CS stage with resistive load.
$-V_{in} \lt V_{Th}$ -> $M_1- OFF$

$\therefore\, V_{out}=V_{DD}$

As, $V_{in}$ increases to $V_{Th},\,\,M_1$ begins to turn ON.

$V_{out}=V_{DD}-I_D\,R_D$

$V_{out}=V_{DD}-R_D(\frac{1}{2}\mu_n\,C_{ox} \frac{W}{L}(V_{in}-V_{Th})^2 )$

Differentiating w.r.t $V_{in}$, we get,

$A_v=\frac{\partial V_{out}}{\partial V_{in}}=0-R_D(\frac{1}{2}\mu_n\,C_{ox}\,2\, \frac{W}{L}(V_{in}-V_{Th})\,(1)) $

$A_v=-gmR_D \hspace{3cm}$

$.....gm=\mu_n\,C_{ox}\frac{W}{L}(V_{in}-V_{Th}) $

$A_v$ can be increase => increase W/L or decrease $I_D$. …