How $\nabla_x f(x,y) = -2y$ holds when $f(x,y) = 2x \cdot y$

Question

Let $x = (x_1, x_2, x_3), y = (y_1, y_2, y_3)\in\mathbb R^3$ and $f(x,y) = 2x \cdot y$. It is written in the paper that $$ \nabla_x f(x,y) = -2y. $$ I cannot understand how it holds. I think it is just $2y$.

Answer 1

For arbitrary functions $g(x,y)$ which is differentiable in $x$ we have $$ \nabla_xg = \left( \begin{matrix} \frac{\partial g}{\partial x_1} \\ \frac{\partial g}{\partial x_2} \\ \frac{\partial g}{\partial x_3}\end{matrix} \right)$$ As $f(x,y) = 2\sum_{i=1}^3 x_i y_i$, we have $$\frac{\partial f}{\partial x_j} = 2\sum_{i=1}^3 \frac{\partial}{\partial x_i} x_j y_j = 2 y_i$$ So $\nabla_x f = 2y$, unless somewhere in the paper they define $\nabla_x$ to be minus the 'normal' $\nabla_x$.