Use the diode equation I = I_s * (e^(V/Vt) - 1) for what follows. Consider a diode with a bias current I and voltage V (these are related by the diode equation). Consider now adding a small signal voltage, v, across the diode which will in turn drive a small additional current, i. Show using differential calculus that the differential resistance relating these is given by r = (dV/di) or if the bias current is measured in mA, r = 1/(mA).