electric field and potential of an infinite flat surface charge

another derivation question from griffith's intro to electrodynamics.

this is the equation for which i'm trying to follow the derivation for:

$$\frac{∂V}{∂n}=\vec{\nabla}V\cdot \hat{n}$$


electric field from an infinite flat surface charge density:

$$\vec{E}_{surface}=\frac{\sigma}{2\epsilon_{0}}\hat{n}$$

the discontinuitiy in the electric field across the surface:

$$\vec{E}_{above}-\vec{E}_{below}=\frac{\sigma}{\epsilon_{0}}\hat{n}$$

in terms of potential:

$$-\vec{\nabla}V_{above}-(-\vec{\nabla}V_{below})=\frac{\sigma}{\epsilon_{0}}\hat{n}$$

more explicitly.. since the only surviving component of the E-field is that which is perpendicular to the surface:

$$-\vec{\nabla}V_{above}+\vec{\nabla}V_{below}=-\frac{∂V_{above}}{∂n}\hat{n}+\frac{∂V_{below}}{∂n}\hat{n}$$



okay, so far so good. i get up to here. but, i don't get how..

$$-\frac{∂V_{above}}{∂n}\hat{n}+\frac{∂V_{below}}{∂n}\hat{n}$$

just gets reduced down to..

$$\frac{∂V}{∂n}$$

...

I remember a time in quantum mechanics when we expressed a discontinuity in this way (via the difference in derivatives), we didn't reduce this to just the derivative. we explicitly left the difference; like this:

$$\frac{∂f_{1}}{∂x}-\frac{∂V_{2}}{∂x}=Δ(\frac{∂f}{∂x})$$



so, in other words, why in this case (for the potential above and below), were we able to leave the delta out?



Thanks again!