Let us set Vin2 to zero and find the effect of Vin1 at X and Y in figure (a). To obtain Vx, we note that M1 forms a common-source stage with a degeneration resistance equal to the impedance seen looking into the source of M2 in figure (b). Neglecting channel length modulation and body effect, we have Rs = 1/gm2 and

$\frac{V_x}{V_{in1}}=\frac{-R_D}{\frac{1}{g_{m1}}+\frac{1}{g_{m2}}}$

To calculate Vy, we note that M1 drives M2 as a source follower and replace Vin1 and M1 by a Thevenin equivalent.

The Thevenin voltage Vt = Vin1 and the resistance Rt = 1/gm1. Here, M2 operates as a common gate stage exhibiting a gain equal to

$\frac{V_y}{V_{in1}}=\frac{R_D}{\frac{1}{g_{m1}}+\frac{1}{g_{m2}}}$

From both the equations, the overall voltage gain for Vin1 is

which, for gm1 = gm2 = gm reduces to

$(V_x-V_y)|_{Due to V_{in1}}=-g_mR_DV_{in1}$

By the virtue of symmetry, the effect of Vin2 at X and Y is identical to that of Vin1 except for a change in the polarities:

$(V_x-V_y)|_{Due to V_{in2}}=g_mR_DV_{in2}$

Adding the two equations to perform superposition, we have

$\frac{(V_x-V_y)_{io1}}{V_{in1}-V_{in2}} = -g_mR_D$

The above equations indicate that the magnitude of the differential gain is equal to gmRd regardless of how the inputs are applied.