Derivative of angular momentum in a rotating frame of reference

Consider total derivative of $\vec{A}$ in an inertial frame, $$\frac{d}{dt}\vec{A}=\hat{i}\frac{d}{dt}A_x+\hat{j}\frac{d}{dt}A_y+\hat{k}\frac{d}{dt}A_z+A_x\frac{d\hat{i}}{dt}+A_y\frac{d\hat{j}}{dt}+A_z\frac{d\hat{k}}{dt}.$$ In an inertial system $\frac{d\hat{i}}{dt}=\frac{d\hat{j}}{dt}=\frac{d\hat{k}}{dt}=0$. Thus we have, $$\frac{d\vec{A}}{dt}_{inertial}=\hat{i}\frac{d}{dt}A_x+\hat{j}\frac{d}{dt}A_y+\hat{k}\frac{d}{dt}A_z.$$ Now consider in a rotating frame one has $\vec{A}=A_x^\prime\hat{i}^\prime+A_y^\prime\hat{j}^\prime+A_z^\prime\hat{k}^\prime$. Taking the total derivative, $$\frac{d}{dt}\vec{A}=\hat{i}^\prime\frac{d}{dt}A_x^\prime+\hat{j}^\prime\frac{d}{dt}A_y^\prime+\hat{k}^\prime\frac{d}{dt}A_z^\prime+A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}.$$ Note that in a rotating frame the quantities like $\frac{d\hat{i}^\prime}{dt}$ do not vanish. Therefore the total time derivative in a rotating frame is,

$$\frac{d\vec{A}}{dt}=\frac{d\vec{A}}{dt}_{rotating}+A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}$$ $\frac{d\vec{A}}{dt}_{rotating}=\hat{i}^\prime\frac{d}{dt}A_x^\prime+\hat{j}^\prime\frac{d}{dt}A_y^\prime+\hat{k}^\prime\frac{d}{dt}A_z^\prime$ is the apparent time derivative of the vector in a rotating frame and the combination $A_x^\prime\frac{d\hat{i^\prime}}{dt}+A_y^\prime\frac{d\hat{j^\prime}}{dt}+A_z^\prime\frac{d\hat{k}^\prime}{dt}$ captures the effects of rotation. Now, we know the relation between the linear velocity and angular velocity, $\vec{v}=\frac{d\vec{r}}{dt}=\omega\times\vec{r}$. Choosing $\vec{r}$ to be equal to $\hat{i}^\prime$, $\hat{j}^\prime$, and $\hat{k}^\prime$ respectively we have, $$\begin{align}\frac{d\hat{i}^\prime}{dt}&=\omega\times\hat{i}^\prime\\\frac{d\hat{j}^\prime}{dt}&=\omega\times\hat{j}^\prime\\\frac{d\hat{k}^\prime}{dt}&=\omega\times\hat{k}^\prime\end{align}$$ Using these we get in a rotating frame, $$\frac{d\vec{A}}{dt}=\frac{d\vec{A}}{dt}_{rotating}+\omega\times\vec{A}$$ Choosing $\vec{A}=\vec{L}$ we get, $$\frac{d\vec{L}}{dt}=\frac{d\vec{L}}{dt}_{rotating}+\omega\times\vec{L}=\vec{\tau}+\omega\times\vec{L}.$$

I prefer to use this notation.

The components of a arbitrary vector $\vec{x'}$ in rotating system are transformed to inertial system by this equation:

$$\vec{x}=R\,\vec{x'}\tag 1$$

where R is the transformation matrix between the rotating system and inertial system

The time derivative of equation (1) is:

$$\vec{\dot{x}}=R\,\vec{\dot{x}'}+\dot{R}\,\vec{x'}\tag 2$$

with

$\dot{R}=R\,\tilde{\omega}\quad$ and $\tilde{\omega}=\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] $ thus: $$\vec{\dot{x}}=R\,\vec{\dot{x}'}+R\,\tilde{\omega}\,\vec{x'}\tag 3$$

multiply equation (3) from the left with $R^T$

$$R^T\,\vec{\dot{x}}=\vec{\dot{x}'}+\vec{\omega}\times \vec{x'}$$

thus

$$\boxed{\left(\vec{\dot{x}}\right)_R=\left(\vec{\dot{x}'}\right)_R+ \left(\vec{\omega}\times \vec{x'}\right)_R}$$

where index R means the components are given in the rotating system.

I like to understand what I am doing, but in this instance I have no idea why the derivative gives this extra term.

The extra term is pretty clear to understand on physical grounds as follows:

Consider a gyroscope in the absence of any external torque. The gyroscope maintains a fixed angular momentum with respect to the axes of an inertial frame.

The axes of the rotating frame rotate with respect to the inertial frame, so the gyroscope must rotate with respect to the rotating frame. This torque-free rotation is the additional term.