How do you define the derivative of a function without an argument? 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 $x$?
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.
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 $$f''(x) = \lim_{h\to0}\dfrac{f(x+2h)-2f(x+h)+f(x)}{\phantom{f}((x+h)\,-\,(x))^2}$$