If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer and longer. Am I on the right track? Thanks!
Best Answer
We can write:
$$1+ x + x^2 = \frac{1-x^3}{1-x}$$
Therefore:
$$f(x) = \frac{1-x}{1-x^3} $$
We can then expand this in powers of $x$:
$$f(x) = (1-x)\sum_{k=0}^{\infty}x^{3 k}$$
which is valid for $\left|x\right|<1$. The coefficient of $x^{36}$ is thus equal to $1$, so the 36th derivative at $x = 0$ is $36!$ .