# Nabla on fraction $\nabla\left(\frac{1}{r}\right)=\left(\frac{d}{dr}\frac{1}{r}\right)\cdot(\nabla r)$

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?

## 1 Answer

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. $$\vec{v}=-\frac{\vec{r}}{r^3}=\operatorname{grad}\left(\frac{1}{r}\right)$$

• But you've implied the part I don't understand again - why is $\partial_ir^{-1}=-r^{-2}(\partial_ir)$? – MetaColon Apr 2 at 21:12
• @MetaColon It's the chain rule: $\frac{\partial r^{-1}}{\partial x_i}=\frac{\partial r^{-1}}{\partial r}\frac{\partial r}{\partial x_i}$ – Botond Apr 2 at 21:12
• Oh I see, it's because you firstly derive the outer and then the inner part. Thanks! – MetaColon Apr 2 at 21:13
• @MetaColon Yes. You're welcome! – Botond Apr 2 at 21:14
• @Botond very nice answer (+1) – the_candyman Apr 2 at 21:28