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# Use the notation $\mathbf{r}=\langle x, y\rangle$ and $r=\|\mathbf{r}\|=\sqrt{x^{2}+y^{2}}$. Show that $\nabla(\ln r)=\frac{\mathbf{r}}{r^{2}}$.

Solution

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The trick here is to use the fact that:

$r^2=x^2+y^2$

that is, we treat $r$ as a function $r(x,y)$. Now simply take the partial derivative with respect to $x$ of $r^2$ to get:

$\dfrac{\partial}{\partial x}r^2=2r\dfrac{\partial r}{\partial x}$

Taking the partial derivative with respect to $x$ of the right side we have:

$2r\dfrac{\partial r}{\partial x}=2x\Rightarrow \dfrac{\partial r}{\partial x}=\dfrac{x}{r}$

The same reasoning gives us:

$\dfrac{\partial r}{\partial y}=\dfrac{y}{r}$

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