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I solved a problem in Calculus 2 exercise lesson and I do not fully understand it.

The task is to show that cross derivates differ in sign ( $f_{\text{xy}}(0;0)=-1$ and $f_{\text{yx}}(0;0)=1\, $ )

The function is: $ f(x,y)=\left\{ \begin{array}{cc} \frac{x y \left(x^2-y^2\right)}{x^2+y^2} & x^2+y^2\neq 0 \\ 0 & x^2+y^2=0 \end{array} \right. $

Short solution is:

Plot of function

I was told that lesson, what I should take away from this is, that cross derivates in this case

  • are not continuous
  • do not equal

Can someone explain:

  • why it is not continuous
  • why can't I just take $f_{xy}=\begin{cases} \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3} & x^2+y^2\neq 0 \\ 0 & x^2+y^2 = 0 \end{cases} $ and where do I deduce, that I need to use limits at first place.
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up vote 1 down vote accepted

The cross-derivative as you have worked out when $(x,y) \neq (0,0)$ is $\displaystyle \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3}$

So why is this function not continuous at the origin?

What does continuity at a point $(x_0,y_0)$ in $2D$ mean?

For a function to be continuous at $(x_0,y_0)$, we need that irrespective of the way you approach the point $(x_0,y_0)$, the value of the limit should not change. (Note that in $2D$ we have infinite directions of approaching the point $(x_0,y_0)$. This is in contrast to $1D$ where we have only $2$ directions to approach a specific point.)

For the function $\displaystyle \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3}$, let us see what is the value of the limit as we approach the origin $(0,0)$ along different lines passing through the origin $(0,0)$ i.e. along lines $y=mx$ for different values of $m$.

$$\lim_{y=mx; x \rightarrow 0} \frac{x^6+9 x^4 y^2-9 x^2 y^4-y^6}{\left(x^2+y^2\right)^3} = \lim_{x \rightarrow 0} \frac{x^6+9 x^4 (mx)^2-9 x^2 (mx)^4-(mx)^6}{\left(x^2+(mx)^2\right)^3} = \frac{1+9 m^2-9 m^4-m^6}{\left(1+m^2\right)^3}$$

Hence, the limit depends on $m$ i.e. the direction in which you approach.

(Note that $\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial y} \right)$ is obtained by plugging in $m=0$ and $\frac{\partial }{\partial y} \left(\frac{\partial f}{\partial x} \right) $ is obtained by plugging in $m= \infty$)

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The function $f_{xy}$ that you wrote down (presumably by using various standard derivative formulas) is not continuous at $0$. As you asserted above, the limit as $(x,y) \to (0,0)$ depends on the path taken, so the limit does not exist.

Perhaps an analogous example from single-variable calculus will clear this up. The function $g(x) = |x|$ defined for all reals has derivative $ g'(x) = \dfrac {|x|} {x} = \left\{ \begin {array} {l l} 1 & x>0 \\ -1 & x<0 \end {array} \right. $

We can extend the domain of definition to include $x = 0$ by defining the value however we like, but the function will not be continuous.

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