This derivation was adapted from the book "Tunnel Diodes" by Sylvester P. Gentile Appendix A:
The variable \$ R \$ is the load. The variable r is the differential negative resistance of the diode.
Take the change of voltage source \$ V \$ and place it across the load:
$$ V = i_1R $$
This is equation 1.
Then take the negative differential resistance and place it in series with the load. Then:
$$ V = i_2(r+R) $$
This is equation 2.
Now divide equation 1 by equation 2
$$ 1 = \frac{R}{r + R} ( \frac{i_1}{i_2} ) $$
Solve for \$ \frac{i_1}{i_2} \$
$$ \frac{i_2}{i_1} = \frac{R}{r + R} $$
Let
$$ v_{in} = i_1R $$
This is equation 3.
$$ v_{out} = i_2R $$
This is equation 4.
Divide equation 4 by equation 3:
$$ \frac{v_{out}}{v_{in}} = \frac{i_2}{i_1} $$
then
$$ \frac{v_{out}}{v_{in}} = \frac{R}{R+r}$$
Multiply both sides by \$ v_{in} \$
Then:
$$ v_{out} = \frac{R}{R+r} (v_{in}) $$