How to scale a device for the second derivative of Amps/Volt?
First of all, you said that for a device that is twice the size of the original device, then the current is twice the amount for the same voltage. That is simply a definition of your choosing. (For example, for many things, twice as big would be defined as twice the mass.)

Given that as the definition:
\$ i_1 = f(v) \$ for device1
if device2 is twice the size of device1 then \$ i_2 = 2f(v) \$

So the second derivative of \$ i_1 \$ with respect to \$ v \$ is
\$ i_1^" = \frac{d^2 f(v)}{dv^2} \$

And the second derivative of \$ i_2 \$ with respect to \$ v \$ is
\$ i_2^" = 2\frac{d^2 f(v)}{dv^2} \$

Therefore, the scaling is the same, because derivative is a linear operator.