Since the current through C_X is equal to −V_{out}C_Y s − I_{in}, we have V_X = −(V_{out}C_Y s + I_{in})/(C_X s), and the smallsignal drain current of M_2 is −g_{m2}(−V_{out}C_Y s − I_{in})/(C_X s). The current through r_{O2} is then equal to −V_{out}C_Y s − g_{m2}(V_{out}C_Y s + I_{in})/(C_X s). Noting that V_X plus the voltage drop across r_{O2} is equal to V_{out}, we write

−r_{O2} \left[(V_{out}C_Y s + I_{in})\frac{g_{m2}}{C_X s} + V_{out}C_Y s\right] − (V_{out}C_Y s + I_{in}) \frac{1}{C_X s} = V_{out}                                                    (6.82)

That is

\frac{V_{out}}{I_{in}}= \frac{−g_{m2}r_{O2} + 1}{C_X s} · \frac{1}{1 + (1 + g_{m2}r_{O2})\frac{C_Y}{C_X}+ C_Y r_{O2}s}                             (6.83)

which, for g_{m2}r_{O2} \gg 1  and  g_{m2}r_{O2}C_Y /C_X \gg 1 (i.e., C_Y > C_X ), reduces to

\frac{V_{out}}{I_{in}}≈ \frac{− g_{m2}}{C_X s} \frac{1}{\frac{C_Y}{C_X}g_{m2} + C_Y s}                                                    (6.84)

and hence

\frac{V_{out}}{V_{in}}=\frac{−g_{m1}g_{m2}}{C_YC_X s}\frac{1}{g_{m2}/C_X + s}                                                         (6.85)

The magnitude of the pole at node X is still given by g_{m2}/C_X . This is because at high frequencies (as we approach this pole), C_Y shunts the output node, dropping the gain and suppressing the Miller effect of r_{O2}.