What is the physical argument for $d(G)_s = d(G)_b + d(G)_{rot} \quad ?$ In the book of Goldstein, Classical Mechanics, at the end of the page 171, it is stated that

A relation between the two differential changes in $G$ can be derived
  on the basis of physical arguments.We can write that the only
  difference between the two is the effect of rotation of the body axes:
$$d(G)_s = d(G)_b + d(G)_{rot},$$

but the does not give any argument; just states that it can be given. Moreover, in the next page, it argues that

Now consider a vector fixed in the rigid body.As the body moves there
  is of course no change in the components of this vector as seen by the
  body  observer, i.e relative to body axes. The only contribution to
  $d(G)_s$ is then the effect of the rotation ofthe body. But since the
  vector is fixed in rhe body system, it rotates with it
  counterclockwise ,and the change in the vector. as observed in space is that given by  $$dr' = d\Omega \times r,$$ hence  $$d(G)_{rot} = d
 \Omega \times G$$

However, in general, $r'$, i.e $r+ dr$, is not fixed in the body frame, and it also moves, so this argument might fail because in general there will be contributions from the rotation of the point itself wrt to the body frame, so how such an argument can be valid ?
Question:


*

*What is the physical argument the existence of the relation $$d(G)_s = d(G)_b + d(G)_{rot} \quad ?$$ 

*How can the argument given for $d(G)_{rot} = d \Omega \times G$ be valid for any vector $G$ ? 

 A: $G$ must be a vector:
$$d(\vec{G})_s=d(\vec{G})_b+d(\vec{G})_r$$
the components of the vector $\vec{G}$ can be given in space coordinate system 
$$\vec{G}_s=R(\,\vec{\phi})\,\vec{G}_b$$
where $R$ is the transformation matrix between body coordinate system and space system
and $\vec{\phi}$ is then vector containing the euler angles $(\alpha\,,\beta\,,\gamma)$
thus:
$$\vec{\dot{G}}_s=R(\,\vec{\phi})\,\vec{\dot{G}}_b+\dot{R}(\,\vec{\phi})\,{{G}}_b$$
with $\dot{R}=R\,\tilde{\omega}$
$$\vec{\dot{G}}_s=R(\,\vec{\phi})\,\vec{\dot{G}}_b+R(\,\vec{\phi})\,\tilde{\omega}\,\vec{{G}}_b\tag 1$$
multiply equation (1) from the left with $R^T$ 
$$R^T\,\vec{\dot{G}}_s=\vec{\dot{G}}_b+\tilde{\omega}\,\vec{{G}}_b\tag 2$$
$$R^T\,\vec{d{G}}_s=\vec{d{G}}_b+\tilde{\omega}\,\vec{{G}}_b\,dt\tag 3$$
$$d(\vec{G})_s=d(\vec{G})_b+d(\vec{G})_{\text{rot}}$$
thus:
$$d(\vec{G})_s=R^T\,\vec{d{G}}_s$$
$$d(\vec{G})_{\text{rot}}=\tilde{\omega}\,\vec{{G}}_b\,dt=\left(A(\vec{\phi})\,d\vec{\phi}\right)\times \vec{G}_b$$
comment
$\tilde{\omega}\,\vec{G}=\vec{\omega}\,\times\,\vec{G}$
with $\tilde{\omega}$ a skew  anti symmetric matrix
$\begin{bmatrix}
   0 & -\omega_z & \omega_y \\
   \omega_z & 0 & -\omega_x \\
   -\omega_y & \omega_x & 0 \\
 \end{bmatrix}$
From:
$\dot{R}=R(\vec{\phi})\,\tilde{\omega}\quad$ 
we get
$\quad\vec{\omega}=A(\vec{\phi})\,\vec{\dot{\phi}}$
