The last part in the derivation of the Euler-Lagrange equations for a stationary action has me confused. It's about the order of differentiation and evaluation, and whichever comes first.
I'll highlight the derivation and state my question at the end.
Derivation
For Lagrangian $L(q,\dot q)$, denote a variation from the stationary path by $$ q(t) \to q(t) + \epsilon \eta(t) \equiv \tilde q(t), $$ from which it follows that $$ \dot q(t) \to \dot q(t) + \epsilon \dot \eta(t) \equiv \dot{\tilde{q}}(t). $$
For a stationary action $S=\int L(q,\dot q)d t$, we require that $$ \delta S \equiv \lim_{\epsilon\to 0}\frac{\Delta S}{\epsilon} = \int \frac{1}{\epsilon}\left( L(\tilde q,\dot{\tilde{q}}) - L(q,\dot q) \right)dt =0. $$ Since $L(\tilde q,\dot{\tilde{q}})$ is a function of $\epsilon$, it can be Taylor expanded: $$ L(\tilde q,\dot{\tilde{q}}) = L(\tilde q,\dot{\tilde{q}})\big|_{\epsilon=0} +\epsilon\frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial\epsilon}\big|_{\epsilon=0} +O(\epsilon^2) $$ Then, using the chain rule, $$ \delta S = \int \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial\epsilon}\big|_{\epsilon=0} dt = \int \left( \left[ \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial \tilde q} \frac{\partial \tilde q}{\partial \epsilon} \right]\bigg|_{\epsilon=0} + \left[ \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial \dot{\tilde{q}}} \frac{\partial \dot{\tilde{q}}}{\partial \epsilon} \right]\bigg|_{\epsilon=0} \right)dt =0 $$ and after partial integration (with vanishing boundary term) $$ \delta S = \int \left( \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial \tilde q} \bigg|_{\epsilon=0} -\frac{d}{dt} \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial \dot{\tilde{q}}} \bigg|_{\epsilon=0} \right)\eta dt =0. $$ Given the arbitrary character of $\eta$, it follows that $$ \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial \tilde q} \bigg|_{\epsilon=0} -\frac{d}{dt} \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial \dot{\tilde{q}}} \bigg|_{\epsilon=0} =0. \quad (1) $$
Question
The mathematics say that in Eq. (1) we now first should calculate the partial derivatives, and then evaluate at $\epsilon=0$, to get the Euler-Lagrange equation $$ \frac{\partial L(q,\dot{q})}{\partial q} -\frac{d}{dt} \frac{\partial L(q,\dot{q})}{\partial \dot{q}} =0. $$ If it were the order way around (first evaluating $\epsilon=0$), I would understand it, but not when first evaluating the derivative.
In other words, is it justified to say that $$ \frac{d}{dt} \frac{\partial L(\tilde q,\dot{\tilde{q}})}{\partial \dot{\tilde{q}}} \bigg|_{\epsilon=0} = \frac{d}{dt} \frac{\partial \left( L(\tilde q,\dot{\tilde{q}}) |_{\epsilon=0} \right)}{\partial \left( \dot{\tilde{q}} |_{\epsilon=0} \right) } \qquad \left( = \frac{d}{dt} \frac{\partial L(q,\dot{q})}{\partial \dot{q}} \right) ? $$