Rules of calculus for non-complete ordered fields Suppose that $K$ is a totally ordered field. We can express the properties of continuity and differentiability just as in the real numbers. Now suppose we have a function $f: K \to K$. I was wondering whether normal rules of calculus apply. In particular I am interested in the chain rule and the fundamental theorem of calculus, that is, the following two statements:


*

*Suppose that $f$ and $g$ are differentiable in $g(x)$ and $x$ respectively. Is it then true that $(f(g(x)))' = f'(g(x))g'(x)$?

*Suppose that $f$ is differentiable on $K$. Is it true that $\int_{a}^b f'(x) \mathrm{d}x = f(b)-f(a)$? Does it even make sense to define integrals of continuous functions, e.g. as the limit of the sums of the area below lower staircase function?
As far as I remember, the proof for the chain rule can directly be inferred from the definitions of the derivative. I am less sure about integrals, as we might need the existence of certain limits.
Intuitively my question arises from the fact that a totally ordered field always contains a copy of the rational numbers. Since these are dense in the real numbers, continuous functions on $K$ cannot have any "jumps". In other words, between any two elements of $K$ there are infinitely other elements of $K$, so continuity and differentiability should have similar properties as in $\mathbb{R}$.
Obviously statements like the intermediate value theorem, which heavily rely on completeness, fail in general ordered fields.
 A: The classical proof of $1)$ in $\mathbb{R}$ applies in any ordered field. As for integration, the limit you speak of need not exist if $K$ doesn't have the least upper bound property so you can't define the integral for every continuous function. Actually I am not sure whether the existence of primitives for every continuous map implies the least upper bound property, but to answer your question by the netagive, observe that $x > 0 \mapsto \frac{1}{x}$ does not have a primitive over the ordered field of rationnal numbers.
Continuous functions on $K$ can have jumps. Actually, continuous functions not having jumps implies that $K$ is isomorphic to $\mathbb{R}$. If $K$ isn't, then there is a non-empty bounded initial subset $I$ of $K$ ($^*$) (initial subset: a subset such that $\forall x \leq y, y \in I \rightarrow x \in I$) with no upper bound. Define $f(x) = -1$ for $x \in I$, and $f(x) = 1$ for $x \notin I$, and see that $f$ is continuous. 
($^*$): Take a non empty bounded set $A$ without a least upper bound and define $I:= \{x \in K \ | \ \exists a \in A, x \leq a\}$.
