How is the derivative of a potential with respect to an outward normal equal to the grad of the potential

Question

If we first consider Gauss' Law $$\oint_s \boldsymbol{E\cdot} \,d\boldsymbol{A} = Q_{enclosed\\in\ surface}$$ We know from physics that $\boldsymbol{E}=-\nabla V$, but I want to know is it mathematically equivalent to say $-\partial V/\partial n = \nabla V$ -- and if so, how? Here, $n$ is the outward normal from the enclosed surface.