# 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?

• The period is 4. Divide 99 by four, and see if its remainder is 0, 1, 2, or 3. – Pedro Tamaroff Nov 4 '13 at 2:59

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 $\sin(x)$.

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 $$99 = 4 \cdot 24 + 3$$

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 $$f^{(99)}(x) = \frac{d^{99}}{dx^{99}} \frac12 (e^{ix} + e^{-ix}) = \frac12 ( i^{99} e^{ix} + (-i)^{99} e^{-ix} ) = \frac12 ( -i e^{ix} + i e^{-ix} ) = \sin(x)$$

• yikes!!!!!!!!!!! – Jessica Nov 4 '13 at 3:14
• @Jessica It's actually not that "yikes" once you know complex numbers... – Tobias Kienzler Nov 4 '13 at 8:58

$f = \cos x , f' = -\sin x , f'' = -\cos x , f''' = \sin x , f^{(4)} = \cos x$

Therefore, the derivatives of sine cycle every $4$. In particular since $99 = 3 (mod 4 )$, so $f^{(99)} = f''' = \sin x$