Differential resistance vs. $R = \frac{U}{I}$ Resistance is defined as $R = \frac{U}{I}$. If Ohm's law holds, $U(I)$ is a linear function and the slope of the curve is equal to the resistance $R$.
I realized only recently that if $U(I)$ is a non-linear function, the slope of the function is no longer equal to the resistance as defined above. We then have two kinds of resistances:


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*the differential resistance $r(I) = \frac{d U}{d I} (I)$ (which is equal to the slope of the curve at the point $I$)

*"ordinary" resistance $R(I) = \frac{U(I)}{I}$.
My questions:


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*Is the quotient $R(I) = \frac{U(I)}{I}$ used to describe devices with non-linear current-voltage characteristics at all? (I am aware of the fact that $R$ isn't a property of the device in this case but it may still be a sensible concept)

*If both $R(I)$ and $r(I)$ are used, when do I use which one?

*What can we say about the relationship of the two quantities in general?


Example: Let's have a look at the current-voltage characteristic of a tunnel diode (see image below, taken from Wikipedia). I am not really familiar with how this device operates but it illustrates my question. For "ordinary resistance", we have $R(i_1) = \frac{v_1}{i_1} \ll R(i_2) = \frac{v_2}{i_2} $ while for the differential resistance we have $r(i_1) = r(i_2) = 0$ (/edit: As freecharly noted, this is wrong. Actually, we have $r(i_1) = r(i_2) = \infty)$
So there's some justification to say that if we apply the constant voltage $v_2$ that the resistance is bigger than the resistance if we apply the constant voltage $v_1$, and there's also some justification to say that the resistance is zero infinity in both cases. How is the termonology actually used?

/edit: I edited the phrasing of the question and included an example in order to make things more clear.
 A: For any current-voltage characteristic   you can always formally define a resistance $$R=\frac {V}{I}$$ or conductance $G=I/V=1/R$ at any point of the characteristic. The differential resistance $r$ and differential conductance $g$ are defined by $$g=\frac {1}{r}=\frac {dI}{dV}$$  In the tunnel diode characteristic, you have a region of negative differential resistance (or conductance) between the peak and valley voltages $v1$ and $v2$, which is useful for microwave frequency oscillators and other applications. The differential conductance $g=dI/dV$ is zero at the peak and the valley of the $I-V$ characteristic.Thus the differential resistance $r=1/g$ is infinite at these extrema, not zero! At $V=v1$ the resistance $R=V/I$ is smaller than the resistance at $V=v2$.
A: An example. Diodes have a non-linear current-voltage characteristic. In the ideal case:
$$ I = I_0(e^\frac{qV}{kT}-1).$$
When there is a current in the forward direction, one can neglect the unity. So $\frac{1}{r} = \frac{dI}{dV} = \frac{q}{kT}I$. The dynamical resistance $r = \frac{kT}{q}/I \approx \frac{0.025\ {\rm volt}}{I}$ or $25/I$ ohm when current is given in mA.
This is then the impedance for small ac current $i$ when superposed on a DC current $I$ through the diode. The small-signal voltage $v = i r$. 
