In connection with introduction to electro dynamics, we deduced that $$\frac{\vec r}{r^3}=-\nabla\left(\frac{1}{r}\right)$$ This is fairly obvious for three dimensions, but we did it in $n$ dimensions.
In the process of derivation, there was the following step included: $$-\nabla\left(\frac{1}{r}\right)=-\left(\frac{d}{dr}\frac{1}{r}\right)\cdot(\nabla r)$$ Once again, the correctness of this can be easily checked in three dimensions, but it seemed kind of obvious that this is generally true. Is this a commonly known rule? And how does one derive it?
Let $$\vec{v}=\operatorname{grad}\left(\frac{1}{r}\right)$$ Then we have that $$\begin{align} v_i&=\partial_i \frac{1}{r}\\ &= \partial_i r^{-1}\\ &= -r^{-2} (\partial_i r)\\ &= -r^{-2} (\partial_i \sqrt{x_jx_j})\\ &= -r^{-2} \frac{1}{2\sqrt{x_jx_j}} \partial_i (x_jx_j)\\ &= -r^{-2} \frac{1}{2 r} 2(\partial_ix_j)x_j\\ &=-r^{-3} \delta_{ij}x_j\\ &=-\frac{x_i}{r^3} \end{align}$$ i.e. $$\vec{v}=-\frac{\vec{r}}{r^3}=\operatorname{grad}\left(\frac{1}{r}\right)$$