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$$\frac{\partial f}{\partial x}(x+y)=\frac{\partial }{\partial x}f(x+y)$$

I was just wondering what the left-hand side mean. (or how to do the operation based on the notation of the LHS, given a specific function $f$)

In addition, when is such commutation true? Under what conditions?

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I think this is a definition of the LHS. They are telling you that the partial derivative of $f(x + y)$ with respect to $x$ is denoted by what is written on the LHS. Observe that $x + y$ is a function of two variables, while, I presume, $f$ is a function of one variable. So $f(x + y)$ becomes a function of two variables, and you're just taking the partial derivative of this function with respect to $x$; that is, $(x,y)\mapsto f(x + y)$. – William Feb 5 '12 at 21:49
My initial reaction after a little bit of thought is that the LHS is bad notation. – Hurkyl Feb 20 '13 at 8:51

If $f$ is some function, then we may define $f_y : x\mapsto f(x+y)$. It is basically a shifted version of the function $f$.

Your equality says that (with $D$ the derivation operator): $$ Df(x+y) = Df_{y}(x), $$ or, if you prefer $$ (Df)_y = D(f_y). $$ In other words: "shifting" or "translating" a function commutes with taking derivatives. If we denote by $T_y$ the operator $f\mapsto f_y$, then we could write this as $$T_y\circ D = D\circ T_y $$

In the language of systems one says that derivation is a time-invariant system. In my opinion it's also a good example of how the $\frac{\partial}{\partial x}$ can sometimes obscure things.

This is always true (if the derived function exists), because (intuitively) derivation only depends on a neighbourhood of the function, and not on the ordinate. In other words, if I give you a picture of a function but forget to draw the $y$-axis, you could still sketch the derived function. (This would be different if I asked you to, for instance, draw $xf(x)$.)

For a more rigorous derivation of this property: let $f$ be any (differentiable) function, then we have that

$$ \begin{align*} Df_y(x) &= \lim_{h\to 0}\frac{f_y(x+h)-f_y(h)}{h} \\ &= \lim_{h\to 0} \frac{f(x+h+y)-f(x+y)}{h} \end{align*}$$ and $$ \begin{align*} (Df)_y(x) &= Df(x+y)\\ &= (Df)(z)|_{z=x+y} \\ &= \lim_{h\to 0}\left.\frac{f(z+h)-f(z)}{h} \right|_{z=x+y} \\ &= \lim_{h\to 0} \frac{f(x+y+h)-f(x+y)}{h} \end{align*} $$

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Thanks, but I still cannot understand why "shifting a function commutes with taking derivatives" is "always true". Could you give me any more explanation on that? – John Conn Feb 6 '12 at 2:48
@John Conn: I have added a more mathematically rigorous explanation of the fact. – Myself Feb 6 '12 at 11:18

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