Can the derivative be defined game-theoretically? It is often said, by way of intuitive explanation, that the derivative of a function at a point is the slope of the line that “best fits” the function through that point. Can this be pressed into a mathematically-rigorous definition, somewhat along the lines of the prisoners’ dilemma?

That is, two parties, who cannot communicate with each other, are given the same function and the same point, and told that if they both come up with the same line through that point that is best relative to the function, they will be set free, otherwise they will both have to serve time.

Would it then be some kind of theorem in Probability/Logic/Game Theory, that they would, if acting rationally and intelligently, come up with the tangent line “almost surely”?

Regards,

Mike Jones

2.Jun.2011 (Beijing time)
As I stated in my comment I don't think your solution quite works. However, I think I just thought of a nice game-theoretical way of defining the derivative of a function:

Two players are both given a function $f$ and a point $(x,f(x))$. Both players then independently come up with a line through $(x,f(x))$, say $f_1$ and $f_2$.

Now the players announce which line they chose and both come up with some $\epsilon&gt;0$ and the minimum of these is selected.

Then the final stage of our game begins: if $|f(x+\epsilon) - f_1(x+\epsilon)| = |f(x+\epsilon) - f_2(x+\epsilon)|$, the game ties and nothing happens. If $|f(x+\epsilon) - f_1(x+\epsilon)| &lt; |f(x+\epsilon) - f_2(x+\epsilon)|$ player 1 wins and player 2 loses. If $|f(x+\epsilon) - f_1(x+\epsilon)| &gt; |f(x+\epsilon) - f_2(x+\epsilon)|$ player 2 wins and player 1 loses.

To spice the game up a little one should of course introduce prison sentences or death penalties for the loser at this point, but I will leave that as an exercise to the reader.

Any proper game theorist would now wonder about the optimal strategy for both players, and unless I am gravely mistaken that would be choosing the line corresponding to the right-derivative of $f$ at $(x,f(x))$.
You've only talked about the 1st derivative of a function, and the geometric way of looking at "the derivative" as a tangent line only works for the 1st derivative.  It does come as possible to talk about the 2nd, 2.5th, pi-th, the -1th derivative (the integral), and other derivatives via [fractional calculus].1  I don't see how a game-theoretic definition would allow for anything else than the 1st derivative.