Interchange of derivatives Given Euler-Lagrangian equation
$$\frac{d}{dt}\frac{\partial L}{\partial \dot q}-\frac{\partial L}{\partial q}=0$$
Can I equivalently write as
$$\frac{\partial \dot L}{\partial \dot q}-\frac{\partial L}{\partial q}=0\tag 1$$
or
$$\frac{\partial^2 L}{\partial t\partial \dot q}+\frac{\partial^2 L}{\partial q\partial \dot q}\dot q+\frac{\partial^2 L}{\partial \dot q^2}\ddot q-\frac{\partial L}{\partial q}=0\tag 2$$
Why? Thanks.
 A: Euler - Lagrange equations are not as simple as they seem.
$L$ is defined as a function of $q \in \mathbb{R}^n$ : $L(q,\dot{q}) = L(q_1,\dots,q_n,\dot{q}_1,\dots,\dot{q}_n)$.
The question is the following : find a curve $\Gamma : \mathbb{R} \rightarrow \mathbb{R}^n ; t\mapsto \Gamma(t)$ such that the action $ \displaystyle S(\Gamma) = \int_\mathcal{I} L\left(\Gamma(t), \dfrac{d \Gamma(t)}{dt}\right)$. In other words, for all applications $\delta: \mathbb{R} \rightarrow \mathbb{R}^n ; t\mapsto \delta(t)$, you have $S(\Gamma) \le S(\Gamma+\delta)$. The Euler-Lagrange equations is a necessary condition for this extremum conditions to be satisfied, given some tedious regularity assumptions.
In the Euler Lagrange equations, $\dfrac{\partial L}{\partial q}$ denotes the vector which coordinates consist in the first n partial derivatives of the function $L$, and $\dfrac{\partial L}{\partial \dot{q}}$ denotes the vector which coordinates consist in the last n partial derivatives of the function $L$. Those are KNOWN ! This means that the Euler Lagrange equations are actually a set of ordinary differential equations (ODE) and not partial differential equations (PDE), such as the Hamiltion Jacobi equations (which adress the very same problem !!)
Ordinary here means that the unknown functions (noted $\Gamma$ above) is defined for a real variable (i.e.  $\Gamma : \color{red}{\mathbb{R}} \mapsto \mathbb{R}^n$). The dot notation $\dot{\Gamma}$ stands for the (total) derivative with respect to the only variable (often noted $t$ : $\dot{\Gamma} = \dfrac{d\Gamma}{d t}$). Since $L$ is not a function of a single variable (unlike $\Gamma$), the notation $\dot{L}$ is meaningless.
Consequently, only your second proposal is meaningful. It is moreover correct (thanks to the so-called chain rule for derivation) though I personnaly prefer the more classical, compact and aesthetically pleasing original version (keeping in mind that $L$ is known, there is no need to extend the partial derivatives)
Hope this helps.
