Let's suppose that we have a three dimensional function $f(\vec{x})$ which is the integral of some another function $g(\vec{x},\vec{y})$, i.e
$f(\vec{x})=\int_{\mathbb{R}^3}g(\vec{x},\vec{y})d^3 \vec{y}$
What is the gradient of the $f(\vec{x})$? Can the operator pass inside the integral?
$\nabla f(\vec{x})=\nabla\int_{\mathbb{R}^3}g(\vec{x},\vec{y})d^3 \vec{y}=\int_{\mathbb{R}^3}\left[\nabla g(\vec{x},\vec{y})\right]d^3 \vec{y}$
The quantity $\nabla g(\vec{x},\vec{y})$ is a vector and it doesn't make sense to me integrating a vector.
In the case of the Laplacian operator $\nabla^2$ can it pass inside the integral?
Edit: The question was inspired from a physics problem where we have a potential $V(\textbf{x})=-\int_{\mathbb{R}^3}\frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})d^3\textbf{y}$ and we take a gradient to find the accelaration: $g(\textbf{x})=-\nabla V(\textbf{x})=\nabla\int_{\mathbb{R}^3}\frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})d^3\textbf{y}$.