Consider the standard rigid body rotation problem of a rigid body B(Oxyz) rotating G(OXYZ). Let $\mathbf e_i$ represent the unit vectors of the B frame and let $\mathbf E_i$ represent the unit vectors of the G frame {i: 1 to 3}. The direction cosine matrix (DCM) from the B frame to the G frame,$^GR_B$, has ij elements $\mathbf E_i$ $\cdot$ $\mathbf e_j$ where {i, j: 1 to 3}.
If each element of either DCM is left in scalar product form (e.g. for $^GR_B$, $\mathbf E_i$ $\cdot$ $\mathbf e_j$) rather than an evaluated form (e.g. for $^GR_B$, cos($\mathbf E_i$, $\mathbf e_j$)), we can differentiate the DCM, with respect to time and with respect to either frame (using the fact that the unit vectors of a given frame are fixed in the same frame; using the left superscript on d to denote the frame for which the derivative is taken with respect to).
Equation (1):
$$\frac{^Gd}{dt}(^GR_B) = \frac{^Gd\mathbf E_i}{dt} \cdot\mathbf e_j + \mathbf E_i \cdot\frac{^Gd\mathbf e_j}{dt} = \mathbf E_i \cdot\frac{^Gd\mathbf e_j}{dt} $$
Equation (2):
$$\frac{^Bd}{dt}(^GR_B) = \frac{^Bd\mathbf E_i}{dt} \cdot\mathbf e_j + \mathbf E_i \cdot\frac{^Bd\mathbf e_j}{dt} = \frac{^Bd\mathbf E_i}{dt} \cdot\mathbf e_j $$
When the scalar product is evaluated, however, the B and G derivatives reduce to a simple time derivative of a scalar.
Equation (3):
$$\frac{^Bd}{dt}cos(\mathbf E_i, \mathbf e_j) = \frac{^Gd}{dt}cos(\mathbf E_i, \mathbf e_j) = \frac{d}{dt}cos(\mathbf E_i, \mathbf e_j) = \frac{d}{dt}(^GR_B) $$
It would seem that equation (3) indicates that equation (1) is equal to equation (2). However, if such holds then I am having trouble seeing it just from the form of equation (1) and (2). Any help would be greatly appreciated.