Derivative of square root of unitary matrix Given a (special) unitary matrix $U(x) \in SU(2)$ which is a function of $x\in\mathbb{R}$, along with its unitary square root $u(x) = \sqrt{U(x)}$ such that $u(x)^2 = U(x)$ and $u(x)^\dagger u(x) = \mathbb{1}_{2\times2}$, what is the derivative $$\frac{d}{dx}u(x)=\frac{d}{dx}\sqrt{U(x)}?$$
Can it be expressed purely in terms of products and sums of $u(x)$, $u(x)^\dagger$, $U(x)$, $U(x)^\dagger$, $\frac{d}{dx}U(x)$ and $\frac{d}{dx}U(x)^\dagger$? Naively, it seems like it should be something like (suppressing arguments) $$\frac{du}{dx} = \frac{1}{2}\left(\frac{1}{2} u^\dagger \frac{dU}{dx} + \frac{1}{2}\frac{dU}{dx}u^\dagger \right),$$
following from a "symmetrized" form of the regular chain rule $\tfrac{d}{dx}\left[f(x)\right]^{1/2}=\frac{1}{2} \left[f(x)\right]^{-1/2}f'(x)$ with the identification of $u\leftrightarrow \left[f(x)\right]^{1/2}$ and $u^\dagger = u^{-1} \leftrightarrow \left[ f(x)\right]^{-1/2}$.
Note: Here $U^\dagger$ is the conjugate transpose.

Edit: I am actually interested in the quantity
$$ u^\dagger \frac{du}{dx} + u \frac{du^\dagger}{dx}, $$
so even if there is not a way to write $\tfrac{du}{dx}$ simply, an expression for this would be sufficient.
 A: $
\def\R#1{{\mathbb R}^{#1}}
\def\o{{\tt1}}
\def\mbrace#1{\left\lbrace\begin{array}{r}#1\end{array}\right\rbrace}
\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}
\def\LR#1{\left(#1\right)}
\def\op#1{\operatorname{#1}}
\def\trace#1{\op{tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
$Given an $\R{4}$ unit vector
$$p=\mbrace{a\\b\\c\\d},\qquad\quad \|p\|^2=\o$$
Construct an $SU(2)$ matrix (corresponding to a unit quaternion)
$$\eqalign{
U &= \m{a+ib & c+id\\-c+id & a-ib} \\
U^\dagger U&=I \\
\det(U) &= (a^2+b^2)+(c^2+d^2) = \o \\
\trace{U} &= 2a \\
\trace{I+U} &= \LR{2+2a} \\
}$$
The square root of a quaternion has a known formula
$${\sqrt U} = \frac{\qquad I+U}{\sqrt{\trace{I+U}}} \qiq
 u = \LR{2+2a}^{-1/2}\LR{I+U}$$
Differentiation yields
$$\eqalign{
\dot u
 &= \LR{2+2a}^{-1/2}\dot U
 \;\:-\; \LR{2+2a}^{-3/2}\LR{I+U}\dot a \\
\dot u^\dagger
 &= \LR{2+2a}^{-1/2}\dot U^\dagger
   \;-\; \LR{2+2a}^{-3/2}\LR{I+U^\dagger}\dot a \\
}$$
where a dot is used to denote the derivative with respect to $x$.
The variable $a$ can be replaced by the trace
and $U$ can be expanded in terms of Pauli matrices (if those are your preferences).
