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Given a $F(x,y)=\frac {y \sqrt {x^2+y^2} }{|y|} $ when $y ≠ 0$ else $F(x,0)=0 $ We have to prove that 1) all the directional derivatives exist at $(0,0)$ 2) but $F$ is not differentiable at $(0,0) $ .

For the first part it is easy to prove just by choosing a unit vector say $<a,b>$ and by definition of directional derivative I can find that the derivative evaluates to $sgn(b)$ . Hence I am done with this part.

Now we also know that $F_{x}$ and $ F_{y} $ are nothing but directional derivatives along $<1,0>$ and $<0,1>$ using these i tried evaluating the derivative to show that it doesn't exist but I am not being able to solve the limit to conclude that it is not defined ! Please help !

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    $\begingroup$ What is $F(x,0)$? $\endgroup$ – José Carlos Santos Sep 25 '19 at 10:14
  • $\begingroup$ F(x,0)=0 sorry forgot to mention that ! $\endgroup$ – Aditya Garg Sep 25 '19 at 10:16
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You have$$\frac{\partial F}{\partial x}(0,0)=0\text{ and }\frac{\partial F}{\partial y}(0,0)=1.$$Therefore, if $F$ is differentiable at $(0,0)$, $F'(0,0)$ is the linear map $(x,y)\mapsto y$. So, we should have$$\lim_{(x,y)\to(0,0)}\frac{F(x,y)-y}{\sqrt{x^2+y^2}}=0.$$But we don't have that since$$\frac{F(x,y)-y}{\sqrt{x^2+y^2}}=\frac y{\lvert y\rvert}-\frac y{\sqrt{x^2+y^2}}=\begin{cases}0&\text{ if }x=0\\1-\frac1{\sqrt2}&\text{ if }x=y>0.\end{cases}$$

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  • $\begingroup$ Thanks for the help sir $\endgroup$ – Aditya Garg Sep 25 '19 at 10:31

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