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Let $$f(x,y) = \begin{cases} \dfrac{xy(x^2-y^2)}{x^2+y^2}, & (x,y) \neq (0,0), \\ 0, & (x,y)=(0,0). \end{cases}$$ Show that

(A) $f_{xy}(0,0) \neq f_{yx}(0,0)$.

(B) $f$ is differentiable at $(0,0)$.

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The function is continuous, the only point of contention is at $(0,0)$ where its easy to see that $|f(x,y)| \leq |x||y|$. –  copper.hat Jul 18 '12 at 2:41
@copper.hat The continuity of $f$ at $(0,0)$ is not useful. –  blindman Jul 18 '12 at 16:35
@blindman: A previous comment, which has been deleted, asserted that $f$ was not continuous. I was just pointing out that it is. –  copper.hat Jul 18 '12 at 16:43
@blindman: Also, differentiablity of $f$ at $(0,0)$ follows almost trivially from the fact that $|f(x,y)|\leq |x||y|$, so, in fact, it is useful in the context of this problem. –  copper.hat Jul 18 '12 at 16:52
To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it is a command ("Show that..."), not a request for help, so please consider rewriting it. –  Zev Chonoles Jul 19 '12 at 4:19

1 Answer 1

up vote 3 down vote accepted

We have seen that

  • $\displaystyle f_x(0,0)=\lim_{\Delta x\rightarrow 0}\frac{f(0+\Delta x, 0)-f(0,0)}{\Delta x}=0$

  • $\displaystyle f_x(a,b)=\frac{(3a^2b-b^3)(a^2+b^2)-2a^3b(a^2-b^2)}{(a^2+b^2)^2} \quad \forall (a,b)\in \mathbb{R}^2\setminus\{(0,0)\}$

  • $\displaystyle f_y(0,0)=\lim_{\Delta y\rightarrow 0}\frac{f(0, 0+\Delta y)-f(0,0)}{\Delta y}=0$

  • $\displaystyle f_y(a,b)=\frac{(a^3-3ab^2)(a^2+b^2)-2ab^3(a^2-b^2)}{(a^2+b^2)^2} \quad \forall (a,b)\in \mathbb{R}^2\setminus\{(0,0)\}$

Therefore, $$ f_{xy}(0,0)=\lim_{\Delta y\rightarrow 0}\frac{f_x(0, 0+\Delta y)-f_x(0,0)}{\Delta y}=\lim_{\Delta y\rightarrow 0}\frac{-(\Delta y)^5}{(\Delta y)^5}=-1, $$ $$ f_{yx}(0,0)=\lim_{\Delta x\rightarrow 0}\frac{f_y(0+\Delta x, 0)-f_y(0,0)}{\Delta x}=\lim_{\Delta x\rightarrow 0}\frac{(\Delta x)^5}{(\Delta x)^5}=1. $$ From the formula of $f_x$ and $f_y$ we deduce that $f$ has partial derivatives which are continuous in a neighborhood of $(0,0)$ and so $f$ is differentiable at $(0,0)$. Indeed, it is clear that $f_x$ and $f_y$ are continuous on $\mathbb{R}^2\setminus\{(0,0)\}$. Moreover, \begin{eqnarray*} |f_x(a,b)-f_x(0,0)|&\leq& \frac{|3a^4b|}{(a^2+b^2)^2}+\frac{|2a^2b^3|}{(a^2+b^2)^2}+\frac{|b^5|}{(a^2+b^2)^2}+ \frac{|a^5b|}{(a^2+b^2)^2}+\frac{|2a^3b^3|}{(a^2+b^2)^2}\\ &\leq&\frac{|3a^4b|}{a^4}+\frac{|2a^2b^3|}{2a^2b^2}+\frac{|b^5|}{b^4}+\frac{|a^5b|}{a^4}+ \frac{|2a^3b^3|}{2a^2b^2}\\ &=&3|b|+|b|+|b|+|a||b|+2|a||b|, \end{eqnarray*} which implies $\displaystyle\lim_{(a,b)\rightarrow(0,0)}f_x(a,b)=f_x(0,0)$, and so $f_x$ is continuous at $(0,0)$. Similarly, $f_y$ is continuous at $(0,0)$.

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