Let $f:\mathbb{R}^3\to\mathbb{R}$, $f(x,y,z)=\begin{cases}\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2},\quad\text{if $(x,y,z)\neq (0,0,0)$}\\ 1,\quad\quad\quad\quad\quad\,\,\,\text{if $(x,y,z)=(0,0,0)$}\end{cases}$
Calculate the differential in every point $(x,y,z)\neq (0,0,0)$.
Calculate the differential of $f$ in the point $(0,0,0)$.
To solve the first question, I sum the partial derivatives. [I write (and abuse notation) $\|x\|=x^2+y^2+z^2$].
So I get:
$\dfrac{2x\cos(\|x\|)\cdot\|x\|-2x\sin(\|x\|)+2y\cos(\|x\|)\cdot \|x\|-2y\sin(\|x\|)+2z\cos(\|x\|)\cdot\|x\|-2z\sin(\|x\|)}{\|x\|^2}$
$=\dfrac{(2x+2y+2z)\|x\|\cos(\|x\|)-(2x+2y+2z)\sin(\|x\|)}{\|x\|^2}$
$=\dfrac{(2x+2y+2z)(\|x\|\cos(\|x\|)-\sin(\|x\|))}{\|x\|^2}$
How do I calculate now $Df(0,0,0)$?
I have to look at the limit:
$\lim_{(x,y,z)\to 0\\ (x,y,z)\neq 0} \dfrac{1}{\|(x,y,z)\|}\left(f(x,y,z)-f(0,0,0)-Df(x,y,z)\right)$
right?
I do not see an obvious way on how to approach this limit, as the expression
$\dfrac{1}{\|x\|}\left(\dfrac{\sin(\|x\|)}{\|x\|}-1-(2x+2y+2z)\dfrac{\|x\|\cos(\|x\|)-\sin(\|x\|)}{\|x\|}\right)$
is rather tedious.
Before I get lost in calculation, I would like to know if I am correct so far (you do not have to check my calculation, I am more interested in the method)
Thanks in advance.