Verified answer

The incremental resistance (R=dV/di) changes as a functon current. This is because I=Is e-(Vd/(n Vt)).

Depends on the diode. An "ideal" diode (one that follows the ideal diode equation), the resistance will be about (kT/q)/I, where I is the current through the diode, q is the electron charge (constant) and kT is the thermal energy (k is Boltzmann's constant, and T is the temperature in absolute scale). Put in more current, get a lower diode resistance. At standard temperature, kT/q is about 0.026 volts (so a voltage divided by current is a resistance).

Real diodes are not ideal. They have some resistance other than the variation in the I-V curve as described above, but they can be modeled as an ideal diode in series with some contact/bulk resistance.

The forward voltage tends to remain constant.

no, it varies, and it varies logarithmically.

For example, a typical signal diode, from the 1N4148 data sheet:

at 5 mA, R = 130 ohms

at 10 mA, R = 80 ohms

at 100 mA, R = 10 ohms

yes, except in the case of Zenier diodes.