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.

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 $m= \infty$)


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.