= Solution
Take the <electric potential> to vanish at infinity and work at $r>0$. For a positive <point charge> $q$ in vacuum,
$$
\boxed{V(r)=\frac{q}{4\pi\epsilon_0r},\qquad\mathbf E(r)=\frac{q}{4\pi\epsilon_0r^2}\hat{\mathbf r}}.
$$
The field is radial and outward. Since a radial scalar function has <gradient> $\nabla V=(dV/dr)\hat{\mathbf r}$,
$$
-\nabla V=-\left(-\frac{q}{4\pi\epsilon_0r^2}\right)\hat{\mathbf r}=\mathbf E.
$$
Thus the two expressions have the required vector relationship. Adding an arbitrary constant to $V$ would leave the <electric field> unchanged; the chosen reference at infinity fixes that constant. The point-charge singularity itself is excluded from the ordinary <derivative> calculation.
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