Calculation of the normal derivative on a sphere Fix $y\in B(0,R)$ (open ball) and define for $x\in B(0,R)\setminus \{y\}$ with
$$
N(x;y):=\begin{cases}
\dfrac{-1}{2\pi}\log\dfrac{rr^*\rho}{R^3},& y\not=0\\
\dfrac{-1}{2\pi}\log\dfrac{r}{R},&y=0
\end{cases}
$$
where $r=|x-y|$, $r^*=|x-y^*|$, $y^*:=\frac{R^2}{|y|^2}y$ and $\rho=|y|$. I want to calculate the normal derivative of $N$ on the boundary of $B(0,R)$ with $y\in B(0,R)$, i.e.
$$
\nabla_xN(x;y)\cdot \nu_x
$$
where $\nu_x$ is the unit normal at $x\in\partial B(0,R)$.
I'm expecting to get $1/(2\pi R)$, but I don't get it. The following is my calculation (for the case $y\not=0$):
I first get 
$$
\frac{\rho}{R}r^*=r
$$
for $x\in\partial B(0,R)$. Then
$$
-2\pi N(x,y)=\log\dfrac{r^2}{R^2}
$$
and I get
$$
-2\pi\nabla_xN(x,y)=\frac{2}{r^2}(x-y)
$$
and $\nu_x=x/R$.
What goes wrong with my calculation here?
 A: To find a derivative on the boundary, one should first take the derivative and then use the equality $|x|=R$ to simplify the result. Plugging before differentiating will lead to an incorrect result. 
E.g., let $u(x)=|x|^2$; then the normal derivative of $u$ on the sphere $|x|=R$ is $2R$. But if I first rewrite the function as $u(x)=R|x|$ on the sphere $|x|=R$, and then take the normal derivative of that, the result comes out to be $R$. 

You are essentially considering 
$$\log |x-y|+\log |x-y^*|\tag{1}$$
whose gradient with respect to $x$ is 
$$\frac{x-y}{|x-y|^2}+\frac{x-y^*}{|x-y^*|^2} \tag{2}$$
Now you want to show that the inner product of $(2)$ with $x$ has constant value when $|x|=R$... I fail to come up with an elegant argument for that. Until someone else does, here is a  brute force approach:


*

*Observe that the problem is really two-dimensional (consider the span of $x,y$). Moreover we may assume $y=(y_1,0)$ by rotational symmetry.

*Scale to $R=1$ for simplicity. 

*Write out 
$$x\cdot \frac{x-y}{|x-y|^2}+x\cdot \frac{x-y^*}{|x-y^*|^2}$$
in components $x=(x_1,x_2)$. Observe that only $x_2^2$ is present in the result, no $x_2$. Replace $x_2^2$ by $1-x_1^2$ and do a bunch of cancellations. 

