What rule can I use to compute $\frac{d^{107}}{dx^{107}} \sin x$? Did I miss something in my calculus class? I don't remember anything concerning this type of problem: 

Compute $$\frac{d^{107}}{dx^{107}} \sin x.$$

So what is the rule here?
 A: One important thing to remember is that
$$\frac{d^4}{dx^4}(\sin x)=\sin x$$
So...
$$\frac{d^{4n}}{dx^{4n}}(\sin x)=\sin x$$
Now, what is the remainder when $107$ is divided by $4$ (a.k.a. $107\mod 4$)?
Basically, the derivative will "cycle" around and around (through $\sin$, $\cos$, etc.), so just the very last few derivatives after it "stops cycling" are important.
A: HINT: Calculate the first four or five derivatives, spot a pattern, and use that pattern to figure out the $107$-th derivative.
A: First of all, in case you don't know what the notation $\frac{d^n}{dx^n}\sin x$ means? It means "the $n$-th derivative of the function $\sin x$". So, for example, $\frac{d^2}{dx^2} x^2 = 2$ because the derivative of $x^2$ is $2x$ and the derivative of $2x$ is $2$.
That said, for solving the question, here is a Hint:
Try to manually calculate $\frac{d^1}{dx^1}\sin x, \frac{d^2}{dx^2}\sin x, \frac{d^3}{dx^3}\sin x, \frac{d^4}{dx^4}\sin x, \frac{d^5}{dx^5}\sin x$ and see if you can find a pattern.
A: Every answer here is pointing you in the right direction, but I would encourage you to simply take a step back for a moment and consider "the bigger picture." What if your teacher asked you, instead, to calculate
$$
\frac{d^{1035}}{dx^{1035}}(\sin x)\,?
$$
How would you do this? Clearly your teacher does not expect you to compute over one thousand derivatives! Similarly, with your own example, your teacher does not expect you to compute over one hundred derivatives. So what do you do? Your main task here is to make the problem simpler. In this vein, how might you make your problem easier to deal with? 
It should be fairly clear that there must be some sort of "trick" to whittle down or reduce the enormous number of $107$ that you have to deal with. This is where it is useful, first, to observe that you are dealing with only $\sin(x)$ and not something like $\sin(\tan(x)\log(\sqrt{x}))$; that is, you know that $\frac{d}{dx}\sin x=\cos x$ and that $\frac{d}{dx}\cos x=-\sin x$. 
Hence, you can see that your first derivative will involve $\cos x$ and every odd derivative thereafter (because the derivative keeps alternating between $\cos$ and $\sin$). But what about the sign (i.e., positive or negative) of the alternating derivatives? This is really the main part of the problem and requires the most ingenuity to solve. You can see your first derivative is positive (i.e., $\cos x$), your second one is negative (i.e., $-\sin x$), your third one is negative (i.e., $-\cos x$), your fourth one is positive (i.e., $\sin x$), and then this process repeats itself. That is, you have found a pattern! Now use this to solve your problem.
First off, you are considering the $107$th derivative of $\sin x$; since $107$ is odd, you already know that the $107$th derivative will involve $\cos x$. But is the sign positive or negative? Well, look back at the pattern you discovered above where "everything returns to normal" after a cycle of having taken four derivatives. Well, note that $107=4\cdot 26+3$. Hence, we will have gone through a total of $26$ cycles of taking four derivatives where we end up back at $\sin x$ at the beginning of each cycle. Since $107=4\cdot 26+3$, we see that we may start at $\sin x$ and need only consider three more derivatives, the third of which you know from above to be $-\cos x$. Hence,
$$
\frac{d^{107}}{dx^{107}}(\sin x)=-\cos x.
$$
Do you see how all of that worked (both the mechanics and the reasoning)? 
Similar reasoning (the exact same basically) can be used to solve the "more complicated" problem at the beginning. That is, how would you compute $\frac{d^{1035}}{dx^{1035}}(\sin x)$? You should note that $1035=4\cdot 258+3$. Thus, this time you will have gone through $258$ cycles of taking four derivatives before beginning the last cycle. And you will end up with
$$
\frac{d^{1035}}{dx^{1035}}(\sin x)=-\cos x,
$$
just as before.
A: Notice, cyclic order of derivative of $\sin x$ is $4 $ hence the power (order of derivative) should be divided by $4$ & remainder is checked as follows
$$\begin{cases}\frac{d}{dx}(\sin x)=\cos x\ \text{(remainder}=1)\\ 
\frac{d^2}{dx^2}(\sin x)=-\sin x \ \text{(remainder}=2)\\
\frac{d^3}{dx^3}(\sin x)=-\cos x\ \text{(remainder}=3)\\
\frac{d^4}{dx^4}(\sin x)=\sin x\ \text{(remainder}=0)\\
\end{cases}$$
Now, we divide power (order) of derivative $107$ by $4$ to find remainder as follows $$107=4\times 26+3$$
In this case, we have $\text{(remainder}=3)$ hence case (3) is the answer
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\frac{d^{107}}{dx^{107}}(\sin x)=-cos x}}$$
