Prove the given formula ($r=||{\textbf{r}}||$ is the length of the position vector field $\textbf{r}(x,y,z)=x\textbf{i}+y\textbf{j}+z\textbf{k}$).
$$\nabla \dfrac{1}{r} = \dfrac{-\textbf{r}}{r^3}$$
Since $r=||{\textbf{r}}||$, $\frac{1}{r}=\frac{1}{\sqrt{x^2+y^2+z^2}}$.
Then the gradient of this would be $(\frac{-2x}{2(x^2+y^2+z^2)^{\frac{3}{2}}}, \frac{-2y}{2(x^2+y^2+z^2)^{\frac{3}{2}}}, \frac{-2z}{2(x^2+y^2+z^2)^{\frac{3}{2}}})$,
which can be simplified to $\frac{-(x,y,z)}{(x^2+y^2+z^2)^{\frac{3}{2}}}$.
The denominator is now $r^3$, but is the numerator correct? Is this the end of the proof, or is there another step?
Thanks.
