Order of derivatives in Euler-Lagrange equations The Euler-Lagrange equations are
$$\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}_i} = \frac{\partial L}{\partial q_i}.\tag{1}$$
Is it equivalent to switch the derivatives on the left hand side,
$$\frac{\partial}{\partial \dot{q}_i}
\frac{\mathrm{d}L}{\mathrm{d}t} =\frac{\partial L}{\partial q_i}~?\tag{2}$$
 A: No, OP's eq. (2) should be 
$$\frac{\partial}{\partial \dot{q}^i}
\frac{\mathrm{d}L}{\mathrm{d}t} ~=~\color{red}{2}\frac{\partial L}{\partial q^i} ,$$
which is known as Nielsen's form of the EL equations, cf. e.g. this Math.SE post.
A: No, they are not equivalent.  Assuming the Lagrangian has the form $L(q, \dot{q}, t)$, the chain rule can be used to evaluate the two expressions.  The left hand side of the first, correct expression evaluates to
$\dot{q}\frac{\partial^2 L}{\partial q \partial \dot{q}} + \ddot{q}\frac{\partial^2 L}{\partial \dot{q}^2} + \frac{\partial^2 L}{\partial t \partial \dot{q}}$.
The left hand side of the second, incorrect expression evaluates to
$\frac{\partial}{\partial \dot{q}} \left( \dot{q}\frac{\partial L}{\partial q} + \ddot{q}\frac{\partial L}{\partial \dot{q}} + \frac{\partial L}{\partial t} \right)$
$\quad = \frac{\partial L}{\partial q} + \dot{q}\frac{\partial^2 L}{\partial q \partial \dot{q}} + \ddot{q}\frac{\partial^2 L}{\partial \dot{q}^2} + \frac{\partial^2 L}{\partial t \partial \dot{q}}$.
