The Euler-Lagrange equation is typically written in this form :
\begin{equation}
\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q}}\right)-\frac{\partial}{\partial q}\mathcal{L}=0
\end{equation}
Since the Lagrangian is a functional form of $t$, $q$, $\dot{q}$, why do we use a full derivative for the time, instead of a partial derivative ?
Best Answer
@Arthur's advice is to revisit the standard proof of the ELE, in which an integration by parts with vanishing boundary terms gives $\int\frac{\partial L}{\partial\dot{q}}\delta\dot{q}dt=-\int\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\delta qdt$. So the real question is why this IBP has the form$$\int_a^bu\frac{dv}{dt}dt=[uv]_a^b-\int_a^b\frac{du}{dt}vdt.$$By the FTA, this reduces to asking why the product rule should be $\frac{d(uv)}{dt}=u\frac{dv}{dt}+\frac{du}{dt}v$, instead of something similar in which a partial time derivative appears at least once.
In the case at hand $q$ is a function of time alone, since the action is a single integral. Then $\frac{\partial L}{\partial\dot{q}}$, being a function of $q,\,\dot{q}$, reduces to a function of time only.
Your question concerns classical mechanics, so we should contrast it with an ELE from classical field theory. The role of time is replaced with multiple variables $x^\mu$; the role of $q$ is replaced with a field, say $\phi$. An action of the form $S=\int\mathcal{L}(\phi,\,\partial\phi/\partial x^\mu)d^nx$ has ELE$$\frac{\partial}{\partial x^\mu}\frac{\partial\mathcal{L}}{\partial(\partial\phi/\partial x^\mu)}-\frac{\partial\mathcal{L}}{\partial\phi}=0$$(with implicit summation over $\mu$). You'll notice the generalization of $\frac{d}{dt}$, namely $\frac{\partial}{\partial x^\mu}$, is a partial derivative.