I know that 

$$U= \frac{1}{2} \sum_{j,k} A_{jk} q_j q_k \quad \quad  T= \frac{1}{2} \sum_{j,k} m_{jk} \dot{q}_j \dot{q}_k $$

and the lagrangian is 

$$ \frac{\partial U}{\partial q_k} - \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_k} = 0$$

If I derivate 

$$ \frac{\partial U}{\partial q_k} = \frac{\partial}{\partial q_k} (\frac{1}{2} \sum_{j,k} A_{jk} q_j q_k) = \frac{1}{2} \sum_{j,k} A_{jk} (\frac{\partial q_j}{\partial q_k} \delta_{jk} \quad q_k + q_j \quad \frac{\partial q_k}{\partial q_k} \delta_{kk} )$$

and

$$ \frac{\partial T}{\partial \dot{q}_k} = \frac{\partial}{\partial \dot{q}_k} (\frac{1}{2} \sum_{j,k} m_{jk} \dot{q}_j \dot{q}_k) = \frac{1}{2} \sum_{j,k} m_{jk} (\frac{\partial \dot{q}_j}{\partial \dot{q}_k} \delta_{jk} \quad \dot{q}_k + \dot{q}_j \frac{\partial \dot{q}_k}{\partial \dot{q}_k} \delta_{kk})$$

but, the results is 

$$ \frac{\partial U}{\partial q_k} = \sum_{j} A_{jk} q_j $$

and 

$$ \frac{\partial T}{\partial \dot{q}_k} = \sum_{j} m_{jk} \dot{q}_j $$

I don't understand this, how this happened ? Where is $\frac{1}{2}$?

This is a passage from Marion (Classical Mechanics), I found it curious and tried to solve it, because the derivative was meaningless to me, and the subindexes are very confusing to understand.

The lagrangian is $$ \frac{\partial L}{\partial q_k} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_k} = 0$$ but it uses this other relation with $U$ and $T$, How can this be valid?