The following expression is often found in a derivation of the Lagrangian equations: $$ \dfrac{\partial \bf{v}}{\partial \dot{q_j}} = \dfrac{\partial \bf{r}}{\partial q_j},$$
where $\bf{v}=\bf{\dot{r}}$. Since $$\textbf{v} = \sum_{j = 1}^{n} \dfrac{\partial \bf{r} }{\partial q_j} \dot{q_j},$$
with $n$ the number of degrees of freedom, the partial derivative of $\bf v$ with respect to $\dot q_j$ is simply the coefficient of $\dot q_j$ of the $j$-th term in the sum. Which is $$\dfrac{\partial \bf{r}}{\partial q_j}.$$ This is possible because all the $q_j$'s for all $ 1 \leq j \leq n$ are independent, and so are their time-derivatives. $\blacksquare$
I ask this here because I recently saw a video on Youtube where a guy proved this, but he did it differently and his method seemed longer and more complicated than what I'm doing here. So I wonder if I was too blunt somewhere in the proof.
