Stay $\|f\| \in W^{1,1}(0,T;\mathbb{R})$ provided $f \in W^{1,1}(0,T;H)$

Question

Supose that $f \in W^{1,1}(0,T; H)$ where $H$ is a Hilbert space. Can we conclude that $\|f\| \in W^{1,1}(0,T;\mathbb{R})$? Since $f \in L^1(0,T;H)$ its clear that $\|f\|:(0,T) \to \mathbb{R} \in L^1(0,T;\mathbb{R})$ but if $f' \in L^1(0,T;H)$ i can't justify that $\|f\|' \in L^1(0,T;\mathbb{R})$. First, i think that is important understant $\|f\|'$ and $f'$, looking the Evans book I see that the weak derivative $f'=v \in L^1(0,T;H)$ is the function that satisfy $$\int_0^T\phi'(t) f(t) dt=-\int_0^T \phi(t) f'(t) dt$$ for all scalar test function $\phi \in C_c^{\infty}(0,T)$. So the work is now prove that $$\int_0^T\phi'(t) \|f\|(t) dt=-\int_0^T \phi(t) \|f\|'(t) dt$$ for all $\phi \in C_c^{\infty}(0,T)$. One important remark is that $\|f'\|$ is not necessary equal to $\|f\|'$. Can somebody give me anyone hint?