# Show that both mixed partial derivatives exist at the origin but are not equal

$$f(x,y) = \begin{cases} \displaystyle \frac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y) \neq (0,0), \\ 0 & \text{if } (x,y) = (0,0). \end{cases}$$

I tried finding both mixed partial derivatives but they ended up being the same for that function. I must be failing to take into account something dealing with the fact that it is piece-wise. I still need to show the mixed partial derivatives exist. How can I do all of this?

Use the definition of partial derivative: $$f_x(0,0) ~=~ \lim_{h\to 0} \frac{f(h,0)-f(0,0)} h ~=~ \lim_{h\to 0} \frac{\frac{h\cdot 0(h^2-0^2)}{h^2+0}-0} h ~=~ \lim_{h\to 0} \frac{0}{h} ~=~ 0$$ A similar computation shows that $$f_y(0,0)=0$$ too, so that $$(0,0)$$ is a critical point (i.e. $$\nabla f(0,0)=\binom 00$$).
Now that we know the values of $$f_x(0,0)$$ and $$f_y(0,0)$$ we can compute $$f_{xy}(0,0)$$ and $$f_{yx}(0,0)$$: $$f_{xy}(0,0) ~=~ \lim_{k\to 0} \frac{f_x(0,k)-f_x(0,0)}k ~=~ \lim_{k\to 0} \frac{f_x(0,k)}k$$ and $$f_{yx}(0,0) ~=~ \lim_{h\to 0} \frac{f_y(h,0)-f_y(0,0)}h ~=~ \lim_{h\to 0} \frac{f_y(h,0)}h$$ First, note that for $$(x,y)\neq(0,0)$$ you have $$f_x(x,y)=\frac{y\big(x^4+4x^2y^2-y^4\big)}{\big(x^2+y^2\big)^2}$$ and $$f_y(x,y)=\frac{x\big(x^4-4x^2y^2-y^4\big)}{\big(x^2+y^2\big)^2}$$ so that for $$h,k\neq 0$$ $$f_x(0,k)=-k \quad\text{and}\quad f_y(h,0)=h$$ Putting all together: $$f_{xy}(0,0) ~=~ \lim_{k\to 0} \frac{f_x(0,k)}k ~=~ \lim_{k\to 0}\frac{-k}{k} ~=~ -1$$ and $$f_{yx}(0,0) ~=~ \lim_{h\to 0} \frac{f_y(h,0)-f_y(0,0)}h ~=~ \lim_{h\to 0} \frac{f_y(h,0)}h ~=~ \lim_{h\to 0}\frac{h}{h} ~=~ 1$$ so $$f_{xy}(0,0)~=~-1 ~~\neq~~ 1~=~f_{yx}(0,0)$$
• The mixed partial derivatives are $f_{xy}$ and $f_{yx}$. – TonyK Apr 12 '13 at 16:15
• @AndreasT Can $f_{xy}$ exist without $f_y$ existing? – Prince Kumar Dec 10 '16 at 11:11
• @PrinceKumar yes of course, $f$ need not even be continuous. Consider $$f(x,y) = \begin{cases}1&\text{if } y\neq 0\\0&\text{if }y=0\end{cases}$$ – AndreasT Dec 24 '16 at 8:53