Forgive me if this question has already been asked. I was unable to find anything relevant to this question.
The $n$th derivative of a function, $f^n(x)$ is well-defined for $n\in\mathbb{Z}^+$. As well, it can be defined for the case where $n\in\mathbb{R}^+$, (i.e. fractional derivatives). What about in the case where $n\in\mathbb{C}$? What would it mean to take the imaginary derivative of a function?
Similarly, let $n$ be a function of some variable, $x$, (i.e. $n=\cos(x)$). Could you define $f^n(x)=f^{\cos(x)}(x)$? And again, what would it even mean to take the $\cos(x)$th derivative of a function?
As well, if $F'(x)=f(x)$, does $f^{-1}(x)=\int\,f(x)\,dx=F(x)$, where $f^{-1}$ is the negative derivative, and not the inverse?
Edit: Duplicates do not answer the final part of this question (i.e. the $\cos(x)$th derivative of a function).
Edit 2: It's interesting because, let's say that $f^{-1}(x)=\int\,f(x)\,dx=F(x)$. Operator theory tells us that when we apply two half derivatives in succession to a function (i.e. $D^{\frac{1}{2}}D^{\frac{1}{2}}f(x)$, that we obtain the first derivative of that function $f(x)$. This is consistent with the idea that the antiderivative is nothing more than $D^{-1}f(x)=D^{-1}F'(x)$.
But consider the complex derivative $f^i(x)$. Following operator theory, one would claim that the antiderivative of this function could be written as $f^{-i}(x)$. However, this is obviously not the case as this implies that $f(x)=f^0(x)=f'{x}$, recalling that $i^2=-1$, such that $(i)(-i)=1$.