# 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.

• Your initial problem can be simplified to 1/x. The second I would attack with partial fractions: $\frac{1}{x^2-3x+2}=\frac{-1}{x-1}+\frac{1}{x-2}$ and each term is easy to take lots of derivatives of. Mar 17, 2011 at 20:03
• @Ross It does not ask to evaluate derivative at any point, so how does it reduce to 1/x? Regarding the second your method sounds better and i'll try it, but now I'm also curious why my first approach is wrong.
Mar 17, 2011 at 20:06
• I misread it, I thought there was a plus sign in the denominator. You can still do partial fractions as the denominator factors into linear terms. For your approach, unless you are asked for the value of the derivative at 0, you must keep all the terms past the $50^{th}$ because they will not go to zero-they will have $x$'s in them. You also seem to have flipped a sign twice-once when you did the MacLauren series and went from $x^2-3x$ to $x^2+3x$ and again going to the line with the $\sum$ signs, where the line before has alternate terms negative and this line has all positive. Mar 17, 2011 at 20:21
• I guess I didn't misread it, there was a plus sign, but it got changed after my comment. Mar 17, 2011 at 20:23
• Lol! I just realized I had automatic university subscription. Here is a free link. (The questions start after two pages or so) scribd.com/doc/50636877/V-I-Arnold-A-mathematical-trivium
Mar 17, 2011 at 20:36

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

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

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

• I was wondering why my answer did not match up to yours, and it seems both are correct. I made a sign error while stating the problem. Sorry for that and thanks.
Mar 17, 2011 at 21:27

HINT: Try using partial fraction decomposition.

• I did, for the first question. It still amounted to differentiating two different functions 100 times, and for the individual functions, have you tried differentiating and see if they fit a pattern?
• @Approximist: You should be able to find a pattern for the derivatives of $1/(x-c)$, and the partial fraction expansion is a linear combination of such terms. Mar 17, 2011 at 20:07
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.
• Define the function: f[x_]:=(x^2+1)/(x^3-x); 100th derivative: f100[x_]:=D[f[x],{x,100}]; To calculate 100!, Log[10,N[Factorial[100]]]