Derivative of function that includes norm I was solving the problem: find the derivative of a function f : H →
R, $f (x) = \sin ||x||^3$ (H is Hilbert space).  I got the answer $f'(x)=3\cos||x||^3 x||x||$. Is this correct or I am doing something wrong?
 A: Let's start with finding the derivative of $N\colon x \mapsto \def\norm#1{\left\|#1\right\|}\norm x^2$, the chain rule will do the rest. For $x, h\in H$, we have
\begin{align*}
  \def\sp#1{\left<#1\right>} N(x+h) &= \norm{x+h}^2\\
          &= \norm x^2 + 2\sp{x,h} + \norm h^2\\ 
          &= N(x) + 2\sp{x,h} + \norm h^2
\end{align*}
As $h \mapsto 2 \sp{x,h}$ is real-linear, and $\norm h^2 = o(\norm h)$ for $h \to 0$, we have that $N'(x)h = 2\sp{x,h}$.
Denote by $g \colon [0,\infty) \to \mathbb R$ the map $\xi \mapsto \sin \xi^{3/2}$, then $f = g \circ N$ and $g'(\xi) = \cos \xi^{3/2} \cdot \frac 32\xi^{1/2}$ and the chain rule gives that for $x,h \in H$:
\begin{align*}
  f'(x)h &= g'\bigl(N(x)\bigr)N'(x)h\\
         &= \cos N(x)^{3/2} \frac 32 N(x)^{1/2} \cdot 2\Re\sp{x,h}\\
         &= 3\cos \norm x^3 \cdot \norm x \cdot \sp{x,h}
\end{align*}
If you identify $H'$ and $H$ via the Riesz isomorphism $x \mapsto \sp{x,\cdot}$, we indeed have $f'(x) = 3\cos\norm x^3 \cdot \norm x \cdot x$, so you are right.
