Finding the derivative of $f(x)=x/||x||$ for $x \in \mathbb{R}^3$ I'm trying to find the derivative of the function $f:\mathbb{R}^3 \to \mathbb{R}^3$ given by $f(x)=x/||x||$ for $x \in \mathbb{R}^3 \setminus \{0\}$ and $f(0)=0$.
I've actually already been given the following expression $$f'(x)(h)=\frac{h}{||x||} - \frac{\langle x,h \rangle x}{||x||^3}$$ but cannot find a nice way to prove that this is in fact the derivative of $f$ on $\mathbb{R}^3 \setminus \{0\}$.

What I've tried is to use the definition $$f(x+h)=f(x) + T(h) + ||h|| \varepsilon(h)$$ where $T(h)$ is a linear map that ends up being the derivative if it exists, and $\varepsilon(0)=0$ and is continuous at $0$. Plugging the function and the given derivative in, it seems you have to show that for $x,h \in \mathbb{R}^3$, $$\lim_{h \to 0} \frac{1}{||h||} \left(\frac{x+h}{||x+h||} - \frac{(x+h)}{||x||} + \frac{\langle x,h \rangle x}{||x||^3} \right)=0$$ and I can't see a nice way of showing this. I don't know if doing this is the intended way or if there is a much easier way. I've tried looking at the components of $f$ since $f$ is differentiable at $a$ if and only if its component functions are also differentiable at $a$, but those expressions aren't any nicer than the one above.
How do you show that the given function is the derivative of $f$?
 A: $$\|x+h\|-\|x\|=\sqrt{x^2+2\langle h,x\rangle+h^2}-x^2=\frac{2\langle h,x\rangle+h^2}{\sqrt{x^2+2\langle h,x\rangle+h^2}+x^2}\sim\frac{\langle h,x\rangle}{\|x\|}$$
and from this the "derivative" of $x\dfrac1{\|x\|}$:
$$h\frac1{\|x\|}-x\frac{\dfrac{\langle h,x\rangle}{\|x\|}}{\|x\|^2}$$
A: This kind of computations is best handled using differentials.
It holds
$$
df = 
\frac{\|\mathbf{x}\| d\mathbf{x}-(d\|\mathbf{x}\|) \mathbf{x}}{\|\mathbf{x}\|^2}
$$
The differential of the norm is
$
\frac{1}{\|\mathbf{x}\|} \mathbf{x}^T d\mathbf{x}
$
yielding
$$
df(\mathbf{x})[d\mathbf{x}]= 
\frac{1}{\|\mathbf{x}\|} d\mathbf{x} -
\frac{1}{\|\mathbf{x}\|^3} \mathbf{x} \mathbf{x}^T d\mathbf{x}
$$
This is the desired expression when replacing $d\mathbf{x}$ by $\mathbf{h}$.
A: Sometimes it helps to introduce a "velocity" term. Here $d\vec{x}/dt$ plays the role of $h$ above.
$\vec{f}=\vec{x}/\sqrt{\vec{x}\cdot\vec{x}}=\nabla r.$
$\frac{d\vec{f}}{dt}=\frac{1}{\sqrt{\vec{x}\cdot\vec{x}}}\frac{d\vec{x}}{dt}-\vec{x}(\frac{1}{2})(2\vec{x}\cdot\frac{d\vec{x}}{dt})(\vec{x}\cdot{\vec{x}})^{-3/2}$
$\frac{d\vec{f}}{dt}=\frac{\vec{x}\cdot\vec{x}}{(\vec{x}\cdot\vec{x})^{3/2}}\frac{d\vec{x}}{dt}-\frac{\vec{x}\cdot\frac{d\vec{x}}{dt}}{(\vec{x}\cdot\vec{x})^{3/2}}\vec{x}=\vec{x}\times(\frac{d\vec{x}}{dt}\times \vec{x})/(\vec{x}\cdot\vec{x})^{3/2}$
So $\frac{d\vec{f}}{dt}=\vec{f}\times (\frac{1}{(\vec{x}\cdot\vec{x})^{1/2}}\frac{d\vec{x}}{dt}\times \vec{f})$
So the vector change of $\vec{f}$ changes in a direction perpendicular to itself and the change $d\vec{x}$.
Take the dot product with both sides we find that $\vec{f} \cdot d\vec{f}/dt=0$ everywhere, telling us the modulus of $\vec{f}$ is constant. Of course this can be found by dotting it with itself $\vec{f}\cdot\vec{f}=1$.
