Differentiating the map $x\mapsto x/\langle x,x\rangle$ on an inner product space 
Let E be a vectorial space of finite dimension. $E^*=E\setminus \{0\}$. $$f:E^*\rightarrow E^*$$
  $$x\rightarrow \frac{x}{\langle x,x\rangle}$$
  Is this function $C^1$?  

How should I proceed?
My guess is evaluate $f(x+h)-f(x)$ for a small increment $h$ and obtain the differential, then check continuity.
But then I start to operate and the norm gets in the way, I really don't know how to deal with it.
Do I need to go through $\langle x+h,x+h \rangle = \langle x,x \rangle + \langle x,h \rangle + \langle h,x \rangle + \langle h,h\rangle$ ?  
I know that the continuity is given by the dimension of the space E, since the differential is a linear map and the space is finite, it's continuous. So that's done, but again, how do I find the differential?
 A: $f$ is indeed $\mathcal{C}^1$.
To prove it, apply the chain rule.
$\varphi_1 : x \mapsto \Vert x \Vert^2 = \langle x,x \rangle$ is $\mathcal{C}^1$ and you have $d\varphi_1(x).h = 2 \langle x,h \rangle$.
The real function $\varphi_2 : t \mapsto \frac{1}{t}$ is also $\mathcal{C}^1$ with $d \varphi_2(t)= -\frac{1}{t^2}$.
Now, you have $$f(x)=(\varphi_2 \circ \varphi_1)(x) \cdot x$$ Applying the chain rule, you get $$df(x).h=-2\frac{\langle x,h \rangle}{\langle x, x\rangle^2} x +\frac{h}{\langle x,x \rangle}$$
The result can also be achieved looking at the coordinates as you mentioned that $\dim E$ is finite, let's say equal to $n \ge 1$. You have:
$$f(x_1, \dots, x_n)=(\frac{x_1}{x_1^2+ \dots +x_n^2}, \dots , \frac{x_n}{x_1^2+ \dots +x_n^2})=(f_1(x_1, \dots,x_n), \dots,f_n(x_1, \dots,x_n)).$$ From there you can compute $\frac{\partial f_i}{\partial x_j}(x_1,\dots,x_n)$. However, on my side, I much prefer the first way which enables to stay at the vector level and is also valid for infinite dimensional spaces.
