Am I allowed to write $(\mathbf e_1\cdot\frac{d}{dt}\mathbf v)_\varepsilon=(\frac{d}{dt}[\mathbf e_1\cdot\mathbf v])_\varepsilon$? Let $\varepsilon$ be the Euclidean space with basis $(\mathbf e_1,\mathbf e_2,\mathbf e_3)$.
For a rigid body in $\varepsilon$ suppose we have $$\mathbf a=\left(\frac{d}{dt}\mathbf v\right)_\varepsilon,$$ where $\mathbf a$ and $\mathbf v$ are the acceleration and velocity of the rigid body.
We want to find out the first component of the acceleration and so one way would be to take the dot product for both sides with $\mathbf e_1$, that is $$\mathbf e_1\cdot\mathbf a=\mathbf e_1\cdot\left(\frac{d}{dt}\mathbf v\right)_\varepsilon.$$
But am I allowed to write $\left(\mathbf e_1\cdot\frac{d}{dt}\mathbf v\right)_\varepsilon$ and so $\left(\frac{d}{dt}[\mathbf e_1\cdot\mathbf v]\right)_\varepsilon$?
If so, why?

Yes, the equivalence
$\mathbf e_i \cdot \dfrac{d\mathbf v}{dt} = \dfrac{d(\mathbf e_i \cdot \mathbf v)}{dt} \tag 1$
is valid, and the reason is that the $\mathbf e_i$ are constant with respect to $t$; then the ordinary Leibniz rule for product differentiation yields
$\dfrac{d(\mathbf e_i \cdot \mathbf v)}{dt} = \dfrac{d \mathbf e_i}{dt} \cdot \mathbf v + \mathbf e_i \cdot \dfrac{d \mathbf v}{dt}; \tag 2$
but
$\dfrac{d \mathbf e_i}{dt} = 0; \tag 3$
from this and (2), the desired result (1) is immediate
