3
$\begingroup$

Why is it that if $f(x,y)=f(y,x)$ then $\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(y,x)}{\partial y}$ for all $x,y$ in $\Bbb R^2$. My lecturer just went over it like it was obvious but I cant seem to come up with a proof with why it is so. I thought maybe starting from the limit definition of a partial derivative would get me somewhere then I could jumble it till they were equal but this got me nowhere.

Any help would be very much appreciated.

$\endgroup$
1
  • $\begingroup$ Perhaps when we derivate $f(x,y)$ about first coordinate, it's the same which we derivate about first coordinate of $f(y,x)$ $\endgroup$
    – Nosrati
    Feb 8, 2017 at 15:29

2 Answers 2

2
$\begingroup$

The formula is correct as long as you interpret it as $$ \partial_1 f(x,y) = \partial_2 f(y,x). $$ Indeed, $$ \begin{align} f(x,y) & =f(y,x) \\ f(x+h,y) &= f(y,x+h). \end{align} $$ Now subtract and divide by $h$.

$\endgroup$
1
  • $\begingroup$ Yes, that interpretation question is key here. This formula is a bit of sleight of hand. $\endgroup$
    – Paul
    Feb 8, 2017 at 15:34
0
$\begingroup$

Not true at all. For example, try $f(x,y) = xy$.

$\endgroup$
2
  • 1
    $\begingroup$ As I read it, the LHS is taken at the point $(x,y)$, the RHS at the point $(y,x)$. $\endgroup$
    – Wojowu
    Feb 8, 2017 at 15:32
  • 1
    $\begingroup$ I do not think you are right... Here $(x,y)$ is the point where we compute the derivative, not the variables... $\endgroup$
    – Siminore
    Feb 8, 2017 at 15:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .