$\| X'(t) \| = (\| X(t) \|)'$ 
Let $X : \mathbb{R} \to \mathbb{R}^n$ be a $C^1$ function. Let $\| .\|$ be the norm : $\| v \| = \max_{1 \leq i \leq N} \mid v_i \mid$. Then is it true that : 
  $$\| X'(t) \|  = (\| X(t) \|)'$$ ?

I am wondering if in general if I have any function $f : \mathbb{R}^n \to \mathbb{R}^p$ and a norm $N$ on a : $\mathbb{R}^p$ then is it always possible to invert the norm and the differential operator or the norm and in the integral?
Thank you. 

To show that this is not always the case, let's give a clear example in two dimension. To that end, we interpret $X'(t)$ as the velocity of a trajectory in $\mathbb R^n$, so that $\vert\vert X'(t)\vert\vert$ represents the speed. 
Consider now trajectories moving along the surface of a sphere, i.e. $\vert\vert X(t)\vert\vert = 1$ at all times. They may attain varous speeds, i.e. different and even growing $\vert\vert X'(t)\vert\vert$'s.
Example:
$X(t)=(\cos(\omega(t)), \sin(\omega(t)))$
$\implies\vert \vert X(t)\vert \vert = \cos(\omega(t))^2+\sin(\omega(t))^2=1$
$\implies\vert \vert X(t)\vert \vert' = 0$
while
$X'(t)=\omega'(t)\cdot(-\sin(\omega(t)), \cos(\omega(t)))$
$\implies\vert \vert X'(t)\vert \vert = \omega'(t)$
You can also play this the other way around, e.g. going with $\omega(t)=c\cdot t$ and move the circle away from the origin: In this case, $\vert \vert X(t)\vert \vert' $ will vary and $\vert \vert X'(t)\vert \vert$ actually wont.
