Find $\frac{\mathrm d^{100}}{\mathrm d x^{100}}\frac{x^2+1}{x^3-x}=$? $$f(x)=\frac{x^2+1}{x^3-x}$$
$$f^{(100)}(x)=?$$
I tried differnetiating once and twice, but did not see any pattern emerging and can't guess what the 100th derivative should be. 

EDIT
so decomposing this as $$f(x)=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}
$$  does the job. Thanks for the hints! (edit: Sivaram has a complete calculation) Although a similar approach would greatly simplify this (next) problem can someone tell me what is wrong with my approach

My usual line of attack is to use Taylor expansion. For example the next problem in the same list asks for the $100^{th}$ derivative of $$\frac{1}{x^2-3x+2}$$ at $x=0$ within 10% relative error. 

NOTE:The above is a mistype, the following attempt is for $\frac{1}{x^2+3x+2}$. A better general approach, which is what I was looking for is described in the answer posted below.

I know I can expand in a Maclaurin series $$\frac{1}{x^2+3x+2}=\frac{1}{2} (1+\frac{x^2+3x}{2} + (\frac{x^2+3x}{2})^2 +\cdots)$$
After taking 100 derivatives I would be left to differentiate the following. 
$$\frac{1}{2}((\frac{x^2+3x}{2})^{50}+(\frac{x^2+3x}{2})^{51}+\cdots)$$
$$=\frac{1}{2}\left(\frac{\sum_{k=0}^{50}{{50}\choose{k}}3^{50-k} x^{50+k}
}{2^{50}}+\frac{\sum_{k=0}^{51}{{51}\choose{k}}3^{51-k} x^{51+k}
}{2^{51}}+\cdots\frac{\sum_{k=0}^{100}{{100}\choose{k}}3^{100-k} x^{100+k}
}{2^{100}}\right)$$
Because anything on either side of these values would disappear when i take the hundreth derivative at $x=0$ . And it is also easy to sea that I will get exactly one term from each of the sums, so I get an answer, 
$$=100!\sum_{k=0}^{50}\frac{3^{2k}}{2^{50+k}}$$
Which is wrong, well because the answer is too huge and Im to find a number within 10%. Can someone tell me where I went wrong, and if there is a cleaner way to approach these problems. 
 A: HINT: Try using partial fraction decomposition.
A: $$\frac{x^2+1}{x^3-x} = -\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-1} = -x^{-1} + (x+1)^{-1} + (x-1)^{-1}$$
$$\frac{d^n(y^{-1})}{dy^n} = \frac{(-1)^n n!}{y^{n+1}}$$
Hence, the $n^{th}$ derivative is
$$\frac{(-1)^{n+1} n!}{x^{n+1}} +  \frac{(-1)^n n!}{(x-1)^{n+1}} +  \frac{(-1)^n n!}{(x+1)^{n+1}} = (-1)^n n! \times \left( \frac{-1}{x^{n+1}} + \frac{1}{(x-1)^{n+1}} + \frac{1}{(x+1)^{n+1}}\right)$$
Similarly,
$$\frac{1}{x^2-3x+2} = \frac{1}{x-2} - \frac{1}{x-1}$$
Hence, the $n^{th}$ derivative is
$$\frac{(-1)^n n!}{(x-2)^{n+1}} -  \frac{(-1)^n n!}{(x-1)^{n+1}} = (-1)^n n! \times \left( \frac{1}{(x-2)^{n+1}} - \frac{1}{(x-1)^{n+1}}\right)$$
A: Doesn't look like there's a pattern to it. Mathematica gives me an answer that's like 4 pages long. And I don't think your answer is too large, because plotting it out gave values of the order $10^{200}$ and $100!\approx 10^{158}$, and the sum is approx $10^{17}$. So, the numbers aren't all that off. However, the 100th derivative evaluated at 0, was 0.
