Q.Derive the relation μ = r_d x g_m and g_m = g_mo [1 - v_GS/v_P]

The first relation comes from the definition of the amplification factor, and the second from differentiating Shockley's equation. …

[!TLDR]

$\mu = r_d \times g_m$ follows from the definition of $\mu$ by introducing $\Delta I_D$; $g_m = g_{mo}(1 - V_{GS}/V_P)$ follows by differentiating Shockley's equation.

Derivation of $\mu = r_d \times g_m$:

The amplification factor is defined as

Multiplying and dividing the right side by the small change in drain current $\Delta I_D$,

Since $r_d = \dfrac{\Delta V_{DS}}{\Delta I_D}$ and $g_m = \dfrac{\Delta I_D}{\Delta V_{GS}}$, this becomes

Derivation of $g_m = g_{mo}\left(1 - \dfrac{V_{GS}}{V_P}\right)$:

Transconductance is the rate of change of drain current with gate voltage, $g_m = \dfrac{\Delta I_D}{\Delta V_{GS}}$. The drain current follows Shockley's equation,

Differentiating with respect to $V_{GS}$ using the chain rule, …