Fréchet Derivative of $\frac{1}{\|x\|}$ In a Hilbert Space $V$, for function $f:V\to\mathbb{R}$, if $f$ is Fréchet differentiable at $x_0$, the Fréchet Derivative $\nabla f(x_0)$ is  $v$ such that $$ \lim_{x\to x_0} \frac{|f(x) - f(x_0) - \langle v, x-x_0 \rangle|}{\|x-x_0\|} = 0$$
As an example, for $f(x) = \|x\|$, the Fréchet derivative is $\nabla f(x) = \frac{x}{\|x\|}$, as shown in this Wikipedia proof.
Now my question is, what is the Fréchet derivative of $f(x) = \frac{1}{\|x\|}$. My gut says something like $-\frac{x}{\|x\|^3}$ following normal derivative rules, but I can't seem to be able to solve it.
 A: You can apply the product rule on the relation
$$1 = \|\cdot\| \cdot \frac1{\|\cdot\|}$$
to obtain
$$D1(x_0) = \frac1{\|x_0\|}D(\|\cdot\|)(x_0) + \|x_0\|D\left(\frac1{\|\cdot\|}\right)(x_0)$$
$$0 = \frac{x_0}{\|x_0\|^2} + \|x\|D\left(\frac1{\|\cdot\|}\right)(x_0) \implies D\left(\frac1{\|\cdot\|}\right)(x_0) = -\frac{x_0}{\|x_0\|^3}.$$
A: There are two ways to solve this.
The first way is to use the chain rule.
This requires the knowledge that a chain rule holds also in Hilbert spaces for Fréchet
derivatives and not just in $\mathbb R^n$.
Or you could just prove the chain rule yourself.
The second way is to guess the Fréchet derivative (you already have a conjecture)
and then try to verify that the $\lim$ in the definition is indeed $0$.
A: Let $f:H\to \mathbb R: f(x)=\|x\|.$ Then, if $x\neq 0,\ f$ is differentiable at $x$. Note that $f(x)\neq 0$ whenever $x\neq 0.$ Let $g:\mathbb R^+\to \mathbb R^+:g(x)=x^{-1}.$ If $x\neq 0,$ then $g$ is differentiable at $x$. By the chain rule, we have
$$(g\circ f)'(x)h=g'(f(x))\circ f'(x)h$$  Calculating, we get
$$(g\circ f)'(x)h=g'(f(x)(\langle\frac{x}{\|x\|},h\rangle)=\langle\frac{x}{\|x\|},h\rangle g'(f(x))=\langle\frac{x}{\|x\|},h\rangle\left(\frac{-1}{\|x\|^2}\right)=\langle\frac{-x}{\|x\|^3},h\rangle$$
