Proving the composition of a function with an $H^1$ function is in $H^1$.

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Let $\Omega \subseteq \mathbb{R}^n$ be open, bounded with smooth boundary. Does $u \in H^1(\Omega)$ imply that $(1+u^2)^{\frac{1}{2}}\in H^1(\Omega)$?

If yes, I would expect a more general result holds about composing $H^1(\Omega)$ with $C^1$ sufficiently slowly increasing functions with sufficiently slowly increasing derivatives. I am not really sure about how one goes about proving this kind of result.

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Define $f(x)=(1+x^2)^{1/2}$. Note that $f \in C^1(\mathbb{R})$ and $f'(x)=x/(1+x^2)^{1/2}$ is bounded in the sense $|f'(x)|\leq 1$ for all $x \in \mathbb{R}$. Then by the result in this question we have $f(u) \in H^1(\Omega)$, exactly as you wanted.