# If $f(x,y)=f(y,x)$ then $\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(y,x)}{\partial y}$

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.

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

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 $h$.
Not true at all. For example, try $f(x,y) = xy$.
• As I read it, the LHS is taken at the point $(x,y)$, the RHS at the point $(y,x)$. Feb 8, 2017 at 15:32
• I do not think you are right... Here $(x,y)$ is the point where we compute the derivative, not the variables... Feb 8, 2017 at 15:33