Derivative of $(-1)^x$, several cases How would you differentiate the function $(-1)^x$ w.r.t. $x$.
Is the formula for $a^x$ valid here? Wolfram says that the derivative of $(-1)^x$ is $iπ(-1)^x$. So, does this mean than $ln(-1)=iπ$ ?

Another thing that intrigued me was that its domain and range are given to be $\emptyset$. Is it correct?(or does Wolfram Mathematica wants to say that a real domain and range do not exist?)

Update:

There arise 2 cases

1) $f_1 : \mathbb{R} \rightarrow \mathbb{R}$    

$f_1(x)=(-1)^x$

2) $f_2 : \mathbb{C} \rightarrow \mathbb{C}$  

$f_2(x)=(-1)^x$

Then as per the comments, $f_1$ is not differentiable and its domain and range cannot be specified, while $f_2$ is a differentiable function.(Right?)
Since it seems you are insisting in asking for the domain of $(-1)^x$ for $x \in \mathbb R$, then 
note that
$$
\left( { - 1} \right)^{\,x}  = \left( { - 1} \right)^{\,\left\lfloor x \right\rfloor  + \left\{ x \right\}}  = \left( { - 1} \right)^{\,\left\lfloor x \right\rfloor } \left( { - 1} \right)^{\,\left\{ x \right\}} 
$$
$(-1)$ elevated at the fractional part of $x$ is not defined in the reals, unless of course $\left\{ x \right\}$ be null
$$
\left( { - 1} \right)^{\,\left\{ x \right\}} \text{n}\text{.}\,\text{d}\text{.}\quad \left| {\;0 &lt; \left\{ x \right\}\; \wedge \;x \in R} \right.
$$

so we can write
$$
y = \left( { - 1} \right)^{\,x} \quad \left| {\;x \in R} \right.\quad  = \left\{ {\begin{array}{*{20}c}
   {\left( { - 1} \right)^{\,x}  = \left( { - 1} \right)^{\,\bmod \left( {x,2} \right)} } &amp; { \Rightarrow \quad y \in \left\{ { - 1,1} \right\}} &amp; {\left| {\;x \in Z\,} \right.}  \\
   {n.\,d.} &amp; { \Rightarrow \quad y \in \emptyset } &amp; {\left| {\;x \notin Z} \right.}  \\
 \end{array} } \right.
$$
$$f(z) = (-1)^{z} = \mathrm{e}^{z\ln(-1)}$$

\begin{align}
\ln(-1) &amp;= \ln|-1| + i\mathrm{Arg}(-1) + i2\pi n \\
&amp;= 0 + i\pi +i2\pi n \\
&amp;= i\pi (2n+1)
\end{align}
where


$n \in \mathbb{Z}$
We designate the principal branch of the logarithm as $-\pi \lt \mathrm{Arg}(z) \le \pi$
On the principal branch, $n=0$ and $\ln(-1) = \mathrm{Ln}(-1) = i\pi$


Thus
$$\frac{\mathrm{d}f}{\mathrm{d}z} = \ln(-1)\,(-1)^{z} = (-1)^{z}[i\pi (2n+1)]$$

On the principal branch
$$\frac{\mathrm{d}f}{\mathrm{d}z} = i\pi (-1)^{z}$$