Rotation systems. Problem interpreting an equation In this equation: 
$$
\mathbf a_i\overset{\rm def}{=}\left(\frac{d^2\mathbf r}{dt^2}\right)_i=\left(\frac{d\mathbf v}{dt}\right)_i=\left[\left(\frac{d}{dt}\right)_r+\boldsymbol\Omega\times\right]\left[\left(\frac{d\mathbf r}{dt}\right)_r+\boldsymbol\Omega\times\mathbf r\right]
$$
(from Wikipedia), why is 
$$\left(\frac{d}{dt}\right)_r \boldsymbol{\Omega} \times \mathbf{r}=\frac{d\boldsymbol{\Omega}}{dt}\times\bf{r}+\bf{\Omega}\times \bf{V_r}$$
In particular, I have qualms with the term 
$$\frac{d\bf{\Omega}}{dt}\times\bf{r}$$
Why are we deriving the angular velocity? Why is it a derivative not of the rotational type (Namely $(\frac{d}{dt})_r$ )
Other sources do not point out that term I have problems with. In any case, I want to know how you evaluate that derivative.
 A: I see now your problem and I believe that I can help.
Let's begin from the velocity formula
$$v_i = v_r + Ω \times r .\tag{1}$$
Let's take the derivative of $v_i$ IN THE INERTIAL frame,
$$a_i = \left(\frac{dv_r}{dt}\right)_i + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times v_i .$$
Here we use as much as we can our formula $(dF/dt)_i = (dF/dt)_r + Ω \times F$, and also substitute $v_i$ in the last term, by its formula (1). So,
$$a_i = \left(\frac{dv_r}{dt}\right)_r + Ω \times  v_r + \left(\frac{dΩ}{dt}\right)_i \times  r + Ω \times  v_r + Ω \times  Ω \times  r .$$
Gathering identical terms,
$$a_i = a_r + 2Ω \times v_r + \left(\frac{dΩ}{dt}\right)_i \times r + Ω \times Ω \times r .$$
This is what they got in Wikipedia.
A: 
Other sources do not point out that term I have problems with.

Other sources explicitly assume a constant angular velocity and thus ignore that component. The wikipedia article you cited is correct.

In any case, I want to know how you evaluate that derivative.

Given any vector quantity $\mathbf q$ that is the same (other than component representation) in the inertial and rotating frame, the time derivative of that vector from the perspective of an inertial versus rotating observer is
$$\left(\frac{d\mathbf q}{dt}\right)_I = \left(\frac{d\mathbf q}{dt}\right)_R + \boldsymbol\Omega \times \mathbf q$$
In dynamics, this is sometimes called the transport theorem (but there are a number of other things called the transport theorem).
Applying the transport theorem to the angular velocity vector yields
$$
\left(\frac{d\boldsymbol\Omega}{dt}\right)_I =
  \left(\frac{d \boldsymbol\Omega}{dt}\right)_R
  + \boldsymbol\Omega \times \boldsymbol\Omega
= \left(\frac{d \boldsymbol\Omega}{dt}\right)_R
$$
In other words, angular acceleration is fundamentally the same vector in the inertial and rotating frame.
Applying the transport theorem instead to angular momentum yields
$$\left(\frac{d\mathbf L}{dt}\right)_I = \left(\frac{d\mathbf L}{dt}\right)_R + \boldsymbol\Omega \times \mathbf L$$
The rotational analog of Newton's second law provides an alternative representation of the left-hand side of the above: $\frac {d\mathbf L}{dt} = \boldsymbol \tau_{\text{ext}}$ where the derivative is calculated from the perspective of an inertial frame and $\boldsymbol \tau_{\text{ext}}$ is the external torque on the system. If the system is a rigid body, the angular momentum is given $\mathbf L = \mathrm I \boldsymbol\Omega$ where $\mathrm I$ is the object's inertia tensor. Since the inertia tensor of a rigid body is constant in a frame rotating with the body, the time derivative of the angular momentum vector from the perspective of an observer rotating with the object simplifies to $\left(\frac{d\mathbf L}{dt}\right)_R = \mathrm I\left(\frac{d \boldsymbol\Omega}{dt}\right)_R$. Putting all of the above together yields
$$
\boldsymbol\tau_{\text{ext}} =
  \mathbf I \frac{d \boldsymbol\Omega}{dt}
  + \mathbf \Omega \times (\mathrm I \, \boldsymbol\Omega)$$
or
$$
\frac {d \boldsymbol\Omega}{dt} =
  {\mathbf I}^{-1}
  \left(
    \boldsymbol\tau_{\text{ext}}
    - \boldsymbol \Omega \times (\mathrm I \, \boldsymbol \Omega)
  \right)
$$
This yields a way to calculate $\frac {d\boldsymbol \Omega}{dt}$ at any point in time for a rigid body. Whether this is integrable via elementary methods is a different question. In most cases, it isn't. It's rather challenging to find a non-trivial rotational system that has an analytic solution. One typically has to revert to numerical methods to determine the rotational behavior of an object.
A: If you just follow your nose, then...
$$\left(\frac{d}{dt}\right)_{rotating} {\bf{\Omega}} =\frac{d\bf{\Omega}}{dt}+\bf{\Omega}\times {\bf{\Omega}}$$
Do you know what the second term is equal to? Hopefully this clears up the problem you have.
A: $$ \mathbf V = \mathbf v + \boldsymbol \Omega \times \mathbf r $$
The derivative of $\mathbf V$ in the inertial frame is indeed,
$$ \mathbf A = \frac{\mathrm D \mathbf v}{\mathrm Dt} + \frac{\mathrm D \boldsymbol \Omega}{\mathrm D t} \times \mathbf r + \boldsymbol \Omega \times \frac{\mathrm D \mathbf r}{\mathrm Dt}.$$
You are right, both $\mathbf v$ and $\mathbf r$ are described according to the rotating axes. Though, for the observer in the inertial frame they rotate together with those axes. So, for doing the derivative according to the inertial frame, you derivate again according to the rotating axes and then derivate the movement of the rotating axes. Let's write that:
$$ \begin{align}
 \mathbf A &= \frac{\mathrm d \mathbf v}{\mathrm dt} + \boldsymbol \Omega \times \mathbf v + \frac{\mathrm D \boldsymbol \Omega}{\mathrm Dt} \times \mathbf r + \boldsymbol \Omega \times (\mathbf v + \boldsymbol \Omega \times \mathbf r) \\
          & = \mathbf a + 2\boldsymbol \Omega \times \mathbf v + \frac{\mathrm D \boldsymbol \Omega}{\mathrm Dt} \times \mathbf r + \boldsymbol \Omega \times \boldsymbol \Omega \times \mathbf r .
\end{align} $$
Does my answer remove your doubts?
