# How come $\frac{d}{dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}$ in Lagrangian mechanics? [duplicate]

It is written in the Goldstein's Classical Mechanics text that $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}=\sum_k \frac{\partial^2{r_i}}{\partial {q_j}\partial{q_k}}\dot q_k+\frac{\partial^2{r_i}}{\partial {q_j}\partial t},\tag{1.50b}$$ where $$\dot r_i=\frac{\mathrm d}{\mathrm dt}r_i=\sum_k\frac{\partial{}r_i}{\partial{q_k}}\dot q_k+\frac{\partial {r_i}}{\partial t}.\tag{1.46}$$ But it seems to me that there is another term in $$\frac{\partial {\dot r_i}} {\partial {q_j}}$$ because of product rule which is $$\sum_k \frac{\partial{r_i}}{\partial{q_k}}\frac{\partial{\dot q_k}}{\partial{q_j}},$$ which I think is equal to $$\frac{\partial{r_i}}{\partial{q_j}}\frac{\partial{\dot q_j}}{\partial{q_j}}$$ since $$q_j$$'s are independent among themselves.

Then how come $$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}~?\tag{1.50b}$$ Does $$\frac{\partial{\dot q_j}}{\partial{q_j}} = 0\ ?$$

In the Lagrangian formalism position and velocity are considered as independent variables, so indeed $$\frac{\partial \dot{q}_j}{\partial q_j} = 0$$. See Calculus of variations -- how does it make sense to vary the position and the velocity independently?
• But why is $\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial {r_i}}{\partial {q_j}}\right) = \frac{\partial {\dot r_i}}{\partial {q_j}}$ ? Is it simply an interchange of derivatives? – Anjan May 4 '19 at 9:38
• Yes, you can exchange those derivatives. Consider $f(g(x,t),t)$. Then $\dot{f} = \frac{\partial f}{\partial g} \dot{g} + \frac{\partial f}{\partial t}$ and $\frac{\partial \dot{f}}{\partial g} = \frac{\partial^2 f}{\partial g^2} \dot{g} + \frac{\partial^2 f}{\partial g \partial t} = \frac{d}{dt} \frac{\partial f}{\partial g}$, where we used $\frac{\partial \dot{g}}{\partial g} = 0$. – jkb1603 May 14 '19 at 19:04