The way of calculate $\frac{\partial}{\partial x} \ A^x $ 
Problem: 
  The square matrix $A\in\mathbf{R}^{n\times n}$ is not function of $x\in \mathbf{N}$. Then, how do to calculate the following matrix derivative?
  $$\frac{\partial}{\partial x} \ A^x $$

I could not find such calculus on any website and materials as far as I know. Do you know any documents, papers or theorems which are related with this? Or how is derivation shown?
My main interest:
Furthermore, I have already shown a proof when the matrix $A$ is able to diagonalize. But, I cannot find how to approach in other cases or the general case.
 A: Let $A\in GL_n(\mathbb{C}),x\in \mathbb{C}$. By definition, $A^x$ is $e^{x\log(A)}$; yet, how  $\log(A)$ is defined ? We write the Jordan Chevalley decomposition of $A$: $A=D(I+N)$ ($D$ diagonalizable, $N$ nilpotent, $DN=ND$) and we put $\log(A)=\log(D)+N-N^2/2+\cdots+N^{n-1}(-1)^n/(n-1)$; yet, how to define $\log(D)=\log(P\Delta P^{-1})$ where $\Delta=diag((\lambda_i))$ ? We put $\log(D)=Pdiag((\log(\lambda_i))P^{-1}$ and it suffices to define the function $\log$ over $\mathbb{C}$. The simplest way is to consider the principal log defined on $\mathbb{C}\setminus (-\infty,0]$ by $\log(re^{i\theta})=\log(r)+i\theta$ where $\theta\in (-\pi,\pi)$. Note that (even if $x\in \mathbb{R}$) $\log(a^x)\not= x\log(a)$. 
Finally, $\log(A)$ is defined when $A$ has no eigenvalues in $(-\infty,0]$.
It can be proved that $\log(A)$ is well defined, is a polynomial in $A$, is a real matrix when $A$ is a real matrix. In particular, $A^x$ is a polynomial in $\log(A)$, then a polynomial in $A$; for example, if $A\in GL_3(\mathbb{C)}$, $A^x=f_0(x)I_3+f_1(x)A+f_2(x)A^2$ (where the $f_i$ are holomorphic functions). Of course, $\log(A^x)\not= x\log(A)$; If you do not want any such trouble, then consider the following particular case:
The case when $A$ is symmetric $>0$ is straightforward. $A=Pdiag((\lambda_i))P^{-1}$, where $\lambda_i>0$. Then $\log(A)=Pdiag((\log(\lambda_i))P^{-1}$ with the standard $\log$ over $(0,+\infty)$ ($\log(A)$ does not depend on the choice of $P$).
Then our function $A^x=e^{x\log(A)}$ is holomorphic and its complex derivative is $\log(A)e^{x\log(A)}=\log(A)A^x$, the product being commutative.
A: My attempt:
Generally, differential matrix $\mathsf{F}$ including inner function $\mathsf{G}$ can be established the following property:
$$
\cfrac{\partial\mathsf{F}(\mathsf{G}(x))}{\partial x}=
\cfrac{\partial\mathsf{F}}{\partial\mathsf{G}}
\cfrac{\partial\mathsf{G}}{\partial x}
$$
At first, we define the function as $f(a,x)=a^x$ where $a$ is a scalar. Then, multiply logarithm and estimate differential of $x$:
$$
\begin{aligned}
f(a,x)=&a^x \\
\log(f(a,x))=&x\log(a) \\
\cfrac{\partial}{\partial x}\log(f(a,x))=&\cfrac{\partial}{\partial x}x\log(a) \\
\end{aligned}
$$
Then, we use the above property and multiply $f(a,x)$ on both sides so that we can obtain the derivative function:
$$
\begin{aligned}
f(a,x)^{-1}\cfrac{\partial f}{\partial x}=&\log(a) \\
\cfrac{\partial f}{\partial x}=&f(a,x)\log(a) \\
=&a^x\log(a)
\end{aligned}
$$
In matrix case, it needs that the original matrix $\mathsf{A}$ is made to diagonalize and estimate the exponent matrix $\mathsf{A}^x$:
$$
\mathsf{A}^x=\mathsf{P}^{\text{T}}\mathsf{D}^x\mathsf{P}
$$
where, $\mathsf{P}$ is an orthogonal matrix and $\mathsf{D}$ is a diagonal matrix which includes all eigenvalues $\lambda_{i}$. Likewise, $\mathsf{F}(\mathsf{D},x)=\mathsf{D}^x$ case is possible to describe by the above differential calculus way. Therefore my answer is:

$$\begin{aligned} \cfrac{\partial}{\partial x}\ \mathsf{A}^x=&
 \cfrac{\partial}{\partial x}\
 \mathsf{P}^{\text{T}}\mathsf{D}^x\mathsf{P} \\ =&
 \mathsf{P}^{\text{T}}\biggl(\cfrac{\partial}{\partial x} \
 \mathsf{D}^x\biggr)\mathsf{P} \\
 =&\mathsf{P}^{\text{T}}\mathsf{D}'\mathsf{P} \end{aligned} $$
where,
$$ \mathsf{D}'= \begin{cases} \lambda_{i}^x\log(\mathsf\lambda_{i}) &
(i=j) \\ \\ 0 & (\text{otherwise}) \end{cases} $$ However, the matrix
  $\mathsf{A}$ must be satisfied full rank.

