 Gradient of a radial function

For $r=\sqrt{x_1^2+\cdots +x_d^2}>0$, the multivariable chain rule givesIndeed, $\partial r/\partial x_j=x_j/r$. Thus the gradient points radially outward when $g^{\prime }(r)>0$, inward when it is negative, and vanishes when it is zero.

At the origin the displayed quotient does not apply. If $g$ has a finite right derivative at zero, the radial function is differentiable there exactly when $g^{\prime }(0^{+})=0$, in which case its gradient is zero. Necessity follows by comparing opposite directional difference quotients: their linear parts cannot both equal a common nonzero radial slope. Sufficiency follows from $g(r)-g(0)=o(r)$, the definition of differentiability with zero linear part.