Timeline for 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?

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May 14, 2019 at 19:04	comment	added	jkb1603		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 $\frac{\partial \dot{g}}{\partial g} = 0$.
May 4, 2019 at 9:38	comment	added	Anjan		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?
Dec 11, 2018 at 13:49	vote	accept	Sameer Baheti
Dec 11, 2018 at 12:29	history	answered	jkb1603	CC BY-SA 4.0