Computing $n$th derivative at a specific value of $x$ I'm trying to find the Maclaurin Series for the function
$$f(x)=\sin(\sin x)$$
But I'm having a little bit of trouble because the derivatives just get messier and messier, and in order to construct the $f^{(n)}(0)$ needed for each term of the Taylor expansion I have to find some kind of formula that gives the $n$th derivative of $f(x)$.
I would appreciate it if somebody could show me how to find such a formula.
OR:
Is there a way to calculate each $f^{(n)}(0)$ without calculating $f^{(n)}(x)$? This would be highly preferable to just calculating an $n$th derivative formula, since I don't need the derivatives at any point other than at $x=0$.
The first couple terms, if it helps, are
$$f'(0)=1$$
$$f'''(0)=-2$$
$$f'''''(0)=-12$$
I believe that all $f^{(2n+1)}(0), n\in \mathbb Z$ is always $0$, (at least it is for the first couple iterates) so I didn't include them.
Thanks!
 A: Not a complete answer, but here's an idea too long to fit in a comment:
Using Euler's formula we have $f(x) = \sin(\sin(x)) = \frac{1}{2i}\left(e^{i\sin(x)} - e^{-i\sin(x)}\right)$.
Let's study the function $g(x) = e^{i\sin(x)}$. We have $g'(x) = i\cos(x)g(x)$.
Using the Leibniz formula for the $n$-th derivative of a product, we find that $$g^{(n+1)}(x) = i\sum\limits_{k = 0}^{n}\binom{n}{k}\cos^{(k)}(x)g^{(n-k)}(x).$$
Similarly, for the function $h(x) = e^{-i\sin(x)}$ we find that $$h^{(n+1)}(x) = -i\sum\limits_{k = 0}^{n}\binom{n}{k}\cos^{(k)}(x)h^{(n-k)}(x).$$
Plugging in $x = 0$ we obtain, using $\cos^{(k)}(0) = (-1)^{k/2}$ when $k$ is even, and $0$ otherwise:
$$g^{(n+1)}(0) = i\sum\limits_{k\text{ even}}^{n}\binom{n}{k}(-1)^{k/2}g^{(n-k)}(0)$$
$$h^{(n+1)}(0) = -i\sum\limits_{k\text{ even}}^{n}\binom{n}{k}(-1)^{k/2}h^{(n-k)}(0)$$
and rewriting this using $k/2$ instead of $k$:
$$g^{(n+1)}(0) = i\sum\limits_{k = 0}^{\lfloor n/2\rfloor}\binom{n}{2k}(-1)^{k}g^{(n-2k)}(0)$$
$$h^{(n+1)}(0) = -i\sum\limits_{k = 0}^{\lfloor n/2 \rfloor}\binom{n}{2k}(-1)^{k}h^{(n-2k)}(0)$$
These formulas provide recursive expressions to calculate $g^{(n)}(0)$ and $h^{(n)}(0)$, together with the base case $g^{(0)}(0) = 1$ and $h^{(0)}(0) = 1$.
One can then find $f^{(n)}(0) = \frac{1}{2i}\left(g^{(n)}(0) - h^{(n)}(0)\right)$.
