Q.Derive the relation for g_m in terms of V_GS and V_P.

Transconductance is obtained by differentiating Shockley's equation, which gives its dependence on the gate voltage and the pinch off voltage.

[!ANSWER]

Differentiating $I_D = I_{DSS}\left(1 - \dfrac{V_{GS}}{V_P}\right)^2$ with respect to $V_{GS}$ gives $g_m = -\dfrac{2 I_{DSS}}{V_P}\left(1 - \dfrac{V_{GS}}{V_P}\right)$, i.e. $g_m = g_{mo}\left(1 - \dfrac{V_{GS}}{V_P}\right)$ with $g_{mo} = -\dfrac{2 I_{DSS}}{V_P}$.

[!TLDR]

Differentiating Shockley's equation gives $g_m = g_{mo}\left(1 - \tfrac{V_{GS}}{V_P}\right)$, where $g_{mo} = -\tfrac{2 I_{DSS}}{V_P}$.

Transconductance is defined as the rate of change of drain current with gate-to-source voltage,

The drain current follows Shockley's equation,

Differentiating with respect to $V_{GS}$ (treating $I_{DSS}$ and $V_P$ as constants) and using the chain rule,

The factor in front is the transconductance at zero gate bias; denoting it $g_{mo} = -\dfrac{2 I_{DSS}}{V_P}$, the result becomes

This shows that $g_m$ is maximum $(= g_{mo})$ at $V_{GS} = 0$ and falls linearly to zero as $V_{GS}$ approaches $V_P$.

[!ANSWER]

$g_m = g_{mo}\left(1 - \dfrac{V_{GS}}{V_P}\right)$, where $g_{mo} = -\dfrac{2 I_{DSS}}{V_P}$.