Transconductance of a saturated NMOS from its bias point

Step-by-step derivation

1. Start from the square-law drain current in saturation: $$I_D = \tfrac{1}{2}\,\mu_n C_{ox}\frac{W}{L}\,(V_{GS}-V_{th})^2 = \tfrac{1}{2}k_n V_{OV}^2 .$$
2. Transconductance is defined as the small-signal sensitivity of drain current to gate voltage: $$g_m = \left.\frac{\partial I_D}{\partial V_{GS}}\right|_{Q}.$$
3. Differentiate the square law with respect to $V_{GS}$, noting that $V_{th}$ is constant so $\partial V_{OV}/\partial V_{GS} = 1$: $$g_m = \tfrac{1}{2}k_n \cdot 2V_{OV} = k_n V_{OV}.$$
4. Eliminate the unknown $k_n$ using the bias equation. From $I_D = \tfrac{1}{2}k_n V_{OV}^2$ we get $k_n = 2I_D/V_{OV}^2$, so $$g_m = \frac{2I_D}{V_{OV}^2}\cdot V_{OV} = \frac{2I_D}{V_{OV}}.$$
5. Substitute the given bias point: $$g_m = \frac{2 \times 1\ \text{mA}}{0.5\ \text{V}} = \mathbf{4\ mA/V}.$$
6. Contrast with the BJT. There $g_m = I_C/V_T$, linear in current. For the MOSFET $g_m \propto \sqrt{I_D}$ at fixed $W/L$, so doubling the current only improves $g_m$ by $\sqrt{2}$ - the reason MOS amplifiers need far more bias current for the same gain.