For a system of multiple particles with conservative forces: $\mathbf{F}_i = - \nabla_i V$, with $V \equiv V(\mathbf{r}_1,\dots,\mathbf{r}_N)$ the potential in function of the position of the $N$ particles.
When considering constraints, we can transform our Cartesian coordinates to generalized coordinates. This results in the potential being a function of these generalized coordinates: $V(\mathbf{r}_1(q_k),\dots,\mathbf{r}_N (q_k))$.
For the generalized force $Q_k$ we find that $Q_k = - \frac{\partial V}{\partial q_k}$. Now my book says that
$$ \frac{\partial V}{\partial q_k} = \sum_i (\nabla_i V) \cdot \frac{\partial\mathbf{r}_i}{\partial q_k} \quad (1)$$
I understand where this formula comes from (chain rule). I'm also aware of what $\nabla_i$ means: $\nabla_i = (\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_i},\frac{\partial}{\partial z_i})$ for partial derivation w.r.t. coordinates of the $i$-th particle. Let's write the full sum (1):
$$\frac{\partial V}{\partial q_k} = (\nabla_1 V) \cdot \frac{\partial \mathbf{r}_1}{\partial q_k} + \dots + (\nabla_N V) \cdot \frac{\partial \mathbf{r}_N}{\partial q_k}.$$
I don't understand what the $\nabla_i$ look like when written full out. For example $\nabla_1 V$, what is this equal to? Do I just take $\mathbf{r}_1$? But the nabla operator has three components, so what's up with them? How do I write these $\nabla_i$ out?
