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Let $f(x,y) = \begin{cases} xy,\,\,\, |y| \leq |x| \\ -xy,\,\,\, |y| > |x| \end{cases}$

Here is what I've done.

$f^{\prime}_x = \cases{y,\,\,\,|y| \leq|x|\\-y,\,\,\,|y|>|x|}$ $f^{\prime}_y = \cases{x,\,\,\,|y| \leq|x|\\-x,\,\,\,|y|>|x|}$

$f^{\prime\prime}_{xy} = \cases{1,\,\,\,|y| \leq|x|\\-1,\,\,\,|y|>|x|}$ $f^{\prime\prime}_{yx} = \cases{1,\,\,\,|y| \leq|x|\\-1,\,\,\,|y|>|x|}$

And from here the answer seems yes, but the book says otherwise. Now, I think that it is right what I did in the first step and that the problem is in the second step, where I might be calculating the second order partial derivatives wrongly. Is it so? If yes, then what is the correct way of calculating partial derivatives of second order for such functions?

H-a-y-K
  • 661

1 Answers1

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First, let's compute $f_x'(0,y)$:

It's easy to see that $$ f(x,0)=0,\qquad \forall x.$$

Then $$ f_x'(0,0)=0.$$.

When $y\neq0$,it's easy to see that $$ f(x,y)=-xy,\qquad \forall |x|<|y|.$$

So we have $$ f_x'(0,y)=-y,\qquad \forall y\neq0. $$

Hence we have $$ f_x'(0,y)=-y,\qquad \forall y.$$

So we have $$ f_{xy}''(0,0)=-1.$$

Secondly, let's compute $f_y'(x,0)$.

It's easy to see that $$ f(0,y)=0,\qquad \forall y.$$

Then $$ f_y'(0,0)=0.$$

When $x\neq0$,we have $$ f(x,y)=xy,\qquad \forall |y|<|x|.$$

Then $$ f_y'(x,0)=x,\qquad \forall x\neq0.$$

Hence $$ f_y'(x,0)=x,\qquad \forall x.$$

So $$ f_{yx}''(0,0)=1.$$

Therefore, $$f_{xy}''(0,0)\neq f_{yx}''(0,0).$$

m-agag2016
  • 934
  • oh, so the overall idea is that the relation between $|y|$ and $|x|$ may not be the same for different orders of $x$ and $y$ when computing the second order derivatives? – H-a-y-K Aug 06 '21 at 16:12
  • @H-a-y-K Yes, you need to compute $f_{xy}''(0,0)$ and $f_{yx}''(0,0)$ only using the definition, independently. – m-agag2016 Aug 06 '21 at 22:59