TITLE: If $f(x) = \cos x$, explain, without taking the derivative, how you would find the $f^{(99)}(x)$?
QUESTION [7 upvotes]: My theory:
derivative of $\cos x = - \sin x $
derivative of $-\sin x = -\cos x $
derivative of $-\cos x = \sin x.$ 
cycle occurs three times but then what do you do??

Is there a good way to solve this?

REPLY [13 votes]: Notice $\quad\displaystyle \cos(x) = \frac12 (e^{ix} + e^{-ix})\quad$ and $e^{\pm i x}$ are eigenfunctions of the operator of taking derivative against $x$ with eigenvalues $\pm i$. i.e.
$\quad\displaystyle\frac{d}{dx} e^{\pm ix} = \pm i e^{\pm ix}.$ We have 
$$f^{(99)}(x) = \frac{d^{99}}{dx^{99}} \frac12 (e^{ix} + e^{-ix})
= \frac12 ( i^{99} e^{ix} + (-i)^{99} e^{-ix} )
= \frac12 ( -i e^{ix} + i e^{-ix} ) = \sin(x)$$