Consider a system of $N$ particles subject to forces $\vec F_i\ (i=1\dots N)$ that derive from a potential $V$. My lecture notes propose a simple proof that
$$Q_j = -\frac{\partial V}{\partial q_j}$$
where the generalized forces are defined as $Q_j = \sum_i \vec F_i\cdot\frac{\partial\vec r_i}{\partial q_j}$. It goes like this:
$$ Q_j = \sum_i \vec F_i\cdot\frac{\partial\vec r_i}{\partial q_j} = -\sum_i\vec\nabla_i V\cdot\frac{\partial\vec r_i}{\partial q_j} = -\frac{\partial V}{\partial q_j} $$
I'm trying to understand the last step in detail, but I get a wrong answer by a factor $N$. For example with two particles, and writing $\vec r_i = (x_i,y_i,z_i)$, I have $$ \begin{aligned} \sum_i\vec\nabla_i V\cdot\frac{\partial\vec r_i}{\partial q_j} &= \vec\nabla_1V\cdot\frac{\partial\vec r_1}{\partial q_j} + \vec\nabla_2V\cdot\frac{\partial\vec r_2}{\partial q_j} \\ &= (\tfrac{\partial V}{\partial x_1}, \tfrac{\partial V}{\partial y_1}, \tfrac{\partial V}{\partial z_1}) \cdot(\tfrac{\partial x_1}{\partial q_j}, \tfrac{\partial y_1}{\partial q_j}, \tfrac{\partial z_1}{\partial q_j}) + (\tfrac{\partial V}{\partial x_2}, \tfrac{\partial V}{\partial y_2}, \tfrac{\partial V}{\partial z_2}) \cdot(\tfrac{\partial x_2}{\partial q_j}, \tfrac{\partial y_2}{\partial q_j}, \tfrac{\partial z_2}{\partial q_j}) \\[1ex] &= \frac{\partial V}{\partial q_j} + \frac{\partial V}{\partial q_j} \\[1ex] &= 2\frac{\partial V}{\partial q_j} \end{aligned} $$ What did I do wrong to get this factor 2?