To understand dynamic resistance one must realise that as EE's we go to great extremes
to pretend all of our problems are linear in nature so we can use the same old techniques
we have allways used to solve them.  This step to linearise problems is a Very important
concept in many areas of engineering and physics.  Without linearisation the problems are
sometimes just to hard to solve in a reasonable amount of time.

Almost all basic linearisations use a Taylor series as their basis.  Assuming you have
some nonlinear function f(x) the Taylor series can be found as follows.

This is the expression for a Taylor series approximation close to some point x=a.  The
more terms that are used in the approximation the more accurate it is.  If this approximation
is taken to extremes (n=infinity) then it becomes exact.  We tend to use only the first
two terms of the approximation and drop the rest.  Because of this this the approximation
is only valid in a very narrow region.  Typically this is called the Q point.  In the above
equation it would be point 'a'.

In general this Q point can be found by looking at the circuit considering
only the DC voltages and currents in the system.  (The why of this will hopefully
become obvious in a minute.)

For the above circuit we would pretend the AC source didn't exist.  In the
case of a AC voltage source we would simply erase it from the diagram.  In the case
of an AC current source we would replace it with a wire.  This leaves.

note:  Since R2 would only be connected on one side we ignore it also.

From our experiments in lab we know that the diode has a non-linear characteristic.
We also know what that characteristic is.  We sketched a picture of it on the curve tracer.
From Pspice I get the following reverse bias characteristic of the diode.

note:  (I manipulated the signs as necessary to get the relevant graph in the first quadrant.)

I have also drawn the load line on the graph.  The point of intersection is the Q point for the circuit.  The
load line combined with the diode characteristic is nothing more than a graphical method for solving two
equations simultaneously.  This method will work for finding the operation point (Q point) for any two
terminal device you know the characteristic of.  To plot the load line we essentially ignore the diode and
ask, "What does the voltage in the circuit at the diode terminals have to be at two different currents?".
These two points give us the ability to construct a line representing how the circuit would behave with
different loads.  (The rest of the circuit is linear.)  The easiest two points are 0 current which gives us the
supply voltage and maximum current (0 resistance load) which is 20V/220ohm=90.09mA.
 
From the taylor series expansion we realise we need the value of the derivative is at the q point.  Since
the derivitive is nothing more than the instantaneous slope of the line we simply draw a line tangent to the
diode characteristic at the q point and measure the slope.  Zooming in on the graph and drawing a tangent
we get.

For the slope we get

note:  This Rd is for the zener diode in Pspice at a particular operation point.  Your results may differ
somewhat.

Slope is typically measured as I did it.  (y=mx+b).  Dynamic resistance has units of ohm so it is necessary
to find the inverse of the slope.  With this we can approximate the diode characteristic with the
following Taylor series expansion.

I(V)=69mA+V*slope

or using Rd

I(V)=69mA+V/Rd

All this has given us is the tangent line in equation form.

(Partially zoomed out diode characteristic illustrating validity of approximation)

Looking at the diode characteristic above we can see that this equation (the tangent line)
is only a Approximation of the real diode characteristic and only valid near the
Q point.  The AC sources in the circuit are generally assumed to be 'small signals'.  This means that
for our calculations the approximation is close enough.  If we wanted a better approximation we could
add more terms to the Taylor expansion and make the approximation better.  (note: We will not go beyond
this level approximation in this class.)

At this point someone is undoubtably wondering  why finding the AC voltage and current through a diode
allows us to find the dynamic resistance (Rd).  This can be explained reasonably simply.  We start by splitting
the current and voltage though the diode into DC terms and AC terms.