consider the Lagrangian 1$/$2 m$\dot{x}$$^{2}$ $-$ V$($x$)$ such that x $=$ f$($t$)$ , $\dot{x}$ $=$ g$($t$)$ and t $=$ f$^{-1}$$($x$)$ , $\dot{x}$ $=$ g$($f$^{-1}$$($x$)$$)$ and, $\frac{\partial{L}}{\partial{x}}$ $=$ $\frac{d}{dt}$ $($ $\frac{\partial{L}}{\partial{\dot{x}}}$ $)$ . Now we know that $\frac{\partial{L}}{\partial{\dot{x}}}$ $=$ m$\dot{x}$ ...$($1$)$.But I can say that $\frac{\partial{L}}{\partial{\dot{x}}}$ $=$ m$\dot{x}$ $\frac{\partial{\dot{x}}}{\partial{x}}$.......$($ 2$)$.
According to me $\frac{\partial{\dot{x}}}{\partial{x}}$ is zero
Because $\partial{\dot{x}}$ is 0 since by the definition of partial differentiation we keep t constant that is t doesn't change ! That is why the above quantity is 0 and we reduce back to 1) from 2) . Now all that I want to know is that whether this is right or not ?
This question is an extension of Why does $\frac{dq}{dt}$ not depend on $q$? Why does the calculus of variations work?
