How does $F=\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n c_{ij}\dot q_i\dot q_j\implies\frac{\partial{\ F}}{\partial\dot{q_r}}=\sum_{j=1}^nc_{q_rj}\dot q_j$?

Question

How does $F=\frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n c_{ij}\dot q_i\dot q_j\implies\frac{\partial{\ F}}{\partial\dot{q_r}}=\sum_{j=1}^nc_{q_rj}\dot q_j$?

I'm not getting how $c_{ij} $ gets transformed to $c_{q_rj}$ and how does the factor $\frac{1}{2}$ vanish...

The original source is https://www.kvi.nl/~scholten/AAM/Friction-Lagrange.pdf

Answer 1

$c_{q_r j}$ makes no sense to me, the correct answer would be $c_{rj}$ since $\frac{\partial \dot q_i}{\partial \dot q_r}=\delta_{ir}$.

For the $1/2$ factor, try to apply the product rule of derivatives.

Answer 2

I think that it is just an abuse of notation to highlight the variable $q_r$ and it's equivalent to

$$ \frac{\partial F}{\partial \dot q_r}=\sum_{j=1}^{n}c_{rj}\dot q_j=-Q^N_r $$