Let $R(t)$ is a rotation matrix: $R(t) = \begin{pmatrix} \cos t & - \sin t \\ \sin t & \cos t \end{pmatrix}$, and $x = (x_1, x_2)^T \in \mathbb{R}^2$ some fixed vector, then
$$R(t)x = \begin{pmatrix} x_1 \cos t - x_2 \sin t \\ x_1 \sin t + x_2 \cos t \end{pmatrix},$$
and $$\frac{d(R(t)x)}{dt} = \begin{pmatrix} - x_1 \sin t - x_2 \cos t \\ x_1 \cos t - x_2 \sin t \end{pmatrix}, $$
then I noticed that $$(R(t)x)^T \Bigg(\frac{d(R(t)x)}{dt} \Bigg) = 0.$$
Is this true for any rotation matrix? Is it true in $\mathbb{R}^3$? Can you prove it?
$R(t)x$ can be interpreted as the path of a curve $f(t)$ on a suitably scaled instance of $S=\mathbb S^{n-1}$ where $n$ is the ambient space dimension. Then for a given $t$, $\frac{d(R(t)x)}{dt}=\frac{df}{dt}$ is a vector in $T_{f(t)}S$, so must be orthogonal to the vector $f(t)$ (which points in the same direction as the normal to $S$ at $t$). Hence your equation is true in any dimension, assuming $R(t)$ is continuous.