How does one make real functions a differentiable field? If you want to apply the results of differential field theory to actual $\Bbb R\to\Bbb R$ functions, then first of all you have to find operations that make these functions a field. The trouble is that with the standard definition of function multiplication, many functions don't have inverses. You can't really say that the inverse of $x$ is $1/x$, because strictly speaking $x\cdot(1/x)$ is only defined on $\Bbb R - \{0\}$.
I imagine the answer is to define multiplication as first multiplying in the traditional sense, and then completing by continuity, but I can't quite work out the details, and either way, I'd like to know what the conventional way of doing it is.


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*Exactly what set of real functions are usually treated as differential fields? The set of differentiable functions defined on all but a set of isolated points of $\Bbb R$? The set of differentiable functions defined on a set dense in $\Bbb R$? What field might we work with if we were trying to prove Liouville's theorem?

*How is multiplication defined on that (those) field(s)?

 A: There is no natural way in which the set of all differentiable (or even smooth) functions $f: \mathbb{R} \rightarrow \mathbb{R}$ forms a differential field.  The problem is that under the natural operations of pointwise addition and mutiplication the set of smooth functions $f: \mathbb{R} \rightarrow \mathbb{R}$ is a ring with many, many zero divisors.
In order to get even an integral domain you need to impose some kind of rigidity property: that a function which is zero on some large subset of the domain is identically zero.  
For this it is good to restrict to the set of real analytic functions $f: I \rightarrow \mathbb{R}$.  By the Principle of Analytic Continuation these functions form a domain.  The fraction field is (mostly by definition) the field of meromorphic functions on $I$.  Note that there is something slightly formal going on here: a meromorphic function on $I$ is not necessarily a function on all of $I$.  It is a function on the complement of a set of isolated points of $I$.  There is also a natural equivalence relation in play here so as to ensure that e.g. $x^2/x$ is the same element as $x$.  All of this seems very necessary in order to define a differential field "of functions"; every such field I've ever seen works in this way.
The field $M(I)$ of meromorphic functions on $I$ is a differential field under differentiation, and is probably the most important example of such a field (of real functions; I think you'll find eventually that it's simpler to consider complex functions first and then think of the real functions inside the complex functions in Galois-theoretic terms).  It is also a very large field.  To build smaller fields one can just choose a subset $S$ of $M(I)$ and build the differential field generated by $S$: namely you take the field generated by $S$ (you could do this either as a naked field or as an $\mathbb{R}$-algebra), then you adjoin the derivatives of all those elements, then you take the field generated by that, then you adjoin the derivatives....You alternate this countably many times and get the differential subfield generated by $S$. 
A: Rational functions in $\Bbb{R}$ as you describe them would form a field if you extended $\Bbb{R}$ to also include the point at infinity. That way, the function $1/x$ is defined everywhere on $\Bbb{R}\cup\{\infty\}$, and has values in $\Bbb{R}\cup\{\infty\}$. This is the same way in which rational functions defined over complex numbers are extended to the Riemann sphere.
Completing the function $1/x$ by continuity cannot work because $1/x$ has no limit at $0$ (unless you think of that limit as $\infty$, in which case it does).
Also, quite often rational functions can be defined as formal constructs, which then later may or may not be evaluated at specific points. Thus you would call $p(x)/q(x)$ a rational function for any polynomials $p,q$ ($q$ not the zero polynomial), and $q(x)/p(x)$ would always be its inverse. You can talk about its derivative $(p'q-pq')/q^2$ even if there are real points where the function is undefined.
The field you are looking for is called the field of rational functions on $\Bbb{R}$.
Edit. In the context of Liouville's theorem, if you have transcendental or algebraic functions $\theta_1,\ldots,\theta_n$ also, then you can consider formal rational functions in $x,\theta_1,\ldots,\theta_n$. The field being closed under differentiation would mean that you can express $\theta_k'$ as a rational function in $x,\theta_1,\ldots,\theta_n$. This also means you can't have a completely "black box" function, because even then you need to be able to differentiate it and get back an element of the same field.
Usually this is used over complex numbers, not real, to avoid questions of where the functions are defined.
Here is a set of lecture notes that should explain this in more detail.
