$f(x,y) = 0$ when $x = y = 0$. Shouldn't $\frac{\partial^2 f}{\partial y \partial x} = 0$?

Question

I'm given the following function:

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

And the task is to prove that $f_{xy}(0,0) = -2$ and that $f_{yx}(0,0) = 2$.

At first glance I'd say that neither of these is true since at $(0,0)$ the function is 0 and so both expressions should equal 0. I thought maybe I was wrong and tried to work with the other piece of the function but I got the following: $$ f_{xy}=\frac{\partial^2 f}{\partial y \partial x} = \frac{2(x^6+9x^4y^2-9x^2y^4-y^6)}{(x^2 + y^2)^3} $$

Which is clearly not $-2$ if we evaluate for $x=0, y=0$. What am I missing? Am I going about it all wrong? Any help is appreciated.

$f'(x)=1$, being zero at one point does not mean that derivatives have to be zero, I thought this was part of the confusion. — clark, Mar 16 '17 at 02:16
The key is that you can’t just use the formula that works for points other than the origin. You have to go back to the basic definition of partial derivatives to compute them at $(0,0)$. — amd, Mar 16 '17 at 02:29
What you naively get from your expression for $f_{xy}$ at $x=y=0$ is $0/0$, i.e. an indeterminate form. Compare what you get if you take $x\to 0$ first, simplify, then take $y\to 0$. — Semiclassical, Mar 16 '17 at 02:36

Mark Viola · Accepted Answer · 2017-03-16T02:26:01.253

Let $f$ be given by

$$f(x,y)=\begin{cases} 2xy\frac{x^2-y^2}{x^2+y^2}&,x^2+y^2>0\\\\ 0&,x^2+y^2=0 \end{cases}$$

For $x^2+y^2>0$, we have

$$\begin{align}\frac{\partial f(x,y)}{\partial x}&=\frac{2y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}\\\\\frac{\partial f(x,y)}{\partial y}&=-\frac{2x(y^4+4x^2y^2-x^4)}{(x^2+y^2)^2}\end{align}$$

To calculate the first partial derivaties at the origin, we revert to using the limit definition of the partial derivatives.

Proceeding, we find that

$$f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=0\\\\ f_y(0,0)=\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}=0$$

We proceed in a similar fashion to find the mixed partial derivatives at the origin. We find that

$$\begin{align} f_{yx}(0,0)&=\lim_{h\to 0}\frac{f_y(h,0)-f_y(0,0)}{h}\\\\ &=\lim_{h\to 0}\frac{-\frac{2h(0^4+4x^20^2-h^4)}{(h^2+0^2)^2}-0}{h}\\\\ &=2 \end{align}$$

and

$$\begin{align} f_{xy}(0,0)&=\lim_{h\to 0}\frac{f_x(0,h)-f_x(0,0)}{h}\\\\ &=\lim_{h\to 0}\frac{\frac{2h(0^4+4x^20^2-h^4)}{(h^2+0^2)^2}-0}{h}\\\\ &=-2 \end{align}$$

as was to be shown!

$f(x,y) = 0$ when $x = y = 0$. Shouldn't $\frac{\partial^2 f}{\partial y \partial x} = 0$?

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