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$\begingroup$

So this equality was given as a part of how to derive the Lagrangian equality from Newton's second law. How it is true?

$$\frac{d\pmb{p}_k}{dt}\cdot\frac{\partial\pmb{r}_k}{\partial q_i} = \frac{d}{dt}\left(\pmb{p}_k\cdot\frac{\partial\pmb{r}_k}{\partial q_i}\right) - \pmb{p}_k\cdot\frac{d}{dt}\frac{\partial\pmb{r}_k}{\partial q_i}$$

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    $\begingroup$ This is equal because $\dfrac{d}{dt}\,\left( x\cdot y\right) =\dfrac{d}{dt}\,\left( x\right) \cdot y+x\cdot \dfrac{d}{dt}\, \left( y\right) $ $\endgroup$ – Eli Oct 2 '19 at 18:28

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