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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,

$$\frac{\partial}{\partial \dot{q}_i} \frac{\mathrm{d}L}{\mathrm{d}t} =\frac{\partial L}{\partial q_i}~?\tag{2}$$

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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.

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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

$\frac{\partial}{\partial \dot{q}} \left( \dot{q}\frac{\partial L}{\partial q} + \ddot{q}\frac{\partial L}{\partial \dot{q}} + \frac{\partial L}{\partial t} \right)$

$\quad = \frac{\partial L}{\partial q} + \dot{q}\frac{\partial^2 L}{\partial q \partial \dot{q}} + \ddot{q}\frac{\partial^2 L}{\partial \dot{q}^2} + \frac{\partial^2 L}{\partial t \partial \dot{q}}$.

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