Partial derivative problem

Question

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 $(0,0)$.

Answer 1

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 $(0,0)$.