Question

Why, in the Euler Lagrange equation isn't d/dt of the partial derivative of L with respect to q' actually the partial derivativ of L with respect to q' and t? Remember that L is L(t,q,q')

Answer 1

Otherwise stated. Why is the equation:\[\frac{ d }{ dt} L _{q'}-L _{t}=0\]instead of:\[L _{tq'}-L _{t}=0\]

Answer 2

Small type, but the Lagrangian is:
\[ \frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial q'} \]

L is a function of two functions of t, q and q', but we only care about the explicit functions and not the implied t when deriving. The time derivative arises from derivative when integrating to minimize

\[ ( \frac{\partial L}{\partial q} \delta q+ \frac{\partial L}{\partial q'} \delta q')\]

but note that q is a only a function of t so
\[ q'(t) = \frac{dq}{dt}\]

Using this, we obtain
\[ ( \frac{\partial L}{\partial q} \delta q+ \frac{\partial L}{\partial q'} \frac{d}{dt}\ \delta q)\]

And more math is done to simplify to our lagrangian and set equal to 0.