I am currently studying calculus of variations (for my classical mechanics course). I have, on multiple occasions, seen the derivative of an infinitesimal quantity defined like below $$\frac{d}{dt} \delta q(t)=\delta \frac{dq(t)}{dt}$$
Now, I do have a geometrical intuition about how this works. By changing a trajectory $q(t)$ between two time instants $t_1$ and $t_2$ by $\delta q(t)$ , I understand it is natural that the derivatives of $q(t)$ will also change, which is reflected in the formula given above. But I do not understand how this equality comes about in a rigorous sense.
Any insight would be really appreciated. Thank you for your time.
You don't need infinitesimals to do calculus of variations. Simply consider adding a finite quantity, $$\delta q(t) = \epsilon \gamma(t),$$ where $\gamma(t)$ is some curve and $\epsilon$ is a small positive number which may be sent to zero later on.
Whilst it is possible to define infinitesimals in a way that makes them rigorous -- it's a not very popular field called "non-standard analysis" and I think few people on this forum will use them.