Prove $\nabla\!\left(\frac{1}{|\vec{r}|}\right) = -\frac{\vec{r}}{|\vec{r}|^3}$

Question

Prove $\nabla\!\left(\frac{1}{|\vec{r}|}\right) = -\frac{\vec{r}}{|\vec{r}|^{3}}$

I was doing simple differentiation, and getting $\nabla\!\left(\frac{1}{|\vec{r}|}\right) = -{|\vec{r}|} ^{-2}\nabla{|\vec{r}|}$ what I am doing wrong as I need another $\vec{r}$ to get the power as 3?

Thank you.

Answer 1

$$\frac{\partial}{\partial x}\frac{1}{r}= -\frac{1}{r^{2}}\cdot\frac{\partial}{\partial x} r= -\frac{1}{r^2}\cdot\frac{\partial}{\partial x}\sqrt{x^2+y^2+z^2}= $$ $$=-\frac{1}{r^2}\frac{2x}{2\sqrt{x^2+y^2+z^2}}=-\frac{1}{r^2}\frac{x}{r}=-\frac{x}{r^3}$$ For $y$ and $z$ we have the same, so: $$\nabla \frac{1}{r}=\left(-\frac{x}{r^3},-\frac{y}{r^3},-\frac{z}{r^3}\right)^\top= -\frac{1}{r^3}(x,y,z)^{\top}=-\frac{\textbf r}{r^3}$$

Answer 2

$$ \nabla\frac{1}{|\vec{r}|} = -\frac{\nabla|\vec{r}|}{|\vec{r}|^2} = -\frac{\vec{r}/|\vec{r}|}{|\vec{r}|^2} = -\frac{\vec{r}}{|\vec{r}|^3} $$