what is the gain of the second stage?

Hi,this circuit is from the book "CMOS analog circuit design by Allen",as allen said,the gain of the second stage is (gm8+gm6)/2*RII where RII=(gm12*rds12*rds11)//(gm7*rds7*rds6),but the output resistance seen from the drain of M8 is (R1+1/gm10),it is relatively small,so the gain from M8 may be small,my question is how does allen's result comes out?Thanks!

Hi afujian,
We can look at it in 2 ways. First consider that M10/M11 and is just a cascoded current mirror. Whatever small signal current you generate from the gm of M8 will get mirrored out to the output. Hence the small signal current will get multiplied by the output resistance to give you vout. Considering you put +vin/2 on gate of M8 and -vin/2 on gate of M6 the total output current is vin*(gm8 + gm6)/2. That multiplied by the output resistance gives you the gain. Hence the result Allen gives.
If you want to look at the resistance on the drain of M8. That is small yes. But then whatever gain you get by multiplying the gm of M8 times the resistance on the drain you have to multiply with the gain from gate of M11 to out. Since M10 is a mirror the resistance on the drain would be roughly 1/gm and the gm gain of M11 will cancel out giving you again the gm8 times the output resistance.

Thanks,I get it.