# Multivariate Function's Differentiability?

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 $$$$ and by definition of directional derivative I can find that the derivative evaluates to $$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 !

• What is $F(x,0)$? – José Carlos Santos Sep 25 '19 at 10:14
• F(x,0)=0 sorry forgot to mention that ! – Aditya Garg Sep 25 '19 at 10:16

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$$\frac{F(x,y)-y}{\sqrt{x^2+y^2}}=\frac y{\lvert y\rvert}-\frac y{\sqrt{x^2+y^2}}=\begin{cases}0&\text{ if }x=0\\1-\frac1{\sqrt2}&\text{ if }x=y>0.\end{cases}$$