Verify normal derivative of a sphere.

I’ve search through many books and went online without any luck, I come up with a way to proof this and I need someone here to verify I am correct. Here is the problem:

Normal derivative is defined as:

$\displaystyle \frac{\partial u}{\partial n} = \nabla u \;\cdot\; \hat{n} $

Where $\displaystyle \hat{n}$ is the unit outward normal of the surface of the sphere and for a small sphere with surface $\displaystyle \Gamma$, the book gave:

$\displaystyle \int_{\Gamma} \frac{\partial u}{\partial n} \;dS \;=\; -\int_{\Gamma} \frac{\partial u}{\partial r} \;dS $

The book claimed on a sphere:

$\displaystyle \frac{\partial u}{\partial n} = \nabla u \;\cdot\; \hat{n} \;=\; -\frac{\partial u}{\partial r} \;$. Where $\displaystyle r $ is the radius of the sphere.

I found the explanation from the PDE book of Strauss.

$\displaystyle \frac{\partial u}{\partial n} \;=\;\hat{n} \;\cdot\; \nabla u \;=\; \frac{x}{r} \frac{du}{dx} \;+\; \frac{y}{r} \frac{du}{dy} \;+\; \frac{z}{r} \frac{du}{dz} \;=\; \frac{\partial u}{\partial r} $

Where $\displaystyle r=\sqrt{x^2+y^2+z^2}$

I don’t get how to go from

$\displaystyle \frac{x}{r} \frac{du}{dx} \;+\; \frac{y}{r} \frac{du}{dy} \;+\; \frac{z}{r} \frac{du}{dz} \;=\; \frac{\partial u}{\partial r} $

This is the way I have to proof the identity:

In spherical coordinates:

$\displaystyle \nabla u \;=\; \frac{\partial u}{\partial r}\hat{r} \;+\; \frac{1}{r}\frac{\partial u}{\partial \theta}\hat{\theta} \;+\; \frac{1}{r sin \theta}\;\frac{\partial u}{\partial \phi} \hat{\phi}$

We know $\displaystyle \hat{r} \hbox { is parallel the the outward unit normal } \hat{n}$ and therefore $\displaystyle \hat{n} \cdot \hat{\theta} = \hat{n} \cdot \hat{\phi}=0$

$\displaystyle \Rightarrow \; \nabla u \cdot \hat{n} \;=\; [\frac{\partial u}{\partial r}\hat{r} \;+\; \frac{1}{r}\frac{\partial u}{\partial \theta}\hat{\theta} \;+\; \frac{1}{r sin \theta}\frac{\partial u}{\partial \phi} \hat{\phi}] \;\cdot\; \hat{r} = \frac{\partial u}{\partial r}$

Where I substute n with r. But I still don't get the "-" sign yet.

Please give me your opinion. and help.

Thanks

Alan