# Are the derivatives of symmetric functions symmetric?

Suppose $f(x,y)=f(y,x)$. Does it follow that $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$?

Intuitively it seems like it must, because taking a "step" in the $x$ direction must be the same as taking one in the $y$. But when I try to prove it I get lost as to which is "really" x or y.

EDIT: I should have been more clear. What I meant was: does $\frac{\partial f(x,y)}{\partial x}=\frac{\partial f(y,x)}{\partial y}$?

(At some intuitive level this is like doing a "find and replace" s/x/y/, but my intuition fails when taking the derivative.)

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Isn't $x^2+y^2$ a counterexample? $2x \neq 2y$. Maybe you mean that if you switch $x$ and $y$ in one partial derivative you get the other? –  only Sep 7 '12 at 22:53

Consider $f(x,y)=xy$. Are $f_x$ and $f_y$ equal?

What is true is that if you interchange $x$ and $y$ in $f_x(x,y)$, you’ll get $f_y(x,y)$.

Added: To be absolutely clear, by ‘interchange $x$ and $y$’ I mean ‘replace each $x$ by a $y$ and each $y$ by an $x$’.

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I think he means $\frac{\partial f}{\partial x}|_{y:=x}=\frac{\partial f}{\partial y}$ but $f(x,y)=x^2+y^2$ doesnt satisfy this. Perhaps $\frac{\partial f}{\partial x}|_{y:=x}=\frac{\partial f}{\partial y}$ OR $\frac{\partial f}{\partial x}|_{x:=y}=\frac{\partial f}{\partial y}$ is satisfied could be something.. –  Seyhmus Güngören Sep 7 '12 at 23:09
@SeyhmusGüngören: you need to interchange the variables, just setting one to the other isn't enough. –  Ben Millwood Sep 7 '12 at 23:20
@BenMillwood Yes I think your conclusion seems correct. E.g.: $2x^2y+2y^2x$ is hopeless according to what I wrote. –  Seyhmus Güngören Sep 7 '12 at 23:37
@Brian It was clear though. –  Seyhmus Güngören Sep 7 '12 at 23:38
@Brian: Thanks, your second one was what I wondered. ($f_x(x,y)=f_y(y,x)$). Do you have a proof? –  Xodarap Sep 8 '12 at 0:33

I believe that if we use a different notation for the variables and the arguments, it will be easier to see.

If $f(a,b)=f(b,a)$, then $$f_x(a,b)=lim_{h\to0}(f(a+h,b)-f(a,b))/(h)\\ =lim_{h\to0}(f(b,a+h)-f(b,a))/(h) =f_y(b,a)$$

The same process could be done to obtain $f_{xx}(a,b)=f_{yy}(b,a)$.

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