The normal derivative of a function $u(x)$ on the surface of a sphere of radius $r$, i.e. $r = |x|$ can be represented in spherical polar coordinates as $$ \begin{align} \frac{\partial u}{\partial \nu} & = \nu \cdot \nabla u \\ & = \hat r \cdot \nabla \\ & = \hat r \cdot \left(\frac{\partial u}{\partial r} \hat r + \frac{1}{r}\frac{\partial u}{\partial \theta} \hat \theta + \frac{1}{r^2\sin \theta}\frac{\partial u}{\partial \phi} \hat \phi\right) \\ & = \frac{\partial u}{\partial r}. \end{align} $$
I want to obtain the same result using a different approach so I did the following: $$ \begin{align} \frac{\partial u}{\partial \nu} & = \nu \cdot \nabla u \\ & = \frac{x}{|x|} \cdot \begin{pmatrix} \displaystyle \frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2}, \frac{\partial u}{\partial x_1} \end{pmatrix} \\ & = \mathbf{\left(\frac{x_1}{r}, \frac{x_2}{r}, \frac{x_3}{r} \right)} \cdot \begin{pmatrix} \displaystyle \frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2}, \frac{\partial u}{\partial x_1} \end{pmatrix} \end{align} $$ So this should be equal to the representation for the normal derivative above, i.e. $\frac{\partial u}{\partial r}$. Using the chain rule on $\frac{\partial u}{\partial r}$ I get $$ \begin{align} \frac{\partial u}{\partial r} & = \frac{\partial u}{\partial x_1}\frac{d x_1}{d r} + \frac{\partial u}{\partial x_2}\frac{d x_2}{d r} + \frac{\partial u}{\partial x_3}\frac{d x_3}{d r} \\ & = \frac{dx}{dr} \cdot \begin{pmatrix} \displaystyle \frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2}, \frac{\partial u}{\partial x_1} \end{pmatrix} \end{align} $$
Then $r = |x| = \sqrt{x_1^2 + x_2^2 + x_3^2}$ means $x_1 = \sqrt{r^2 - x_2^2 - x_3^2}$. This gives $\displaystyle \frac{d x_1}{dr} = \frac{r}{x_1}$ and so $$ \frac{\partial u}{\partial r} = \mathbf{\left(\frac{r}{x_1}, \frac{r}{x_2}, \frac{r}{x_3} \right)} \cdot \begin{pmatrix} \displaystyle \frac{\partial u}{\partial x_1}, \frac{\partial u}{\partial x_2}, \frac{\partial u}{\partial x_1} \end{pmatrix}. $$ So the numerator and denominator are switched when I take this approach! Where am I going wrong?