How $f_{yx} = f_{xy}$?

Question

In this pic, it is given that, " it readily follows that $f_{yx}= f_{xy}$ when $x=0, y≠0$ or $x≠0, y=0$. But I can't get it! How they are equal? Here below is my solution pic,

Answer 1

This is first time, I am answering my own question. Finally after @kernel sir, answer I get the point how to calculate that derivatives! Since after simplification $f(x,y)$ can be written as,

$$ f(x,y) = \begin{cases} \frac{x^{3} y-x y^{3}}{x^2+y^2} &\text{if $x^2+y^2≠0$}\\ 0,&\text{when $x=y=0$}\end{cases}$$

So from here,

$$ f_{x}(x,y) = \frac{3x^2y-y^3}{(x^2+y^2)} - \frac{2x^4y-2x^2y^3}{(x^2+y^2)^2}$$ when $x^2 + y^2≠ 0$

$$f_{yx}(x,y) = \frac{∂}{∂y}(f_{x}(x,y))$$

$f_{yx}(x,y)=\frac{(x^2+y^2)(3x^2-3y^2)-(3x^2y-y^3)(2y)}{(x^2+y^2)^2} -\frac{(x^2+y^2)^2 (2x^4-6x^2y^2)-2(2x^4y-2x^2y^3)(x^2+y^2)(2y)}{(x^2+y^2)^4}$

when $x^2+y^2≠0$

and hence,

$f_{yx}(0,y)= \frac{(0+y^2)(0-3y^2)-(0-y^3)(2y)}{(0+y^2)^2} -0$

$= \frac{-3y^4 +2y^4}{y^4}$

$= -1$

and similarly we can calculate, $f_{xy}(0,y)$ and I found that,

$f_{yx}(0,y)=f_{xy}(0,y)=-1$

Thanks to all of you. The key point is that use quotient rule of derivatives and just calculation. Sorry to trouble you all.

Further, I found ""kernel sir"" answer very Useful, Thank u.

Answer 2

You need to compute the partial derivatives, and then study what happens under the conditions $x=0, y\neq 0$, and $y=0, x\neq 0$. If you actually compute the partial derivatives, we obtain:

$f_x=y(\frac{x^2-y^2}{x^2+y^2}+\frac{4x^2y^2}{(x^2+y^2)^2})$

$f_y=x(\frac{x^2-y^2}{x^2+y^2}-\frac{4x^2y^2}{(x^2+y^2)^2})$

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

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

notice that when $x=0$, $y\neq 0$, we have $f_{xy}=\frac{-y^6}{y^6}=-1$

we have for $y=0$, $x\neq 0$ that $f_{xy}=\frac{x^6}{(x^2)^3}=1$

now let's turn our attention to $f_{yx}$. Notice that we get the same result. So indeed, we have $f_{xy}=f_{yx}$

However, this is to be expected, because any function which is $C^1(\mathbb{R}^2)$ (the first derivative exists and is continuous) automatically has equal mixed partials $f_{xy}=f_{yx}$.

In case you haven't seen the above notation, the function $f(x,y)$ is continuous everywhere except the origin, and in particular, continuous along $x=0, y\neq 0$ and $x\neq 0, y\neq 0$