 Solution

For $r=\sqrt{x^2+y^2+z^2}>0$, the multivariable chain rule gives ${\partial }_{x}g(r)=g^{\prime }(r)x/r$, with the corresponding expressions for $y,z$. Thus the gradient of a radial function isThe origin needs a separate check because $\hat{\mathbf{r}}$ is undefined there. If the one-variable function has a finite right derivative at zero, the radial composite is differentiable at the origin exactly when $g^{\prime }(0^{+})=0$, in which case its gradient is zero. A nonzero radial slope gives incompatible opposite-direction derivatives.