Let $f:\mathbb{R}^{2}\to \mathbb{R}$ be a $C^{1}$ map. Define $u:(0, \infty)\times \mathbb{R}^{2} \to \mathbb{R}$ as
$$u(r,x)=\frac{1}{2\pi} \int _{y\in S^{1}} f(x+ry)dy$$.
Then find $\frac{\partial u}{\partial r}$.
My attempt: $\frac{\partial u}{\partial r}(r,x)= lim_{h\to 0} \frac{u(r+h,x)-u(r,x)}{h}$. This gives :
$\frac{\partial u}{\partial r}(r,x)= lim_{h\to 0}\frac{1}{2\pi} \int_{y\in S^{1}} \frac{(f(x+(r+h)y)-f(x+ry) ) }{h}$. Now taking limit inside we have
$\frac{\partial u}{\partial r}(r,x)= \frac{1}{2\pi} \int_{y\in S^{1}} D_{y}f(x+ry)dy$, where $D_{y}f(x+ry)$ denotes the directional derivative of $f$ at $(x+ry)$, in the direction of $y$.
Is this a correct and complete answer or can I do something more? Thanks in advance for any help!!
