Dot product with a scalar function

Question

Example 5: Find $\nabla\cdot\left(\frac{\boldsymbol r}{r^2}\right)$ where $\boldsymbol r(x,y,z)=x\boldsymbol i+y\boldsymbol j+z\boldsymbol k$, $r=\sqrt{x^2+y^2+z^2}$.

In this case, why can we not bring the scalar function $r^{-2}$ outside the dot product?

When I leave the function in place, I calculate a result of $\frac{1}{r^{2}}$. When I pull it out, I find $\frac{3}{r^{2}}$.

Answer 1

It's not just a dot product, it's a divergence operator, and the variables $x,y,z$ appear in $r^{-2}$, it's not a constant.

However, you may use the product rule:

$$\nabla \cdot(\psi \boldsymbol F)=\psi \nabla\cdot \boldsymbol F+\nabla\psi\cdot \boldsymbol F$$

And with $\psi=r^{-2}$, it's not difficult to prove that

$$\nabla\psi=-\frac{2}{r^4}\boldsymbol r$$

Answer 2

$$\nabla \cdot \left( \frac{\vec{r}}{r^2} \right) = \nabla \cdot \left( \frac{x}{x^2+y^2+z^2}\vec{i} + \frac{y}{x^2+y^2+z^2}\vec{j} + \frac{z}{x^2+y^2+z^2}\vec{k} \right) \\ = \frac{y^2+z^2-x^2}{(x^2+y^2+z^2)^2} + \frac{x^2+z^2-y^2}{(x^2+y^2+z^2)^2} + \frac{x^2+y^2-z^2}{(x^2+y^2+z^2)^2} = \frac{x^2+y^2+z^2}{(x^2+y^2+z^2)^2} \\ = \frac{1}{x^2+y^2+z^2} = \frac{1}{r^2}.$$

Answer 3

It is only a scalar function , not a scalar. It has $x,y,z$ in it. Hence we can apply partial derivatives on $r^{-2}$ as well. So we must not bring it outside.