In order to describe the real motion of an aircraft (not the imagined motion describing its attitude with respect to a reference axis system), we will be interested in the time derivative (with respect to the inertial frame \(\mathbf{A}\)) of all rotational parameters of a rotation formalism. The rotation parameters of the rotation matrix formalism are the entries of the rotation matrix \(B\). We shall therefore be interested in the time derivative of \(B\) with respect to the inertial frame. Without derivation, we state that

\begin{equation}
\frac{dB}{dt} = -[\tilde{\boldsymbol{\omega}}_B] B
\end{equation}

Here \([\tilde{\boldsymbol{\omega}}]\) is the cross-product equivalent matrix of vector \(\boldsymbol{\omega}=(\omega_1, \omega_2, \omega_3)\):
\begin{equation}
[\tilde{\boldsymbol{\omega}}] =
\begin{pmatrix}
0 & -\omega_3 & \omega_2 \\
\omega_3 & 0 & -\omega_1 \\
-\omega_2 & \omega_1 & 0
\end{pmatrix}
\end{equation}
which is a matrix that, when applied to another vector, results in the same result as evaluating the cross product \(\boldsymbol{\omega}\times\) (with some other vector). And \(\boldsymbol{\omega}_B=(p,q,r)\) here is the angular velocity vector expressed with respect to the body frame \(\mathcal{B}\).

The above are common ingredients for the differential equation of motion: we will see that in order to obtain the time derivatives of the parameters of a rotation formalism, we must know the angular velocity vector \(\boldsymbol{\omega}_{\mathcal{B}}=(p,q,r)\) and the current attitude of the aircraft. If we know these two ingredients, this equation can be used to compute attitude \(A\) at the next time step during numerical integration, as one computes the whole time history of the attitude of the aircraft. (The angular velocity information comes from the rotational dynamic equation of motion, while the attitude information comes from the previous time step and initially from a starting attitude.) 

From a mathematical perspective, the differential kinematic equation is a differential equation in the parameters of the rotation formalism, containing the parameters and their first time derivatives, and the \({\boldsymbol\omega_{\{mathcal{B}}\) are time dependent coefficients of the equation (their time dependence being given by another equation).

We can also obtain a similar equation for matrix \(A\):
\begin{equation}
\frac{dA}{dt} = [\tilde{\boldsymbol{\omega}}_A] A
\end{equation}
This time, the angular velocity vector shall be expressed with respect to the inertial frame \(\mathcal{A}\). The angular velocity vector obeys the usual transformation law for vectors under basis transformation, i.e. \(\boldsymbol{\omega}_{\mathcal{A}}= R_B^A \boldsymbol{\omega}_{\mathcal{B}}= A \boldsymbol{\omega}_{\mathcal{B}}\).

The above relation can be inverted. For instance, the angular velocity vector \(\boldsymbol{\omega}_{\mathcal{A}}=(\omega_1, \omega_2, \omega_3)\) of the aircraft (expressed with respect to basis \(\mathbf{A}\)) can be obtained from \(A(t)\) by the following expression:

\begin{equation}
[\tilde{\boldsymbol{\omega}}_A] = \frac{d A}{dt} A^T
\end{equation}

In fact, one can find the following four relations for the first time derivatives of the rotation matrix describing the motion of a rotating object (the derivative always being taken with respect to the inertial axis system \(\mathcal{A}\) and denoted by a dot over the letter here):
\begin{eqnarray}
\dot R_B^A &=& [\tilde{\boldsymbol{\omega}}_{\mathcal{A}}] R_B^A \\
\dot R_B^A &=& R_B^A [\tilde{\boldsymbol{\omega}}_{\mathcal{B}}] \\
\dot R_A^B &=& -R_A^B [\tilde{\boldsymbol{\omega}}_{\mathcal{A}}] \\
\dot R_A^B &=& -[\tilde{\boldsymbol{\omega}}_{\mathcal{B}}] R_{A}^B
\end{eqnarray}
with the one highlighted by us above being the last one. The proof of these relations and further details can be looked up, for instance, in the brief tutorial by Shiyu Zhao. However, to convey the idea, we shall sketch the proof of the first one of these relations here.

Proof: Let us work with only one basis here, \(\mathcal{A}\), and let us remember that in our more recent notation \(A := R_B^A\). Let \(\mathbf{v}(t_0)\) be an arbitrary fixed vector, e.g. (but not necessarily) the value of a vector \(\mathbf{v}(t)\) at some fixed time \(t_0\). For any vector \(\mathbf{v}(t)=A(t)\mathbf{v}(t_0)\) we have \(\dot{\mathbf{v}}=\boldsymbol{\omega}\times\mathbf{v} = [\tilde{\boldsymbol{\omega}}]\mathbf{v}\). On the other hand, we also have
\begin{equation}
\dot{\mathbf{v}}=\frac{dA}{dt}\mathbf{v}(t_0)=\frac{dA}{dt}\overbrace{A^{-1}\mathbf{v}}^{\mathbf{v}(t_0)=}=\frac{dA}{dt}A^{T}\mathbf{v},
\end{equation}
where in the last step we have used that \(A\) is an orthonormal matrix describing a rotation, and therefore \(A^{-1}=A^T\). Equating the two expressions for \(\dot{\mathbf{v}}\) (and removing \(\mathbf{v}\) at the end) yields the result \([\tilde{\boldsymbol{\omega}}_{\mathcal{A}}] = \frac{d A}{dt} A^T\).

Remember that in this proof we have not introduced a second basis, \(\mathcal{B}\), and therefore, if a basis is to be used for the quantities, it has to be the inertial frame basis \(\mathcal{A}\), which is why \(\boldsymbol{\omega}_{\mathcal{A}}\) appears in the final expression (and not \(\boldsymbol{\omega}_{\mathcal{B}}\)). The other three relations can be obtained by further manipulation of this first result.