Understanding why $\frac{d}{dt} \left( \frac{ \partial L}{ \partial \dot{q}}\right) = \frac{\partial L } {\partial q}$

Question

From pg. 96 of No-Nonsense Classical Mechanics:

The author asserts that

$$ \frac{dp}{dt} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) = \frac{\partial L}{\partial q} $$

But why is this so? I know that

$$ \frac{dp}{dt} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) = \frac{d}{dt} \frac{\partial L}{\partial \left( \frac{dq}{dt} \right)} $$

If we could justify in the denominator that

$$ dt \partial \left( \frac{dq}{dt} \right) = \partial \left( \frac{dt dq}{dt} \right) = \partial dq $$

and then further that

$$ \frac{d \partial L}{ \partial dq} = \frac{\partial L}{\partial q} $$

we would get there, but this seems like really sloppy reasoning to me (abusing the notation of $d$ of $\partial$).

Answer 1

Equation (5.2) is Euler-Lagrange equation

$$ \frac{{\rm d}}{{\rm d}t}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0 \tag{*} $$

you can find the derivation of that equation in many different references. What the author is doing in this paragraph is just using the definition of momentum to rewrite the equation, nothing more than that.

You cannot derive (*) just by using the definition of generalized momentum