When $|\Phi(|x_n|)|\geq 1 \implies |\Phi'(r)|\geq c$ for $r\in (|x_n|-\epsilon, |x_n|+\epsilon)$?

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Let $f_n:\mathbb R^3 \to \mathbb C$ be a non-constant functions, such that $\nabla f_n \in L^2(\mathbb R^3)$ for each $n\in \mathbb N.$ (Specifically, $f_n\in \dot{H}^{1}(\mathbb R^3)$ = completion of $C_c^{\infty}(\mathbb R^3)$ with the $\|\nabla f\|_{L^2}$ norm )

Assume that there exists $\Phi_n:[0,\infty) \to \mathbb C$ such that $f_n(x)= \Phi_n(|x|)$ for all $x\in \mathbb R^3, n\in \mathbb N.$ There exits $\{x_n\}\subset \mathbb R^3$ such that $|f_n(x_n)| \geq 1$ for all $n \in \mathbb N$ and $x_n\to \infty.$

Question: Does there exists $\epsilon >0$, $c>0$ and $n_0 \in \mathbb N$ such that $|\Phi'_n(t)|\geq c$ for all $t\in (|x_n|-\epsilon, |x_n|+\epsilon)$ and $n\geq n_0$?

Note: From our assumption $\|\nabla f_n \|_{L^2} <\infty$ implies we have $\int_0^{\infty}|\Phi'_n(t)|^2 t^2 dt < \infty.$ I guess this information might be useful...