How to take taylor polynomial of harder functions?? I'm solving an old calculus exam because my test in less than 10 days and I came against sin(sin(x)) and I am supposed to find the 5th-degree Taylor polynomial when x=0.
I tried to do it with the two ways I know:  
1-normal derivation five times but it took me more than 40 min doing that and re-checking and filling the numbers ... etc.
(2)-I already know what the Taylor polynomial of sin(x) considering it is a famous one and I substituted it with f(x) and also at some point substituted with fُ(x) and so on to eventually re-fill what I know and eliminating what is bigger than power 5 but I barely found any difference in time.
I'm wondering how a mathematician will do it and if there is a way to save time?
 A: To the fifth degree,
$$\sin x=x-\frac{x^3}6+\frac{x^5}{120}.$$
From this,
$$\sin(\sin x)=\left(x-\dfrac{x^3}6+\dfrac{x^5}{120}\right)-\frac{\left(x-\dfrac{x^3}6+\dfrac{x^5}{120}\right)^3}6+\frac{\left(x-\dfrac{x^3}6+\dfrac{x^5}{120}\right)^5}{120}.$$
But you only need the terms to degree five, so you can drop the terms that yield higher powers,
$$\sin(\sin x)=\left(x-\dfrac{x^3}6+\dfrac{x^5}{120}\right)-\frac{\left(x-\dfrac{x^3}6+\cdots\right)^3}6+\frac{(x+\cdots)^5}{120}.$$
By the binomial theorem, the middle term yields
$$-\frac{x^3-3\dfrac{x^5}6+\cdots}6.$$
Collecting these results,
$$\sin(\sin x)=x-\frac{x^3}6+\frac{x^5}{120}-\frac{x^3}6+\frac{x^5}{12}+\frac{x^5}{120}=x-\frac{x^3}3+\frac{x^5}{10}.$$
A: Well, that's not too difficult. The chain rule says $$\frac{d}{dx}\sin(\sin x)=\cos x\cdot\cos(\sin x)\tag1$$ and $$\frac{d}{dx}\cos(\sin x)=-\cos x\cdot\sin(\sin x)\tag2.$$ Obviously, $\sin(\sin x)$ is an odd function, $\cos(\sin x)$ an even one. With 
\begin{align}
\sin(\sin x)&=a_1x+a_3x^3+a_5x^5+\ldots\\
\cos(\sin x)&=b_0+b_2x^2+b_4x^4+\ldots
\end{align} and $$\cos x=1-\frac12x^2+\frac1{24}x^4+\ldots,$$ comparing coefficients in (1) gives
\begin{align}
a_1&=b_0\\
3a_3&=-\frac12b_0+b_2\\
5a_5&=\frac1{24}b_0-\frac12b_2+b_4
\end{align} In the same way, (2) gives
\begin{align}
2b_2&=-a_1\\
4b_4&=\frac12a_1-a_3
\end{align} Clearly, $b_0=1$, so we obtain recursively $a_1=1$, $b_2=-\frac12$, $a_3=-\frac13$, $b_4=\frac5{24}$, $a_5=\frac1{10}$, and your result is
$$\sin(\sin x)=x-\frac13x^3+\frac1{10}x^5+\ldots$$
