The positive square-root branch is smooth

Define $P:(0,\mathrm{\infty })\to \mathbb{R}$ by $P(t)=t^2$. By [F3], $P$ is smooth with $P^{\mathrm{\prime }}(t)=2t$. Fix $q>0$ and put $t=s(q)>0$. Then $P(t)=q$ and the one-dimensional derivative is invertible because $2t\mathrm{\ne }0$.

By [F2], there are open neighborhoods $V\subseteq (0,\mathrm{\infty })$ of $t$ and $W$ of $q$ on which $P{\mathrm{\mid }}_V$ has a smooth inverse $h$. Shrink $W$ inside $(0,\mathrm{\infty })$. For $y\in W$, $h(y)>0$ and $h(y{)}^2=y$, so uniqueness in [F1] implies $h(y)=s(y)$. Thus $s$ agrees with a smooth function near every point of its domain and is smooth.

Differentiate $s(q{)}^2=q$ using [F3] to obtain $2s(q)s^{\mathrm{\prime }}(q)=1$. Since $s(q)>0$, division gives the stated derivative. The domain excludes the zero endpoint; no smooth extension through zero is claimed. The local inverses agree by uniqueness, without any choice of a family of inverses.