How do you define the derivative of a function without an argument?

Question

So the derivative of $f: x\mapsto f(x)$ is defined by $f':x \mapsto \lim_{h\to0}\dfrac{f(x+h)-f(x)}{\phantom{f}(x+h)\,-\,(x)}$. But is there a way to define $f'$ solely in terms of $f$, without explicit reference to $x$?

Answer 1

Here's an attempt. For any $h \in \mathbb{R}$, we define a function $\tau_h$ by $\tau_h(x)=x+h$. (Think of $\tau_h$ as "translation by $h$", if you like.) (Edit: If you don't like the explicit use of variables in the definition of $\tau_h$, you can define $\tau_h = id +\hat{h}$, where $id$ denotes the identity function and $\hat{h}$ denotes the constant function with value $h$.) Now for any function $f$ and any real $h$ we can define a difference quotient as $$\Delta_h(f)=\frac{ f \circ \tau_h - f}{h}$$

Note that $\Delta_h(f)$ is a function defined without explicit reference to variables (although it does of course require an explicit reference to the parameter $h$).

Now to finish the story we can define $f' = \lim_{h \to 0} \Delta_h(f)$, where the limit here is understood in the sense of pointwise-convergence.

Edited to add: One nice advantage of this formulation is that it lends itself naturally to generalizations. Let $\sigma = \{\sigma_h\}_{h \in \mathbb{R}}$ be any parametrized family of functions $\sigma_h: \mathbb{R} \to \mathbb{R}$. Then we can define difference quotients relative to $\sigma_h$ as $$\Delta_h^{\sigma}(f)=\frac{ f \circ \sigma_h - f}{h}$$ and then define the "derivative of $f$ relative to $\sigma$" as $f^{\sigma} = \lim_{h \to 0} \Delta_h^{\sigma}(f)$. In this notation, the usual derivative $f'$ is just $f^{\tau}$, where $\tau = \{\tau_h\}_{h \in \mathbb{R}} $ is the family of translations defined earlier.

Answer 2

Here's another (purely algebraic) attempt, taking a different approach entirely.

Begin by choosing some collection $\mathfrak{F}$ of well-behaved functions (for example, all polynomials with real coefficients, or all functions that can be represented by Taylor series). Then we define a derivative on $\mathfrak{F}$ to be a linear map $\Delta:\mathfrak{F} \to \mathfrak{F}$ satisfying $$\Delta(id) = \hat{1}$$ $$\Delta(fg) = f \Delta(g) + g \Delta(f)$$

(here $id$ denotes the identity function and $\hat{1}$ denotes the constant function $\hat{1}: x \mapsto 1$).

It is not too hard to show that these conditions force $\Delta$ to behave as desired for all polynomials. If we also add some additional conditions about convergence, then we can also prove that $\Delta$ behaves as desired for "friendly" non-polynomial functions like trigonometric and exponential functions (and really any function that can be represented by a Taylor series). I am not sure how far this can be pushed; in particular, I'm not sure if there's a way to make sure that $\Delta$ behaves as desired for a non-analytic but smooth function.

Answer 3

Actually if $g$ is the derivative of $f$, they are related by $$ g = f' $$ where $'$ is interpreted as the operation of taking the derivative, while you seem to use $f'$ as a different name for $g$ primary.

That operator view style shows up in rules like $$ (fg)' = f'g + fg' $$ and so on.

Depending on the properties of $f$ one might think of many ways to define the differentiation operation by its action on the representation of $f$, independent of a specific argument variable, like:

$$ f = \sin \Rightarrow f' = \cos \\ f = \mbox{id}^n \Rightarrow f' = n\, \mbox{id}^{n-1} \\ f = (a_n) \Rightarrow f' = ( (n+1) a_{n+1} ) \\ f' = \mathcal{F}^{-1} \left\{ 2\pi i\, \mbox{id } \mathcal{F}\, \left\{ f \right\} \right\} \\ $$

Answer 4

Note your definition of $f'$ has an argument-you wrote it as $f'(x)$, which is correct. You can apply the same definition: $$f''(x) = \lim_{h\to0}\dfrac{f'(x+h)-f'(x)}{\phantom{f}(x+h)\,-\,(x)}$$ If you want it in terms of $f$, you can substitute in and have $$f''(x) = \lim_{h\to0}\dfrac{f(x+2h)-2f(x+h)+f(x)}{\phantom{f}((x+h)\,-\,(x))^2}$$