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.
 A: 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
$$
\dfrac{d^{n}}{dx^{n}}\left((-1)^{x}\right)=(i\pi)^n(-1)^{x}
$$
A: Mistake 1: The function $(-1)^x$ is not well-defined, even if $x$ is a complex number. 
Mistake 2: You cannot say $\left( e^{\pi i} \right)^x = e^{\pi i x}$. The multiplication of exponent rule does not extend to complex numbers. Indeed, otherwise you have problems like $1^{\pi} = e^{2\pi^2 i}$, which is clearly false.
A: We have that $e^{i\pi}=-1$, so $(-1)^x=e^{i\pi x}$, therefore $\frac{d}{dx}e^{i\pi x}=i\pi e^{i\pi x}=i\pi (-1)^x$.
It's easy to see that $\frac{d^n}{dx^n}(-1)^x=(i\pi)^n(-1)^x$.
A: Observe the function $w=(-1)^z$ This is equivalent to $w=e^{z\ln(-1)}$ Now, recalling that the logarithm is a multivalued function, $$\ln(-1)=\ln\vert(-1)\vert +i\arg(-1)=i(\pi+2\pi n)$$We see that $(-1)^z$ Is equivalent to $w=e^{i\pi (2n+1)z}$ For all integers $n$. This means, except in special cases, that $(-1)^z$ has an infinite number of complex values. If we choose to let n be constant, and define $(-1)^z$ as the resulting single valued function, the complex derivative is equal to$$ i\pi (2n+1)e^{i\pi (2n+1)z}$$
