What is the gradient of an integral?

Question

I have a scalar potential at position $r$ due to a source at position $r'$ $$\phi(\vec r)=-\frac{1}{4\pi}\int_v\frac{\nabla'\vec M}{\vert{\vec {r}-\vec r'} \vert} d^3\vec r'+\frac{1}{4\pi}\int_s\frac{\vec M\cdot \hat n}{\vert{\vec {r}-\vec r'} \vert} d^2\vec r'$$ where $\hat n$ is a direction vector. And I want to derive the corresponding field at position $r$ $$ H=-\nabla\phi(\vec r)$$ where$\nabla'$ refers to the divergence at $r'$ and $\nabla$ refers to the gradient at $r$. According to the reference the field is $$H=-\frac{1}{4\pi}\int_v\frac{(\vec r-\vec r')\nabla'\vec M}{\vert{\vec {r}-\vec r'} \vert^3} d^3\vec r'+\frac{1}{4\pi}\int_s\frac{(\vec r-\vec r')\vec M\cdot \hat n}{\vert{\vec {r}-\vec r'} \vert^3} d^2\vec r' $$ Is this expression correct? How to derive it? Many thanks!

Answer 1

It is correct, the quantity that depends on $\vec{r}$ in the integrals is only $\frac{1}{|\vec{r}-\vec{r}' |}$, so you use $$ \nabla \left(\frac{1}{|\vec{r}-\vec{r}' |}\right) = -\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}' |^3}$$

EDIT: maybe you also want to know that the operator $\nabla$ can be passed inside the integrals