<s>[INST] Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$? 
Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$?

By letting $g(x)=x^3+3x^2+3x+5$, it is easy to see that $g'(-1) = g''(-1) = 0$, and so $f'(-1) = f''(-1) = f'''(-1) = 0$.
However, I have no idea how to compute $f^{(n)}(-1)$ for any positive integer $n$, and in particular for $n=99$.
Any idea would be appreciated.
 [/INST] Let $t=x+1$. Then:
$$ f(x)= {1\over (x+1)^3+4} = {1\over 4}{1\over 1+{t^3\over 4}}=$$
$$ = {1\over 4}\Big(1-{t^3\over 4}+ \big({t^3\over 4}\big)^2-\big({t^3\over 4}\big)^3+...\Big)$$ 
So $$f^{(99)}(x) = -{1\over 4}{99!\over 4^{33}} + t(...)\implies f^{(99)}(-1) = -{99!\over 4^{34}}$$
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