 Gradient in cylindrical coordinates

For a differentiable scalar field and cylindrical coordinates with $r>0$, the gradient has the displayed components in the local orthonormal basis. The chain rule applied to $r=\sqrt{x^2+y^2}$ and a local angle $\theta$ gives $r_x=\cos \theta$, $r_y=\sin \theta$, ${\theta }_x=-\sin \theta /r$, and ${\theta }_y=\cos \theta /r$. Substitution in $\nabla \varphi ={\varphi }_x{\mathbf{e}}_x+{\varphi }_y{\mathbf{e}}_y+{\varphi }_z{\mathbf{e}}_z$ proves the formula. The angular factor $1/r$ appears because an angular change $d\theta$ sweeps arc length $rd\theta$.