To acquire some sense of *differential* calculus over fields other than the standard real or complex (topological) fields, one might start by getting acquainted with this paper or other similar contributions of the same authors. Every topological field can be considered as a "topologized" module over the discrete topological ring $\mathbb Z$ . So it is meaningfull to speak of differentiability and smoothness of maps between any topological fields in the BGN−sense. However, then the important *determination axiom* (see iii, p. 8 in the referred paper) is not satisfied, and so there need not be any uniquely determined derivatives.

The basic definition goes as follows. Consider a map $f:E\supseteq U\to F$ where $E,F$ are topologized modules over the topological ring $\boldsymbol R$ and $U$ is open in $E$ . Writing $g^{[0]}=f$ and $U^{[0]}=U$ and $E^{[0]}=E$ , recursively define $U^{[i+1]}=(U^{[i]})^{[1]}$ where $V^{[1]}=\{(w,z,t):w,w+tz\in V\}$ , and $E^{[i+1]}=(E^{[i]})^{[1]}$ where $G^{[1]}=G\times G\times\boldsymbol R$ and the operation "$\times$" refers to the particular topologized module product inherent in the BGN−axioms. Then $f$ is smooth if there exists a sequence $g^{[i]}:E^{[i]}\supseteq U^{[i]}\to F$ of maps satisfying the property that $g^{[i]}(w+tz)=g^{[i]}(w)+tg^{[i+1]}(w,z,t)$ holds for $(w,z,t)\in U^{[i+1]}$ . The maps $g^{[i]}$ need not be unique when the determination axiom is not satisfied.

notthe same as continuous. – Kevin Ventullo May 7 '11 at 23:27