A friend suggested me a rather tricky problem, namely find the $100^{th}$ derivative of $$ f(x)=\frac{x^2+1}{x^3-x}. $$ I have computed the zeroth derivative $$ \frac{x^2+1}{x^3-x} $$ and the first derivative $$ \frac{2x(x^3-x)-(3x^2-1)(x^2+1)}{(x^3-x)^2}=\frac{1-x^4-4x^2}{(x^3-x)^2} $$ but I don't see any obvious structure.
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1$\begingroup$ Two probably won't be enough. Keep on differentiating, and there might be a pattern. $\endgroup$– H HuangCommented Aug 7, 2019 at 18:17
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$\begingroup$ This may or may not help but $1-x^4 -4x^2 = 2(x^2 -1) -(x^2 +1)^2$ $\endgroup$– fleabloodCommented Aug 7, 2019 at 18:31
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3 Answers
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Write your function as $$ \frac{x^2 + 1}{x^3 - x} = \frac{1}{x-1} + \frac{1}{x+1} - \frac{1}{x} $$
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Hint: since $x^3-x=x(x-1)(x+1)$ you can easily split $f(x)$ into partial fractions.
answered Aug 7, 2019 at 18:18
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$$\frac{x^2+1}{x^3-x}=\frac{x^2-1+2}{x^3-x}=\frac{1}{x}+\frac{2}{x(x-1)(x+1)}=$$ $$=\frac{1}{x}+\frac{2}{x+1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x+1}+2\left(\frac{1}{x+1}-\frac{1}{x}\right)=$$ $$=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}.$$ Can you end it now?
answered Aug 7, 2019 at 18:28