Show that $u_x(x, y) = f′(r) \cos(\theta)$ and hence deduce that $f′(0) = 0$, which implies the Neumann boundary condition $u_r = 0$ when $r = 0$. 
Suppose that $u(x,y)$ is a continuously differentiable, circularly symmetric function, so that when expressed in polar coordinates, $x = r \cos(\theta)$, , $y = r \sin(\theta)$, it depends solely on the radius $r$; that is $u = f(r)$. Show that $u_x(x, y) = f′(r) \cos(\theta)$ and hence deduce that $f′(0) = 0$, which implies the Neumann boundary condition $u_r = 0$ when $r = 0$.

I asked a question related to this problem a while ago, but I never finished the problem itself.
I understood enough to know that I needed to use change of variables:
$$\begin{align}
\frac{\partial u}{\partial r} &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial y} = \cos\theta\frac{\partial u}{\partial x} + \sin\theta \frac{\partial u}{\partial y}
\\ 
\frac{\partial u}{\partial \theta} &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y}
\end{align}$$
But from here I'm unsure of how to proceed. 
Would anyone mind please explaining this problem?
 A: Suppose $u(x,y) = f(r)$. Then
$$u_x(x,y) = \frac{d f(r)}{d x} = f'(r) \frac{\partial r}{\partial x} = f'(r) \frac{x}{r}= f'(r)\cos(\theta)$$
Now since $u(x,y)$ is circularly symmetric, we have $u_x(x,y)=-u_x(-x,y)$. Then $u_x(0,y)=-u_x(0,y)=0$. From the above, we then find
$$f'(0)\cos(\theta)  = u_x(0,0) = 0.$$
Since the function is invariant under a rotation of the axes, we must have $f'(0)=0$.
A: As you say $u(x,y) = f(r)$ since the function $u$ is circularly symmetric and then it depends only on the radius $r$. Now we are going to compute $u_x(x,y)$:
$$\frac{\partial u}{\partial x}(x,y) = \frac{\partial f(r)}{\partial x} =\frac{d f}{d r}\frac{\partial r}{\partial x}$$
by the chain rule, since $\frac{d f}{d r}=f'(r)$ and $r^2=x^2+y^2$, a simple computation gives
$$\frac{d f}{d r}\frac{\partial r}{\partial x}=f'(r) \frac{\partial r}{\partial x} = f'(r) \frac{2x}{2\sqrt{x^2+y^2}}= f'(r) \frac{x}{r}=f'(r)\cos(\theta)$$
because $x=r\cos\theta$; hence $$\color{red}{\frac{\partial u}{\partial x}(x,y) = f'(r)\cos(\theta)}$$
Now: $$f'(0)=\frac{df}{dr}(0)=\lim_{h\to 0} \frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{u(h_1,h_2)-u(0,0)}{h}$$
where $h=\sqrt{h_1^2+h_2^2}$.
This limit does not depend on the direction $(h_1,h_2)$, that is it does not depend on the value of $\theta$, since the function is circularity symmetric, and this implies that $f'(0)=0$. Indeed, you may notice that
$$f'(0)=\lim_{h\to 0^+}\frac{u(h,0)-u(0,0)}{h}=\lim_{h\to 0^-}\frac{u(h,0)-u(0,0)}{h}=-f'(0)$$
because when $h\to 0^-$, then $(h,0)$ approaches $(0,0)$ from the negative half-line of the $x-$axis, so $\theta=\pi$ and then $\cos\theta=-1$.
A: Here's a different approach to the end result.
Lemma: Suppose $g$ is differentiable and even on some symmetric interval $(-a,a).$ Then $g'(0)=0.$
Proof: Set $h(x)=g(-x).$ Then $h'(x) = -g'(-x).$ Since $g$ is even, we also have $h(x) = g(x).$ It follows that $g'(x)= -g'(-x).$ Thus $g'(0)= -g'(0),$ which implies $g'(0)=0.$
Now in your problem, $u(x,0)$ is even, since $u(-x,0) = f(|x|) = u(x,0).$ Thus $u_x (0,0) = 0.$ Now for $x>0,$
$$\frac{f(x)-f(0)}{x} = \frac{u(x,0)-u(0,0)}{x} \to u_x(0,0=0) =0$$
as $x\to 0^+$ Thus $f'(0)=0$ from the right, which is the desired result.
