Partial derivative of $xy\frac{x^2-y^2}{x^2+y^2}$

Question

I am asked to show, that

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

is everywhere two times partial differentiable, but it is still $D_1D_2f(0,0)\neq D_2D_1 f(0,0)$

But this does not make much sense in my opinion. Since $f(0,0)=0$ hence it should be equal?

I calculated $\frac{\partial^2 f(x,y)}{\partial x\partial y}$ and $\frac{\partial^2 f(x,y)}{\partial y\partial x}$ and I got in both cases the same result (for $(x,y)\neq (0,0)$) which is:

$\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}$

Wolframalpha says, that this is correct.

So is there a mistake in the task? Or do I not understand it?

Also, when I want to show, that $f$ is everywhere two times partial differentiable. Is it enough to calculate $\frac{\partial^2 f}{\partial^2 x}$ and $\frac{\partial^2 f}{\partial^2 y}$ and not needed to go by the definition, since we know, that this function is differentiable as a function in $\mathbb{R}$?

What do you think? Thanks in advance.

Answer 1

The limit definition of partial derivative at $(x,y)=(0,0)$: $$f_x(0,0)=\frac{\partial f}{\partial x}(0,0)=\lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h}=\frac{\frac{(0+h)\cdot 0\cdot ((0+h)^2-0^2)}{(0+h)^2+0^2}-0}{h}=0;\\ f_y(0,0)=\frac{\partial f}{\partial y}(0,0)=\lim_{h\to 0} \frac{f(0,0+h)-f(0,0)}{h}=\frac{\frac{0\cdot(0+h)\cdot (0^2-(0+h)^2)}{0^2+(0+h)^2}-0}{h}=0.$$ Note that: $$f_x(x,y)=\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\left(\frac{x^3y-xy^3}{x^2+y^2}\right)=\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2};\\ f_y(x,y)=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial y}\left(\frac{x^3y-xy^3}{x^2+y^2}\right)=\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}.\\ $$ The partial derivative of partial derivative: $$\frac{\partial^2 f}{\partial x\partial y}(0,0)= \lim_{h\to 0} \frac{f_x(0,0+h)-f_x(0,0)}{h}=\\ \lim_{h\to 0} \frac{\frac{(0+h)(0^4+4\cdot 0^2(0+h)^2-(0+h)^4)}{(0^2+(0+h)^2)^2}-0}{h}=\lim_{h\to 0} \frac{-h^5}{h^5}=-1;\\ \frac{\partial^2 f}{\partial y\partial x}(0,0)= \lim_{h\to 0} \frac{f_y(0+h,0)-f_y(0,0)}{h}=\\ \lim_{h\to 0} \frac{\frac{(0+h)^5-4(0+h)^3\cdot 0^2-(0+h)\cdot 0^4}{((0+h)^2+0^2)^2}-0}{h}=\lim_{h\to 0} \frac{h^5}{h^5}=1.$$

Answer 2

First a comment

You wrote But this does not make much sense in my opinion. Since $f(0,0)=0$...

How can you induce an affirmation on second derivatives of a map based on its value at a point?

Second regarding partial second derivatives

You have for $(x,y)\neq(0,0)$

$$\begin{cases} \frac{\partial f}{\partial x}(x,y) = \frac{(x^2+y^2)(3x^2y-y^3)+2x^2y(y^2-x^2)}{(x^2+y^2)^2}\\ \frac{\partial f}{\partial y}(x,y) = \frac{(x^2+y^2)(x^3-3x y^2)+2xy^2(y^2-x^2)}{(x^2+y^2)^2} \end{cases}$$

In particular $$\begin{align*} \frac{\partial f}{\partial x}(0,y) &= -y \text{ for } y \neq 0\\ \frac{\partial f}{\partial x}(x,0) &= x \text{ for } x \neq 0 \end{align*}$$ which implies $$ \frac{\partial^2 f}{\partial x \partial y}(0,0) =1 \neq -1 = \frac{\partial^2 f}{\partial y \partial x}(0,0)$$

You can have a look here for more details.