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I posted this problem earlier today, but now I have some other questions regarding the problem.

$$F(x,y)=\begin{cases} f(x,y) & (x,y)\neq (0,0) \\ 0 & (x,y)=0 \end{cases}$$

where $f(x,y)=\frac{x^2y}{x^2+y^2}$

*1.) Determine if f is continuous in origin.*

The function f(x,y) is not continuous in origin, because it is not defined in origin.

2.) Decide $F_1(x,y)$ and $F_2(x,y)$.

$$F_1(x,y)=\frac{2xy(x^2+y^2)-x^2y(2x)}{(x^2+y^2)^2}=\frac{2xy^3}{(x^2+y^2)^2}$$ $$F_2(x,y)=\frac{x^2(x^2-y^2)}{(x^2+y^2)^2} $$

But this derivative will only apply to $(x,y)\neq(0,0)$. Do I also have to find another expression for $F_1$ and $F_2$ when $(x,y)=(0,0)$, or should that be done in task number 3 ?

3. Determine the partial derivatives in origin and decide if $F$ is differentiable in origin.

So I'm a bit confused. I thought about that finding the partial derivatives in origin by using

$$ F_1(0,0)=\lim_{h\to0}\frac{F(0+h,0)-F(0,0)}{h}=\lim_{h\to 0}\frac{\frac{h^2*0}{h^2+0}}{h}=0 $$

$$F_2(0,0)=\lim_{h\to0}\frac{F(0,0+h)-F(0,0)}{h}=\lim_{h\to 0}\frac{\frac{0^2*h}{h^2+0}}{h}=0$$

I don't know if this is right, but how do I then determine if $F$ is differentiable in origin?

amWhy
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Mathomat55
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    I believe the first question asks if the extended function is continuous at $(0,0)$, not $f$ itself. (otherwise the question makes no sense). And the answer is yes, $F$ is continuous. – Mark Apr 04 '20 at 15:15
  • You know what? I also thought about that, and I even sent a message to my professor about it. But he said that he ment what he wrote. I then saw one of our videolectures about partial differentiation. With another example (not a piecewise-function) he said that the function was not continuous in origo, and then proceeded to make a piecewise function afterwards. I hope you understand what I meant. And thats why I wrote the the $f(x,y)$ is not continuous – Mathomat55 Apr 04 '20 at 15:22
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    Alright then. It's just strange to ask if a function is continuous at a point where it isn't even defined. – Mark Apr 04 '20 at 15:43
  • I know, I also find it strange – Mathomat55 Apr 04 '20 at 15:51
  • origin: $(0, 0)$ is the origin, not the origo. – amWhy Apr 23 '20 at 19:41

1 Answers1

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Your computation of the partial derivatives at $(0,0)$ is fine (and elsewhere too). Since they are both equal to $0$, if $F$ was differentiable at the origin, $F'(0,0)$ would be the null function. In other words, we would have$$\lim_{(x,y)\to(0,0)}\frac{F(x,y)}{\sqrt{x^2+y^2}}=0,$$which is equivalent to$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{(x^2+y^2)^{3/2}}=0.$$But this is not true, because if $y=x>0$,$$\frac{x^2y}{(x^2+y^2)^{3/2}}=\frac{x^3}{2\sqrt2\,x^3}=\frac1{2\sqrt2}\neq0.$$

José Carlos Santos
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