<s>[INST] Taylor series expansion of $f(x) = \sin^3 \left(\ln(1+x) \right)$. How does one use Taylor series expansion to compute $f^{(3)}(0)$ in which $f(x) = \sin^3 \left(\ln(1+x) \right)$.
 [/INST] One starts with
$$\begin{align}
\log(1+x)=&x-{x^2\over 2}+{x^3\over 3}+o(x^3)\\
\sin{X}=&X-{X^3\over 6}+o(X^3)
\end{align}$$
Now make $X=\log(1+x)$ in the development of the sine function to get
$$\begin{align}\sin(\log(1+x))=&x-{x^2\over 2}+{x^3\over 3}-{\left(x-{x^2\over 2}+{x^3\over 3}\right)^3\over 6}+o(x^4)\\
=&x-{x^2\over 2}+{x^3\over 6}+o(x^3)\end{align}$$
Now every Taylor series around $0$ is of the form $f(0)+f'(0)x+f''(0)x^2/2+f'''(0)x^3/6+o(x^3)$ and so $f'''(0)=1$. This is for the sine
Now if we raise it to the cube and do the same type of calculations, noting that we can limit the series of sine and logarithms to the first order, we get
$$\sin^3(\log(1+x))=x^3+o(x)$$ 
And this means $f'''(0)=6$
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