[svantana]

Is it just me or this article pretty naive? The headline's use of the word "exact" would imply integer arithmetic only, but the computations are done with floating point. So basically (s)he is trading one rounding error for another, which seems to be small-ish in some particular cases. What about discontinuities? And why forward derivatives only? I hope noone will use this for any application that actually relies on exact derivatives.

[Leszek]

They're exact in the sense that they give the same value as if you had calculated the value of the analytic derivative. This is different from numerical differentiation, which approximates the derivative with finite differences.