Let $\Omega=\Omega(q)$ be a function of coordinates and define \\ \[L'=L+\frac{d{\Omega}(q)}{dt}.\]\\ For a function \(\Omega(q)\)of coordinates \(q\), we will have $$\frac{d}{dt}{\Omega}(q)={\sum_k} \frac{\partial{\Omega}}{\partial{q_k}}\dot{q_k} $$ Use this to prove that EOM from $L'$ and $L$ are identical
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