Proof: Derivative of $(-1)^{x}$ The derivative for $(-1)^{x}$ is 
\begin{equation}
\frac d{dx}\left[(-1)^x\right]=i\pi(-1)^{x}
\end{equation}
But why?
What happens with higher order derivatives?
Thanks in advance.

Since
$$(-1)^{x}=(e^{i\pi})^x=e^{i\pi x}$$
We have
$$
\dfrac{d}{dx}\left((-1)^{x}\right)=\dfrac{d}{dx}\left(e^{i\pi x}\right)=i\pi e^{i\pi x}=i\pi(-1)^{x}
$$
For higher order derivatives
$$
\dfrac{d^{n}}{dx^{n}}\left((-1)^{x}\right)=(i\pi)^n(-1)^{x}
$$
