Gradient of an integral 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?
$\nabla f(\vec{x})=\nabla_x\int_{\mathbb{R}^3}g(\vec{x},\vec{y})d^3 \vec{y}=\int_{\mathbb{R}^3}\left[\nabla_x g(\vec{x},\vec{y})\right]d^3 \vec{y}$
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: $g(\textbf{x})=-\nabla V(\textbf{x})=\nabla_x\int_{\mathbb{R}^3}\frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})d^3\textbf{y}$.

The operator $\nabla$ can be passed inside the integral if some suitable conditions on $g$  are fulfilled. There are appropriate theorems on differentiating integrals with respect to parameter. It can be done with potential $V\;$ if the function $\rho$ (say) is bounded in $\mathbb  R^3$ and has bounded support, since in this case the integral 
$\int_{\mathbb{R}^3}\nabla{\!\!}_{x}\frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})d^3\textbf{y}$ will be converging absolutely and uniformly.
The Laplace operator cannot be put inside the integral because it would mean that $\Delta V(x)\equiv0$ and for smooth enough $\rho$  actually $\Delta V(x)=\rho(x)$. The theorem aplied above doesn't work because the integral $\int_{\mathbb{R}^3}\left|\Delta \frac{G}{|\textbf{x}-\textbf{y}|}\rho(\textbf{y})\right| d^3\textbf{y}$ may diverge. The expression $\left|\Delta \frac{1}{|\textbf{x}-\textbf{y}|}\right|$  has non-integrable singularity at $y=x$.
