## Calculus (3rd Edition)

Since $r(t)=\lt t,1,1 \gt$, then $r'(t)=\lt 1,0,0 \gt$. $$\|r(t)\|=\sqrt{2+t^2}, \quad \|r'(t)\|=1$$ and moreover $$\|r(t)\|'=\frac{d}{dt} \|r(t)\|=\frac{t}{\sqrt{2+t^2}}.$$ That is $$\|r(t)\|'\neq \|r'(t)\|.$$