I know that the time derivative of some quantity $r(t)$ in a rotating frame which rotates with angular velocity $\Omega(t)$ is related to the derivative in a fixed (i.e. inertial) frame by
$$ \Big(\frac{d}{dt}\Big)_{rot} r(t) = \Big(\frac{d}{dt}\Big)_{in}r(t) - \Omega(t) \times r(t) $$
However, how can the time derivatives of two rotating frames, each rotating with its own angular velocity $\Omega_1(t),\Omega_2(t)$, be related to each other? Is there a similar formula for this case?
$$ (1) \Big(\frac{d}{dt}\Big)_{rot_1} r(t) = \Big(\frac{d}{dt}\Big)_{in}r(t) - \Omega_1(t) \times r(t) $$ where $rot_1$ denotes a coordinate system rotating at $\Omega_1(t)$ relative to the fixed system $in$. $$ (2) \Big(\frac{d}{dt}\Big)_{rot_2} r(t) = \Big(\frac{d}{dt}\Big)_{in}r(t) - \Omega_2(t) \times r(t) $$ where $rot_2$ denotes a coordinate system rotating at $\Omega_2(t)$ relative to the fixed system $in$.
From (1) and (2) $$\Big(\frac{d}{dt}\Big)_{rot_1} r(t) + \Omega_1(t) \times r(t) = \Big(\frac{d}{dt}\Big)_{rot_2} r(t) + \Omega_2(t) \times r(t)$$
Finally, $$ \Big(\frac{d}{dt}\Big)_{rot_2} r(t) = \Big(\frac{d}{dt}\Big)_{rot_1} r(t) + (\Omega_1(t) - \Omega_2(t)) \times r(t)$$
