Newton's $2^{nd}$ Law in a rotational frame I'm having a bit of trouble following the derivation for Newtons $2^{nd}$ law in a rotating frame, mainly just finding the second derivative of position in terms of the inertial fram. I am reading from Taylor's Classical Mechanics.
We start off with the relation between derivatives in a non-inertial & inertial reference frames. Let $S_0$ denote the non-inertial frame, and $S$ denote the inertial frame.
$$\left(\frac{d\vec{r}}{dt}\right)_{S_0}=\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}$$
We can differentiate this again
\begin{equation}
\begin{split}
\left( \frac{d^2\vec{r}}{dt^2}\right)_{S_0}&=\left(\frac{d}{dt}\right)_{S_0} \left[\left(\frac{d\vec{r}}{dt}\right)_{S_0}\right]\\
&=\left(\frac{d}{dt}_{S_0} \right)\left[\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}\right]
\end{split}
\end{equation}
This next step is where I'm confused. The textbook states:

Applying (9.30) to the outside derivative on the right, we find
$$\left( \frac{d^2\vec{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt} \right)_S \left[\left( \frac{d\vec{r}}{dt} \right)_S + \vec{\Omega} \times \vec{r}\right]+ \vec{\Omega} \times \left[ \left( \frac{d\vec{r}}{dt}\right)_S + \vec{\Omega} \times \vec{r}\right]$$

Where (9.30) is the aforementioned relation in its general form:
\begin{equation}
\left(\frac{d\vec{Q}}{dt}\right)_{S_0}=\left(\frac{d\vec{Q}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{Q}
\end{equation}
How do we get to the form shown in the book? I think the wording of the step confuses me.
 A: In the term $(1)\left(\frac{d}{dt}\right) _{S_0}\left[\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}\right]$, treat Q in 9.30 as $\left[\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}\right]$, and apply 9.30 to term (1) to express this term using $\left(\frac{d}{dt}\right) _{S}$.
A: The general rule is
$$ \left(\frac{d\vec{Q}}{dt}\right)_{S_0}=\left(\frac{d\vec{Q}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{Q} $$
where $\vec{Q}$ is any vector defined as riding on the rotating frame.
When you have $\vec{Q} = \vec{r}$ then you arrive at the first equation
$$ \left(\frac{d\vec{r}}{dt}\right)_{S_0}=\left(\frac{d\vec{r}}{dt}\right)_{S}+ \vec{\Omega} \times \vec{r}
$$
But use $\vec{Q} = \left( \frac{d\vec{r}}{dt}\right)_S + \vec{\Omega} \times \vec{r}$ to arrive at the second equation.
$$\left( \frac{d^2\vec{r}}{dt^2}\right)_{S_0}=\left(\frac{d}{dt} \right)_S \left[\left( \frac{d\vec{r}}{dt} \right)_S + \vec{\Omega} \times \vec{r}\right]+ \vec{\Omega} \times \left[ \left( \frac{d\vec{r}}{dt}\right)_S + \vec{\Omega} \times \vec{r}\right]$$
