I am to find a formula for the n-th derivative of $$\frac{1+x}{1-x}$$ I came up with $$\frac{(-1)^{n+1} \times 2 \times n!}{(1-x)^{n+1}}$$

This seems right, but I noticed that WolframAlpha somehow manages to eat the sign for any n.

For example, for $n=2$ I have: $\frac{-4}{(1-x)^{3}}$, but WolframAlpha gives $\frac{4}{(1-x)^3}$. I am not particularly good at calculating, but I would like to think that I have mastered the numbers below $10$ so I'm pretty sure I'm right here and WolframAlpha did something very weird.

EDIT: Yep, my derivative is wrong. BUT this still doesn't answer the question why WolframAlpha responds with a positive fraction all the time - I didn't ask it to calculate a derivate, I just entered my formula.

PS: Who removed my WolframAlpha input? That was sort of essential to the question, I was never that interested in the derivative itself, I just wanted to know why WA responded that way.

(Exact input: (-1)^(n+1)*2*n!/((1-x)^(n+1)), n=2 )

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    $\begingroup$ In the WA output, for $n=2$, they have $4\over (\color{maroon}{x-1})^3$, which is the same as $-4\over (\color{maroon}{1-x})^3$. $\endgroup$ – David Mitra May 6 '12 at 14:55

For your Wolfram Alpha input "(-1)^(n+1)*2*n!/((1-x)^(n+1))", I get the output $4\over (x-1)^3$ (not what you have written, I surmise you just misread it).

Note that $$ {4\over(x-1)^3} ={4\over \bigl(-(1-x)\bigr)^3}={4\over (-1)^3(1-x)^3}={-4\over(1-x)^3}; $$ which is exactly what your formula gave.



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

This makes the problem easier to derivate. I think WA is right, you probably forgot to multiply by the derivative of $(1-x)$ at some point....

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    $\begingroup$ "differentiate", not "derivate" $\endgroup$ – Antonio Vargas May 6 '12 at 15:30

Let $$f(x) := \frac{1+x}{1-x} = \frac{x - 1 + 2}{1-x} = -1 + \frac{2}{1-x}$$ Thus $$f'(x) = -2 (1-x)^{-2} \cdot (-1) = \frac{2}{(1-x)^2}$$and $$f''(x) = 2 \cdot (-2 (1-x)^{-3}) \cdot (-1) = \frac{2 \times 2!}{(1-x)^3}$$

More generally, if we assume that $$f^{(n)}(x) = \frac{2 \cdot n!}{(1 - x)^{n+1}}$$ then $$ f^{(n+1)}(x) = 2 \cdot n! \cdot ({-(n+1)} (1-x)^{-(n+2)}) \cdot (-1) = \frac{2 \cdot (n+1)!}{(1 - x)^{(n+1) + 1}}, $$ which can be proved for all $n$ by induction. I suspect you forgot the $(-1)$ factor in each derivative that comes from the Chain Rule.

  • $\begingroup$ My derivative may be wrong - I will check that right away - but I didn't ask WolframAlpha to calculate the derivative, I asked it to calculate my formula for the derivative for a given n. $\endgroup$ – Cubic May 6 '12 at 14:32

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