If $f(x) = \cos x$, explain, without taking the derivative, how you would find the $f^{(99)}(x)$? 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?

 A: $ f = \cos x , f' = -\sin x , f'' = -\cos x , f''' = \sin x , f^{(4)} = \cos x $
Therefore, the derivatives of sine cycle every $4$. In particular since $99 = 3 (mod 4 )$, so $f^{(99)} = f''' = \sin x $
A: We have
\begin{align*}
f(x) &= \cos{x} \\
f^{(1)}(x) &= -\sin{x} \\
f^{(2)}(x) &= - \cos{x} \\
f^{(3)}(x) &= \sin{x} \\
f^{(4)}(x) &= \cos{x}
\end{align*}
So the cycle has length $4$. In particular, we'll see that
$$f^{(8)}(x) = \left(\cos{x}\right)^{(8)} = \left(\cos{x}\right)^{(4)} = \cos{x}$$
Likewise for $12$, $16$, and so on.
So now use the fact that $$99 = 4 \cdot 24 + 3$$
A: 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 
$$f^{(99)}(x) = \frac{d^{99}}{dx^{99}} \frac12 (e^{ix} + e^{-ix})
= \frac12 ( i^{99} e^{ix} + (-i)^{99} e^{-ix} )
= \frac12 ( -i e^{ix} + i e^{-ix} ) = \sin(x)$$
A: The cycle occurs, not three times but 4 times.  Thus every 4 derivatives taken you return to $\cos(x)$.  Therefore, since $99=96+3$, you need the $3$rd derivative of $\cos(x)$ which is $\sin(x)$.
