Pass the derivative under integral

Question

Suppose we have an operator $T: C^{\infty}(\mathbb{R}^n) \to C^{\infty}(\mathbb{R}^n)$ such that it has a singular integral representation \begin{equation*} Tf = \int_{\mathbb{R}^n}K(x,y)f(y)dy \end{equation*} where $K(x,y)$ is singular when $y = x$ and and smooth when $y \neq x$. Moreover, assume that $K(x,y)$ is rapid decreasing when $\lvert x - y \rvert \to 0$. Can we pass the gradient of $Tf$ inside the integral, i.e., \begin{equation*} \nabla Tf = \int_{\mathbb{R}^n}K(x,y) \nabla f(y)dy \end{equation*} I don't know whether this is true or not. If it is true, I'd also like to see how one invokes Dominated Converence Theorem in this case. In reality, I'm trying to show this is true for a pseudo-differential operator of order 1 on a compact, boundaryless Riemannian manifold. Any insight would be appreciated.