# Does this partial derivative exist at (0,0)?

Question:$$f(x,y)$$ = $$\lbrace\frac{2xy}{x^2+y^2}$$ if $$(x,y)$$ $$\neq (0,0$$), $$0$$ otherwise$$\rbrace$$

Does $$f_{yx}(0,0) exist?$$

Attempt:

$$f_x$$=$$\frac{2y(y^2-x^2)}{(x^2+y^2)^2}$$

$$f_y$$=$$\frac{2x(1-2y^2)}{(x^2+y^2)^2}$$

$$f_{xy}$$=$$\lim_{h\to 0}\frac{f_x(0,0+h)-f_x(0,0)}{h}$$=$$\lim_{h\to 0}\frac{2}{h^2}$$ $$\to$$ +infinity

$$f_{yx}$$=$$\lim_{h\to 0}\frac{f_y(0+h,0)-f_y(0,0)}{h}$$=$$\lim_{h\to 0}\frac{2}{h^4}$$ $$\to$$ +infinity

So the limit does not exist. Is this correct? If both limit goes to infinity it means limit does not exist right?

• Actually the function isn't even continuous at $(0,0)$. Check for example two paths $(x,0)\to (0,0)$ and $(x,x)\to (0,0)$ – Mostafa Ayaz Mar 24 '19 at 7:54

Yes,the limit does not exist. Otherwise also suppose you approach $$(0,0)$$ along $$y=mx$$ so putting y=mx gives, $$f(x,mx)=\frac {2m}{1+m^2}$$. The value the function tends to at $$(0,0)$$ varies with $$m$$. So the limit is non-existent at $$(0,0)$$.
• I forgot to put in the full question, my bad. But i need to make sure if $f_{yx}$ (0,0) exists or not. – crown Mar 24 '19 at 8:00