<s>[INST] 4-th derivative of $(1+x+x^2) / (1-x+x^2) $ using Taylor polynomial for $1/(1-x)$ Using $n$-th Taylor polynomial for $f_1(x)=\frac{1}{1-x}$ with center in $0$, find $4$-th derivative of $f_2(x)=\frac{1+x+x^2}{1-x+x^2}$ in the point $0$ without calculating it's $1$,$2$ or $3$ derivative.  
I'm looking for hints, it's a homework. I've tried using the Taylor expansion with Lagrange rest, but it won't work because of restriction of calculating $1-3$ derivatives. Also I don't see connection between $f_1$ and $f_2$
Thanks in advance for help!
 [/INST] $$f_2(x)=\frac{1+x+x^2}{1-x+x^2}=\frac{(1+x+x^2)(1+x)}{(1-x+x^2)(1+x)}=\\
=\frac{(1+x+x^2)(1+x)}{1+x^3}=
=\frac{1+2x+2x^2+x^3}{1+x^3}=\\
 =1+2\frac{1+x}{1+x^3}=1+2 \frac{1}{1+x^3}+2x\frac{1}{1+x^3}.$$
Added:
From Taylor's polynomial for 
$$\frac{1}{1-x}=\sum\limits_{k=1}^{n}{\;x^k}+o(x^n)$$
we have
$$\frac{1}{1+x^3}=\sum\limits_{k=1}^{n}{(-1)^k (x^3)^k}+o((x^3)^n)=\sum\limits_{k=1}^{n}{(-1)^k x^{3k}}+o(x^{3n}),$$
so
$$f_2(x)=1+2 \sum\limits_{k=1}^{n}{(-1)^k x^{3k}}+2x \sum\limits_{k=1}^{n}{(-1)^k x^{3k}}=\\
=1+2 \sum\limits_{k=1}^{n}{(-1)^k x^{3k}} + 2\sum\limits_{k=1}^{n}{(-1)^k x^{3k+1}}+o(x^{3n+1}).$$
In this expansion coefficient $a_4$ at $x^4$ equals $-2.$ On the other hand, $a_4=\frac{f_2^{(4)}(0)}{4!},$ therefore,
$$f_2^{(4)}(0)=(-2)\cdot 4!=-48.$$
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