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$$f(x)=\frac{x^2+1}{x^3-x}$$

$$f^{(100)}(x)=?$$

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

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My usual line of attack is to use talor 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.

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$$\frac{x^2+1}{x^3-x} = -\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-1} = -x^{-1} + (x+1)^{-1} + (x-1)^{-1}$$

$$\frac{d^n(y^{-1})}{dy^n} = \frac{(-1)^n n!}{y^{n+1}}$$

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

$$\frac{(-1)^n n!}{(x-2)^{n+1}} – \frac{(-1)^n n!}{(x-1)^{n+1}} = (-1)^n n! \times \left( \frac{1}{(x-2)^{n+1}} – \frac{1}{(x-1)^{n+1}}\right)$$

**HINT**: Try using partial fraction decomposition.

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.

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