Derivative of $(-1)^x$ I'm taking a summer calc 2 class and we're getting into alternating patterns. I was interested in seeing the graph of $(-1)^x$ so I typed it into my TI-84 for $y = (-1)^x$. Surprisingly, the graph is blank and dots appear randomly throughout. 
Then I typed it into wolfram alpha and got a pattern I kind of expected. Since I'm learning this calculus stuff I thought hey... I can describe the function with the derivative... So I write down $f(x) = (-1)^x$ and then $f'(x)=\;?$ and realize I have no idea how to get this derivative.
Wolfram alpha tells me it is $i\pi(-1)^x$. The "step by step" on wolfram alpha looks like this infamous bundle of joy:
Step 1: derivation of $-1^x$
Step 2: derivation of $-1^x$ is $i\pi(-1)^x$
Yey! I'm no more smarter having read that. So... how do I (or why can't I) take the derivative of $(-1)^x$? Also why can't my calculator graph this?
 A: This is probably not the type of function you are used to dealing with: $f(x)=(-1)^x=e^{i(2n+1)\pi x}$, for any integer $n$. So this function is actually multi valued, and most of the time, this function takes on complex values. This is an interesting (and often vexing) property of some functions in the complex plane. For more information about the phenomenon, look up branch points. Nevertheless, we can still differentiate it in the normal way: $f'(x)=i(2n+1)\pi e^{i(2n+1)\pi x}$. Again, the derivative is multi-valued, but that's what you get.
A: You can take that derivative.
$\frac{d}{dx}(-1)^{x} = \ln(-1)(-1)^{x} = i \pi (-1)^{x}$.
It's the same method for any $a^{x}$.
Your issue is that you are now working with complex numbers. $\ln(x)$ is not defined in the reals for x less than or equal to zero.
Your calculator can't graph it because it is only showing the values where the expression is purely real. You should know that your function has a value of $i$ when $x = \frac{1}{2}$, for example.
