Prove that $\vec{\nabla} r^n = n r^{n-2} \vec{\tilde{r}}$

Question

I need to prove that $\vec{\nabla} r^n = n r^{n-2} \vec{\tilde{r}}$

I have a demonstration using spherical coordinates in my notebook, but I would like to know how to do it the standard way (cartesian coordinates) if possible.

If not, can someone write it in spherical, to test my notes?

Answer 1

Hint: Observe in Cartesian coordinate system position vector $\vec r:=xi+yj+zk$, so $r=\sqrt{x^2+y^2+z^2}$. Thus $r^n=(x^2+y^2+z^2)^{\frac{n}{2}}$. And $\vec{\nabla}:= \frac{\partial}{\partial x}i+ \frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k$.
Now try to compute $\vec {\nabla}r^n$.