Dirac delta and derivative inside an integral In development of a calculus (in GR but it doesn't matter here) I have seen one dubious substitution: in an integral of the form: 
$$\int dx ~\delta(x-x_0)~\partial_x F(x) $$ 
The author substitutes $\partial_x$ for $-\partial_{x_0}$ and I honestly don't see where this minus sign is coming from. The integral only takes account when $x - x_0 = 0$ so $dx - dx_0 = 0$ implying $dx = dx_0$. Finally giving something of the form $\frac{d}{dx}=\frac{d}{dx_0}$ and the same for partial derivatives.
 A: To expand on the comment by wsc, we could write
$$
\int\,dx\,\delta(x-x_0) \partial_x F(x) = -\int dx \partial_x \delta(x-x_0) F(x)\,,
$$
and then pick our favorite representation of the $\delta$ function.  Let's say
$$
\lim_{\epsilon \searrow 0} \frac{1}{\sqrt{2\pi\epsilon}}\,e^{-(x-x_0)^2/2\epsilon} = \delta(x - x_0)\,,
$$
and then write
$$
\int dx \delta(x-x_0) \partial_x F(x) = -\lim_{\epsilon \searrow 0} \int dx  \frac{1}{\sqrt{2\pi\epsilon}}\,\partial_x e^{-(x-x_0)^2/2\epsilon} F(x)\,.
$$
Since we see that
$$
\partial_x e^{-(x-x_0)^2/2\epsilon} = -\partial_{x_0} e^{-(x-x_0)^2/2\epsilon}\,,
$$
we can write
$$
\int dx \delta(x-x_0) \partial_x F(x) = \lim_{\epsilon \searrow 0} \int dx  \frac{1}{\sqrt{2\pi\epsilon}}\,\partial_{x_0} e^{-(x-x_0)^2/2\epsilon} F(x)\,,
$$
and conclude
$$
\int dx \delta(x-x_0) \partial_x F(x) = \partial_{x_0} \lim_{\epsilon \searrow 0} \int dx  \frac{1}{\sqrt{2\pi\epsilon}} e^{-(x-x_0)^2/2\epsilon} F(x) = \partial_{x_0} \int dx \delta(x-x_0) F(x) = F^\prime(x_0)\,.
$$
