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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.

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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.

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  • $\begingroup$ There's another way to make infinitesimals consistent - smooth infinitesimal analysis, which is different from non-standard analysis. $\endgroup$ – user117644 Jul 11 '15 at 0:06

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