Answer
Using the Chain Rule we compute the derivatives of $u$ with respect to $r$ and $\theta$ and obtain Eq. (8):
$||\nabla u|{|^2} = {u_r}^2 + \frac{1}{{{r^2}}}{u_\theta }^2$
Work Step by Step
In polar coordinates, we have the relations:
$x = r\cos \theta $, ${\ \ }$ $y = r\sin \theta $
So, $u = u\left( {x\left( {r,\theta } \right),y\left( {r,\theta } \right)} \right)$.
Using the Chain Rule we compute the derivatives of $u$ with respect to $r$ and $\theta$:
1. $\frac{{\partial u}}{{\partial r}} = \frac{{\partial u}}{{\partial x}}\frac{{\partial x}}{{\partial r}} + \frac{{\partial u}}{{\partial y}}\frac{{\partial y}}{{\partial r}}$
${u_r} = {u_x}\cos \theta + {u_y}\sin \theta $
2. $\frac{{\partial u}}{{\partial \theta }} = \frac{{\partial u}}{{\partial x}}\frac{{\partial x}}{{\partial \theta }} + \frac{{\partial u}}{{\partial y}}\frac{{\partial y}}{{\partial \theta }}$
${u_\theta } = - r{u_x}\sin \theta + r{u_y}\cos \theta $
Next, we compute ${u_r}^2 + \frac{1}{{{r^2}}}{u_\theta }^2$
${u_r}^2 + \frac{1}{{{r^2}}}{u_\theta }^2 = {\left( {{u_x}\cos \theta + {u_y}\sin \theta } \right)^2} + \frac{1}{{{r^2}}}{\left( { - r{u_x}\sin \theta + r{u_y}\cos \theta } \right)^2}$
$ = {u_x}^2{\cos ^2}\theta + 2{u_x}{u_y}\cos \theta \sin \theta + {u_y}^2{\sin ^2}\theta $
${\ \ \ }$ $ + \frac{1}{{{r^2}}}\left( {{r^2}{u_x}^2{{\sin }^2}\theta - 2{r^2}{u_x}{u_y}\sin \theta \cos \theta + {r^2}{u_y}{{\cos }^2}\theta } \right)$
$ = {u_x}^2{\cos ^2}\theta + 2{u_x}{u_y}\cos \theta \sin \theta + {u_y}^2{\sin ^2}\theta $
${\ \ \ }$ $ + {u_x}^2{\sin ^2}\theta - 2{u_x}{u_y}\sin \theta \cos \theta + {u_y}{\cos ^2}\theta $
$ = {u_x}^2 + {u_y}^2$
Since $\nabla u = \left( {{u_x},{u_y}} \right)$, so $||\nabla u|{|^2} = {u_x}^2 + {u_y}^2$.
Hence, $||\nabla u|{|^2} = {u_r}^2 + \frac{1}{{{r^2}}}{u_\theta }^2$.
