 12:40
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2$\begingroup$ @T.Katsipis $f(x,0)=2x\implies\frac{\partial f}{\partial x}(0,0)=2$ and $f(0,y)=0\implies\frac{\partial f}{\partial y}(0,0)=0$. $\endgroup$ Jun 8, 2018 at 12:45
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$\begingroup$ @T.Katsipis If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ Jun 8, 2018 at 12:50
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$\begingroup$ @JoséCarlosSantos In fact you haven't answer to the question which is "prove that $f$ isn't differentiable". $\endgroup$ Jun 8, 2018 at 13:22
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Proof that $f$ isn't differentiable.
If $f$ was differentiable at $(0,0)$, we would have $$\lim\limits_{(x,y) \to (0,0)} \frac{f(x,y) - f(0,0) -2x}{\sqrt{x^2+y^2}} = 0,$$
as $\frac{\partial f}{\partial x}(0,0)=2$ and $\frac{\partial f}{\partial y}(0,0)=0$.
But $$\frac{f(x,y) - f(0,0) -2x}{\sqrt{x^2+y^2}} = \frac{-2xy^2}{\left(x^2+y^2 \right)^{3/2}} = \pm \frac{1}{\sqrt{2}}$$
for $x=y$. Hence $f$ is not differentiable at $(0,0)$.
answered Jun 8, 2018 at 13:38