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I have to verify that $$f(x,y)=\frac {2x-y}{x+y}$$ indicating possible exceptional points and intestigating those points My attempt:

I calculated

$$f_x=\frac {3y}{(x+y)^2}$$ $$f_y=\frac {-3x}{(x+y)^2}$$ and for $$(x,y)\ne(0,0)$$ $$f_{xy}=\frac {3x-3y}{(x+y)^3}$$ $$f_{yx}=\frac {3x-3y}{(x+y)^3}$$ At $(0,0)$,does $f_{xy}$ and $f_{yx}$ exist? My guess is that it does't exist because $f$ is not defined at $(0,0)$.Am I correct?

PiGamma
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  • Yep, this makes no sense to speak further about derivatives when $f(0,0)$ is not even defined. In this case $f(x,0)=2$ and $f(0,y)=-1$ so $f$ is not continuous in $(0,0)$. – zwim Nov 17 '17 at 04:45

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They indeed do not exist at $(0, 0)$ because of dividing by 0 (so there's a perforation at each set for which $x + y = 0$ is true). However, because at $(0, 0)$ $x = y$, $f_{x,y} \iff f_{y,x}$, so they are interchangeable, so i.e. their limits will be the same. They get very close to each other, intersecting when approaching 0, but they do not intersect at $(0, 0)$ because then $x + y = 0$ is true, which means the function won't exist at all.

SuperSjoerdie
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