I need clarification on how this identity goes about. I understand the first four equalities but am not sure how $\partial u / \partial r$ arises from the second to last equality.
$$ \frac{\partial u}{\partial n} = \mathbf{n}\cdot \nabla u =\frac{\mathbf x}{r}\cdot \nabla u =\frac{x}{r} u_x + \frac{y}{r} u_y + \frac{z}{r} u_z =\frac{\partial u}{\partial r} $$
Thank you for the clarifications.
Note: this is taken from “Partial Differential Equations: An Introduction” by Walter A. Strauss
On
$$ \frac{\partial u}{\partial r} = \frac{\partial x}{\partial r}\frac{\partial u}{\partial x} + \frac{\partial y}{\partial r}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial r}\frac{\partial u}{\partial z} $$ we know that $r = \sqrt{x^2+y^2+z^2}$ thus $$ \frac{\partial r}{\partial x} = \frac{\partial }{\partial x} \sqrt{x^2+y^2+z^2} = \frac{1}{2}\frac{1}{\sqrt{x^2+y^2+z^2}}\cdot 2x = \frac{x}{r} $$ By symmetry we have similar expressions for the other dimensions. This leads us to this. $$ \frac{\partial u}{\partial r} = \frac{x}{r}\frac{\partial u}{\partial x} + \frac{y}{r}\frac{\partial u}{\partial y} + \frac{z}{ r}\frac{\partial u}{\partial z} $$
Let $\xi(r,\theta,\varphi)=(r\sin\theta\cos\varphi,r\sin\theta\sin\varphi,r\cos\theta)$, and consider $u\circ\xi$, we have \begin{align*} \dfrac{\partial(u\circ\xi)}{\partial r}&=\dfrac{\partial u}{\partial x}\dfrac{\partial\xi}{\partial r}+\dfrac{\partial u}{\partial y}\dfrac{\partial\xi}{\partial r}+\dfrac{\partial u}{\partial z}\dfrac{\partial\xi}{\partial r}\\ &=\sin\theta\cos\varphi\cdot u_{x}+\sin\theta\sin\varphi\cdot u_{y}+\cos\theta\cdot u_{z}\\ &=\dfrac{x}{r}\cdot u_{x}+\dfrac{y}{r}\cdot u_{y}+\dfrac{z}{r}\cdot u_{z}, \end{align*} essentially, $\dfrac{\partial u}{\partial r}$ in this context means $\dfrac{\partial(u\circ\xi)}{\partial r}$.
Indeed, if you choose other coordinate system instead of $\xi$, it should lead you to the same result.