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This question was inspired by the Lagrange equation, $\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = 0$. What happens if the partial derivatives are replaced by total derivatives, leading to a situation where a function's derivative with respect to one variable is differentiated by the original function?

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up vote 6 down vote accepted

Write $$\frac{d}{dx}\left(\frac{dx}{dt}\right)=\frac{d}{dx}\left(\frac{1}{\frac{dt}{dx}}\right)$$ and use the chain rule. $$\frac{d}{dx}\left(\frac{dx}{dt}\right)=\left(\frac{-1}{(\frac{dt}{dx})^2}\right)\frac{d^2t}{dx^2}=-\left (\frac{dx}{dt}\right )^2\frac{d^2t}{dx^2}$$

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Alternatively, the chain rule $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}$ gives $$\frac{d}{dx}(\frac{dx}{dt}) = \frac{d}{dt}(\frac{dx}{dt})\frac{dt}{dx} = \frac{d^2x}{dt^2}\frac{dt}{dx}.$$ Of course, this is the same as Ross' answer above. That is, we have the identity $$-\left(\frac{dx}{dt}\right)^2\frac{d^2t}{dx^2} = \frac{d^2x}{dt^2}\frac{dt}{dx},$$ which follows from differentiating the equation $\frac{dx}{dt}\frac{dt}{dx} = 1$ with respect to $t$.

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$\frac{\partial L}{\partial \dot{q}}$ may be understood as the generalized momentum of the system (e.g. $\frac{d}{dv}\left(\frac{1}{2}mv^2\right) = mv$, the derivative of kinetic energy with respect to velocity is momentum). Then $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ is the time derivative of the generalized momentum. Likewise the $\frac{\partial L}{\partial q}$ term behaves like a generalized force. This begins to look like Newton's second law, but for generalized coordinates.

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