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Use the notation r=x,y\mathbf{r}=\langle x, y\rangle and r=r=x2+y2r=\|\mathbf{r}\|=\sqrt{x^{2}+y^{2}}. Show that (lnr)=rr2\nabla(\ln r)=\frac{\mathbf{r}}{r^{2}}.

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Answered 2 years ago
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The trick here is to use the fact that:

r2=x2+y2r^2=x^2+y^2

that is, we treat rr as a function r(x,y)r(x,y). Now simply take the partial derivative with respect to xx of r2r^2 to get:

xr2=2rrx\dfrac{\partial}{\partial x}r^2=2r\dfrac{\partial r}{\partial x}

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

2rrx=2xrx=xr2r\dfrac{\partial r}{\partial x}=2x\Rightarrow \dfrac{\partial r}{\partial x}=\dfrac{x}{r}

The same reasoning gives us:

ry=yr\dfrac{\partial r}{\partial y}=\dfrac{y}{r}

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