Let $f_n:[0,T]\to \mathbb{R}^d$, $f^n \in C_0([0,T];\mathbb{R}^d)$ be a sequence of continuous functions starting at $0$. Denote $\|x\|$ the Euclidean norm in $\mathbb{R}^d$. That is $\|x\|^2=x_1^2+\ldots+x_d^2$.
Let $t\in [0,T]$. If we know $\|f^n(t)\|\to \infty $ as $n\to \infty$, does this imply $\|\frac{d}{dt}f^n(t)\|\to \infty $ as $n\to \infty$?
Note that $\frac{d}{dt}f^n(t)=(\frac{d}{dt}f^n_1(t),\ldots,\frac{d}{dt}f^n_d(t))$.
There is a counterexample: let $B$ be a small neighbourhood of $t$ in $[0,T]$ such that $0\not\in B$. Let $g^n:\mathbb{R}\rightarrow\mathbb{R}$ such that for $x\in B$ $g^n(x)=n$. Let $f^n(x)=(g^n(x),0,\dots,0)$. Then $\frac{d}{dx}f^n(t)=(0,\dots, 0)$ but $||f^n(t)||\rightarrow\infty$.