<s>[INST] Find the $f^{(3)}(0)$ of $f(x) = \sin^{3}(\ln(1+x))$ using Taylor expansion For the Taylor expansions I used:
$\ln(1+x) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^{k}}{k}$
$\sin(x) = \sum_{k=0}^\infty (-1)^{k+1} \frac{x^{2k+1}}{(2k+1)!}$
So $\sin(x) = x - \frac{x^{3}}{3!} + O(x^{5})$
and $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$ 
Then the composite expansion (for rest of question dropping H.O.T.):
$\sin(\ln(1+x)) \approx (x - \frac{x^2}{2} + \frac{x^3}{3}) - \frac{(x - \frac{x^2}{2} + \frac{x^3}{3})^3}{3!} $
Reducing:
$\approx (x - \frac{x^2}{2} + \frac{x^3}{3}) - \frac{x^3}{6} \approx x - \frac{x^2}{2} + \frac{x^3}{6}$
Cube the result and simplify again:
$(x - \frac{x^2}{2} + \frac{x^3}{6})^3 \approx x^3$ keeping the third-order term.
At this point, my understanding is that the coefficient of the third-order term in the formula for the general Taylor series is $f^{(3)}$
$\frac{f^{(3)}(a)}{3!}(x-a)^3$
So I believe I can infer that the coefficient of the composite Taylor expansion, 1, also relates to $f^{(3)}$, and then solving:
$\frac{f^{(3)}}{3!} = 1$
$f^{(3)} = 3! = 6$
Have I understood this question and the right approach for solving it?
 [/INST] This is correct, though take note that you should be more careful with your remainders on each step to ensure you didn't miss anything that might contribute to the $x^3$ coefficient:
\begin{align}\sin(\ln(1+x))&=\sin\bigg(x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)\bigg)\tag{$\star$}\\&=\left[\color{#2255FF}{x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)}\right]-\frac16(\color{#228855}{x+\mathcal O(x^2)})^3+\mathcal O(\color{#FF9944}{x}^4)\\&=x-\frac{x^2}2+\frac16x^3+\mathcal O(x^4)\end{align}
$(\star)$: $\color{#2255FF}{x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)}=\color{#228855}{x+\mathcal O(x^2)}=\color{#FF9944}{\mathcal O(x)}$
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