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Let's consider the function given by: $$ F\left( {x,y} \right) = \begin{cases} \frac{{xy}}{{\sqrt {x^2 + y^2 } }}\text{ when }\,\,\,\,(x,y) \ne (0,0) \\ \\ \\ \\ \,\,\,\,\,\,0\hspace{10ex}\text{ if } \,\,\,\,\,\,\,\,\,(x,y)=(0,0) \end{cases} $$

I proved a lot of things of this function, the partial derivated exist everywhere, in particular at $(0,0)$ , But I want to prove that $F$ is not differentiable at $(0,0)$, I I tried by definition but it does not work. Please help me !!

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marked as duplicate by fgp, user17762, Martin, Hagen von Eitzen, Git Gud May 11 '13 at 19:10

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  • $\begingroup$ "I tried by definition but it does not work" can you show your work? It should have worked, if you took the derivatives correctly. $\endgroup$ – Ataraxia May 11 '13 at 18:24
  • $\begingroup$ This has be asked here in the past, the last time wasn't long ago. $\endgroup$ – fgp May 11 '13 at 18:33
  • $\begingroup$ Show how you "tried by definition", and explain what didn't work. That way, we can show you what you overlooked and/or did wrong! $\endgroup$ – Excluded and Offended May 11 '13 at 18:35
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Correct hint: If you've computed the partial derivatives at $(0,0)$, you know they're both $0$. So, if $f$ were differentiable, you'd have $$\lim_{(x,y)\to (0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$ Is this so?

Alternatively, if $f$ were differentiable, the directional derivative in every direction $\vec v$ would have to be given by $\nabla f(0,0)\cdot\vec v$.

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  • $\begingroup$ Thanks for pointing it out. $+1$ $\endgroup$ – Git Gud May 11 '13 at 19:06

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