If you define f(x,y)= $(x^2+y^2)^\frac{1}{2}$
Then $$f^\prime_x= \frac{x}{f(x,y)}$$and $f^\prime_y= \frac {y}{f(x,y)}$
Now is used the definition of the partials on (0,0) $f^\prime_x= \lim_{h \to 0} \frac {f(0+h,0)-f(0,0)}{h}= \lim_{h \to 0} \frac {h-0}{h}=1$ $f^\prime_x= \lim_{h \to 0} \frac {f(0,0+h)-f(0,0)}{h}= \lim_{h \to 0} \frac {h-0}{h}=1$
Now to test differentiability
$\lim_{(x,y) \to (0,0)} \frac {f(x,y)-f(0,0)- f^\prime_x Δx - f^\prime_y Δy}{(Δx^2+Δy^2)^(\frac{1}{2})}=\lim_{(x,y) \to (0,0)} \frac {(x^2+y^2)^(\frac{1}{2})-0-x-y}{(x^2+y^2)^(\frac{1}{2})}$
Note the partials are set to 1.now I don't know if it can set to any path on differentiability I know it can be done on function,is possible to set a path on any limit?tell me in the questions bellow, but I am going to evaluate as it is possible to set a path.
To y=x
$\lim_{(x,y) \to (0,0)} \frac {(2)^(\frac{1}{2})x-2x}{(2)^(\frac{1}{2})x}=\lim_{(x,y) \to (0,0)} 1-(2)^(\frac{1}{2})= 1-(2)^(\frac{1}{2}) $
Now setting another path y=-x
$\lim_{(x,y) \to (0,0)} \frac {(2)^(\frac{1}{2})x-0}{(2)^(\frac{1}{2})x}=\lim_{(x,y) \to (0,0)} 1=1 $
If it was a function I would say as we approach f from different paths to origin we obtain different values hence discontinuous.not sure what to say next,but something like not diferrentiable.
If is true then even though the function as partials on point (0,0) and the function is continuous as easily notice the function is not diferentiable on (0,0) hence not diferrentiable.is this true? I thought if a functions was continuous and its partials exit was diferrentiable.
As a reminder I am asking 3 questions.(path on diferentiabillity?conclusion of last question? And no differentiability)
The limit to compute for the partial derivative with respect to $x$ at $(0,0)$ is $$ \lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{\sqrt{h^2}}{h} =\lim_{h\to0}\frac{|h|}{h} $$ which doesn't exist. Therefore $f$ is not differentiable at $(0,0)$.
The fact is that $\sqrt{a^2}=|a|$.