For the differential pair as shown in the figure, we have

Vout1 = Vdd - Rd1.Id1 and Vout2 = Vdd - Rd2.Id2

i.e.

Vout1 - Vout2 = Rd2.Id2 - Rd1.Id1 = Rd( Id2 - Id1 ) if Rd1 = Rd2 = Rd.

Thus, we simply calculate Id1 and Id2 in terms of Vin1 and Vin2, assuming the circuit is symmetric, M1 and M2 are saturated, and lambda = 0. Since the voltage at node P is equal to Vin1 - Vgs1 and Vin2 - Vgs2,

Therefore, Vin1 - Vin2 = Vgs1 - Vgs2.

For a square law device, we have:

$(V_{GS}-V_{TH})^2 = \frac{I_D}{\frac{1}{2}\mu_nC_{ox}\frac{W}{L}}$

and, therefore,

$V_{GS}=\sqrt{\frac{2I_D}{\mu_nC_{ox}\frac{W}{L}}}+V_{TH}$

Substituting, we get,

$V_{in1}-V_{in2}=\sqrt{\frac{2I_{D1}}{\mu_nC_{ox}\frac{W}{L}}}-\sqrt{\frac{2I_{D2}}{\mu_nC_{ox}\frac{W}{L}}}$

Our objective is to find differential output current, Id1 - Id2. Squaring the two sides and substituting Id1 + Id2 = Iss, we get,

$(V_{in1}-V_{in2})^2 = \frac{2}{\mu_nC_{ox}\frac{W}{L}}(I_{SS}-2\sqrt{I_{D1}I_{D2}})$

That is,

$\frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{in1}-V_{in2})^2 - I_{SS}=-2\sqrt{I_{D1}I_{D2}}$

Squaring the two sides again and noting that

$4I_{D1}I_{D2} = (I_{D1}+I_{D2})^2-(I_{D1}+I_{D2})^2=I_{SS}^2-(I_{D1}-I_{D2})^2$

we arrive at

$(I_{D1}-I_{D2})^2 = -\frac{1}{4}(\mu_nC_{ox}\frac{W}{L})^2(V_{in1}-V_{in2})^4+I_{SS}\mu_nC_{ox}\frac{W}{L}(V_{in1}-V_{in2})^2$

Thus,

$I_{D1}-I_{D2}=\frac{1}{2}\mu_nC_{ox}\frac{W}{L}(V_{in1}-V_{in2})\sqrt{\frac{4I_{SS}}{\mu_nC_{ox}\frac{W}{L}}-(V_{in1}-V_{in2})^2}$

Denoting, Id1 - Id2 = $\Delta I_D$

and Vin1 - Vin2 = $\Delta V_{in}$

Dividing, we get

$\frac{\delta\Delta I_D}{\delta\Delta I_in}=\frac{1}{2}\mu_nC_{ox}\frac{W}{L}\frac{\frac{4I_{SS}}{\mu_n C_{ox}\frac{W}{L}}-2\Delta V_{in}^2}{\sqrt{\frac{4I_{SS}}{\mu_nC_{ox}\frac{W}{L}}}-\Delta V_{in}^2}$

For, $\Delta V_{in} = 0$

$G_m = \sqrt{\mu_nC_{ox}\frac{W}{L}I_{SS}}$

Moreover, since $V_{out1}-V_{out2} = R_D\Delta I = R_D G_m \Delta V_{in}$

we can write the small signal differential voltage gain of the circuit in the equilibrium condition as

$|A_v| = \sqrt{\mu_nC_{ox}\frac{W}{L}I_{SS}R_D}$