For $A(t)$ differentiable, taking positive matrices as values, how show $\sqrt{A(t)}$ is differentiable? On p. 150 of Lax's Linear Algebra, he mentions that is is not hard to show that if $R(t)$ is a differentiable matrix-valued function of a single variable, whose values are positive matrices, then the square root function $\sqrt{R(t)}$, is differentiable as well. I have been trying to prove this without success.
The definition Lax gives for the derivative of a matrix valued function is that $$\lim_{h\to 0}\left\|\frac{A(t_0+h) - A(t_0)}{h}- \Omega_{t_0}\right\| = 0 $$ for some matrix $\Omega_{t_0}$, where the double bars denotes the matrix norm. I believe this is equivalent to the existence of the $n^2$ entry-wise derivatives. 
I have been trying to manipulate the limit $$\lim_{h\to 0}\frac{\sqrt{A(t_0+h)}-\sqrt{A(t_0)}}{h}$$
In the scalar-valued case, the proof I have seen for the differentiability of the square root function involves rationalizing the numerator by multiplying top and bottom by $\sqrt{x_0+h} + \sqrt{x_0}$. We might try a similar thing with the matrix-valued function, but we would need $\sqrt{A(t_0 + h)}$ to commute with $\sqrt{A(t_0)}$ in order to clear the numerator, and they don't necessarily commute. 
So I'm kind of out of ideas. How might we prove the differentiability of $\sqrt{A(t)}$?
 A: Here is a mostly elementary approach. There are certainly some better or fancier proofs, but this is what immediately came to my mind. For convenience, I shall write $A_t$ for $A(t)$. Without loss of generality, assume that $t_0=0$. Let $X_h=A_h^{1/2}-A_0^{1/2}$. Proving that $A_t^{1/2}$ is differentiable at $t=0$ is equivalent to showing that $\color{green}{X_h=hX+o(h)}$ for some constant matrix $X$ (which, if exists, is the derivative $\left.\frac{d}{dt}A_t^{1/2}\right|_{t=0}$).
As eigenvalues of a matrix vary continuously with matrix entries, we know that $X_h=o(1)$ (this is the only non-elementary result that we shall use in this proof). In the below, we will show further that $X_h=O(h)$. This is stronger than continuity but weaker than differentiability at $t=0$. Let $u$ be a unit eigenvector corresponding to an eigenvalue $\lambda$ of $X_h$. Let $Y_h = A_h-A_0$. Then $A_0+Y_h=A_h=(A_0^{1/2}+X_h)^2$ and hence
\begin{align*}
Y_h &= X_hA_0^{1/2} + A_0^{1/2}X_h + X_h^2,\\
u^\ast Y_hu &= 2\lambda\, u^\ast A_0^{1/2}u + \lambda^2,\\
\lambda &= -u^\ast A_0^{1/2}u \pm \sqrt{(u^\ast A_0^{1/2}u)^2 + u^\ast Y_hu}.\tag{1}
\end{align*}
Since $A_0\succ0$ and $A_t$ is differentiable, we have $0<\lambda_\min(A_0)^{1/2}\le u^\ast A_0^{1/2}u\le\lambda_\max(A_0)^{1/2}$ and $Y_h=h\dot{A}_0+o(h)$. As $X_h=o(1)$, the plus sign in the "$\pm$" in $(1)$ must be taken when $h$ is small. Apply $\sqrt{a+x}=\sqrt{a}\sqrt{1+\frac xa}=\sqrt{a}\left(1+\frac x{2a}+o(x)\right)$ to $(1)$, we get $\lambda=O(h)$ and in turn $\rho(X_h)=O(h)$. As $X_h$ is Hermitian, we infer that $X_h=O(h)$.
We now show that $A_t^{1/2}$ is differentiable at $0$. If $A_t^{1/2}$ is indeed differentiable, then by differentiating $(A_t^{1/2})^2=A_t$ on both sides, we see that the derivative $X=\left.\frac{d}{dt}A_t^{1/2}\right|_{t=0}$ must satisfy the equation $A_0^{1/2}X + XA_0^{1/2} = \dot{A}_0$, or equivalently, $(I\otimes A_0^{1/2} + A_0^{1/2}\otimes I)\,\mathrm{vec}(X) = \mathrm{vec}(\dot{A}_0)$. So, let $X$ be a solution to this equation. Since $A_0$ is positive definite, the solution exists and is unique. Now we can express $A_h- (A_0^{1/2}+hX)^2$ in two different forms:
\begin{align*}
A_h- (A_0^{1/2}+hX)^2
&= A_h - (A_0 + hA_0^{1/2}X+hXA_0^{1/2} + h^2X^2)\\
&= A_h - (A_0 + h\dot{A}_0 + h^2X^2) = \color{red}{o(h)},\tag{2}\\
A_h- (A_0^{1/2}+hX)^2 &= (A_0^{1/2}+X_h)^2 - (A_0^{1/2}+hX)^2\\
&= \color{red}{A_0^{1/2}(X_h-hX) + (X_h-hX)A_0^{1/2} + o(h)},\tag{3}
\end{align*}
where we have used the result $X_h=O(h)$ in deriving $(3)$. Compare $(2)$ and $(3)$, we obtain $A_0^{1/2}(X_h-hX) + (X_h-hX)A_0^{1/2}=o(h)$, i.e. $(I\otimes A_0^{1/2} + A_0^{1/2}\otimes I)\mathrm{vec}(X_h-hX) = o(h)$. Since $I\otimes A_0^{1/2} + A_0^{1/2}\otimes I$ is invertible, it follows that $\color{green}{X_h-hX=o(h)}$. QED
