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I'm not sure if I'm mistaken: but my notes say the following.
A scalar field $f$ is radial if $f({\textbf{x}}) = \phi ( ||{\textbf{x}}||)$ for some $\phi : [0,\infty) \rightarrow \mathbb{R}$.
I understand this definition, but then it goes on to say:

$$\nabla f({\textbf{x}}) = \phi ' (||{\textbf{x}}||) \frac{{\textbf{x}}}{||{\textbf{x}}||}$$ is radial. I don't see how I can pick a function $g$ say that will show that $$\nabla f({\textbf{x}}) = g(||{\textbf{x}}||).$$

Twenty-six colours
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    The gradient is not a scalar field. "Radial scalar field" and "Radial vector field" requires different definitions. If the book hasn't defined radial vector fields yet, then that's bad; it should have. – Arthur May 27 '17 at 11:47
  • To add to the above, a simple definition of a radial vector field is as follows: A vector field $\mathbf{F}(\mathbf{x})$ is radial iff $\mathbf{F}(\mathbf{x}) = k(\mathbf{x}) \cdot \dfrac{\mathbf{x}}{\lVert \mathbf{x} \rVert}$ for some scalar-valued function $k(\mathbf{x})$. Intuitively, in a radial vector field, the vector assigned to any point points directly away from the origin. – cemulate May 27 '17 at 20:56
  • Thanks guys. Isn't the definition that $F$ is radial vector field iff $F(x) = \phi(||x||)x$ for some scalar valued function $\phi$? Or can that be transformed to your definition? – Twenty-six colours May 28 '17 at 03:42
  • Also, about the intuition, does that have anything to do with the normal to a sphere/circle with centre at the origin? Iirc, the normal to a sphere/circle is in the form $\frac{x}{r}$ where $r$ is the radius of the sphere/circle. – Twenty-six colours May 28 '17 at 03:43

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