The Cauchy-Schwarz inequality for regular two electron integrals is given as:
$$\left(\left.ab\right|cd\right)\leq\sqrt{\left(\left.ab\right|ab\right)}\sqrt{\left(\left.cd\right|cd\right)}$$
Now this can be differentiated with respect to $x$:
$$\frac{\partial}{\partial x}\left(\left.ab\right|cd\right)\leq\frac{\partial}{\partial x}\sqrt{\left(\left.ab\right|ab\right)}\sqrt{\left(\left.cd\right|cd\right)}$$
By using the chainrule for $\frac{d\sqrt{f(x)}}{dx}=\frac{df(x)}{dx}\frac{1}{2\sqrt{f(x)}}$, now giving:
$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)+\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)+\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)+\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{2\sqrt{\left(\left.ab\right|ab\right)}}\left[\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)+\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)+\left(ab\left|\frac{\partial a}{\partial x}b\right.\right)+\left(ab\left|a\frac{\partial b}{\partial x}\right.\right)\right]+\frac{\sqrt{\left(\left.ab\right|ab\right)}}{2\sqrt{\left(\left.cd\right|cd\right)}}\left[\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)+\left(\left.c\frac{\partial d}{\partial x}\right|cd\right)+\left(cd\left|\frac{\partial c}{\partial x}d\right.\right)+\left(cd\left|c\frac{\partial d}{\partial x}\right.\right)\right]$$
By using the symmetry of the two electron integrals $(ab|cd)=(cd|ab)$, it can simplified to:
$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)+\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)+\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)+\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left[\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)+\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)\right]+\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left[\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)+\left(\left.c\frac{\partial d}{\partial x}\right|cd\right)\right] \tag{1}$$
Now by comparing terms it can be seen that:
$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)$$
$$\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)$$
$$\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)\leq\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)$$
$$\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left(cd\left|c\frac{\partial d}{\partial x}\right.\right)$$
For geometrical detrivatices $x$ would be a cartesian direction of a nuclie. Now my question is, for geometrical derivatives when would be the right point to use the Cauchy-Schwarz inequlity for screening the integrals?
I think I could either $(1)$ and screen the total derivative $\frac{\partial}{\partial x}(ab|cd)$, or use the last four equations and screen the four different (up to three for geometrical derivatives) contributions individually.
Also what would be a reasonable threshold for derivatives?
For regular integrals $10^{-10}$ or lower is often used.