Derivative of Complex/Irrational Exponents How would you differentiate an equation like this (and show evidence)?
$$\frac{d}{dx}(x^{\pi i})$$
 A: I agree with the other solutions, but I think they are somewhat confusing to someone who has never worked with complex functions before. First, let's take a moment and understand what $x^{\pi i}$ really is:
$$x^{\pi i}=(e^{\ln x})^{\pi i}=e^{i\pi \ln x}$$
Now, Euler's identity tells us that $e^{i\theta}=\cos \theta+i\sin \theta$. If you have never learned about this before, what I am about to do may be confusing, BetterExplained has a good explanation of this. From this formula, we find:
$$x^{\pi i}=\cos(\pi\ln x)+i\sin(\pi\ln x)$$
Now, we have expressed the function in terms of components, so we can take the derivative of each component:
$$\frac{d}{dx}x^{\pi i}=-\sin(\pi\ln x)\cdot\frac{\pi}{x}+i\cos(\pi\ln x)\cdot\frac{\pi}{x}$$
Now, the first thing you should see is that you can factor out a $\frac\pi x$:
$$\frac{d}{dx}x^{\pi i}=\frac{\pi}{x}\left(i\cos(\pi\ln x)-\sin(\pi\ln x)\right)$$
Now, finally, although not strictly necessary, you might notice that the $\cos$ is the imaginary component where as $\sin$ is the real component now. In order to get it back to the way it was before, you can factor out an $i$, which gives us:
$$\frac{d}{dx}x^{\pi i}=\frac{i\pi}{x}\left(\cos(\pi\ln x)+i\sin(\pi\ln x)\right)$$
After all this, you can see that the expression in parenthesis is exactly what we started with, $x^{\pi i}$, which gives us:
$$\frac{d}{dx}x^{\pi i}=\frac{i\pi}{x}\cdot x^{\pi i}=i\pi x^{i\pi-1}$$
Thus, we get the same answer, but it is now easier to see how this type of power rule works even for complex exponents. I hope this gives you a better understanding of power rule in complex analysis!
A: Just like any other real valued expression, the power is just a constant. Thus :
$$(x^{\pi i})' = \pi i x^{\pi i-1}$$
It just stems from the fact that :
$$(x^n)' = nx^{n-1}$$ 
and if you let $n=\pi i$ you just yield your result.
A: The derivative of $x^n$ is well known: $n x^{n-1}$. Let $n= i \pi$ and we get $$\frac{d}{dx}(x^{i \pi}) = i \pi x ^ {i \pi - 1}.$$
$i$ and $\pi$ are just constants, so when you differentiate $x^{\pi i}$, you can treat $\pi i$ just as you'd treat $9$ or $\sqrt 7$ or $e^{4^2}$ (or any other constants) in the exponent.
A: $$y=x^{i\pi}$$
now let $u=i\pi$
$$y=x^u$$
$$\frac{dy}{dx}=ux^{u-1}=i\pi x^{i\pi-1}$$
