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I need to find the nth derivative of $\frac{x^{n}}{(1-x)^{2}}$ for $0<x<1$

So far, I tried the same method used for $\frac{x^{n}}{1-x}$ and here's what I got: \begin{equation} \frac{x^{n}}{(1-x)^{2}}=x^{n}(1+2x+3x^{2}+4x^{3}+....)=(x^{n}+2x^{n+1}+3x^{n+2}+...) \end{equation} Take nth derivative to get: \begin{align} \frac{\partial^n }{\partial x^n} \frac{x^{n}}{(1-x)^{2}} \notag\\ = & (n!+2(n+1)!x+3\frac{(n+2)!}{2!}x^{2}+...) \notag\\ = & n!(1+2(n+1)x+3\frac{(n+2)(n+1)}{2!}x^{2}+4\frac{(n+3)(n+2)(n+1)}{3!}x^{3}+..) \end{align} Next would be finding a a function whose taylor expansion is the series in the parenthesis but I couldn't think of one. Any ideas for a function or for another method to do this? Thanks!

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Now you have two almost complete answers, do you know how to "accept" an answer? Some think it is rude not to accept answers - which is done by a click in the check mark to the left of the answer post. – AD. Jun 24 '13 at 10:42

Write $x = 1-t$, so you're looking at $$(-1)^n \dfrac{d^n}{dt^n} \dfrac{(1-t)^n}{t^2} = \dfrac{d^n}{dt^n} \sum_{j=0}^n {n \choose j} (-1)^{n+j} t^{j-2}$$ The terms with $j \ge 2$ all vanish, and you're left with $$ \dfrac{d^n}{dt^n} \left((-1)^n (t^{-2} - n t^{-1})\right)$$

From there it's easy. You can write the answer as $$\dfrac{n!(nx+1)}{(1-x)^{n+2}}$$

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What a great derivation, thank you so much! – johnny israeli Jun 24 '13 at 10:31

Consider the function $$g(u):={(1-u)^n\over u^2}={1\over u^2}\sum_{k=0}^n{n\choose k}(-u)^k ={1\over u^2}-{n\over u}+\sum_{k=2}^n{n\choose k}(-u)^{k-2}\ .$$ It follows that $$g^{(n)}(u)={(-1)^n (n+1)!\over u^{n+2}}-{n\>(-1)^n\> n!\over u^{n+1}}=(-1)^n n!{n(1-u)+1\over u^{n+2}}\ .$$ Now $f(x)=g(1-x)$, and therefore $$f^{(n)}(x)=(-1)^n g^{(n)}(1-x)=n!{n x+1\over (1-x)^{n+2}}\ .$$

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A related problem. Here is a closed form

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

Added: Here is the $n$th derivative for the wanted function

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

You can apply the technique I used to solve this problem.

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wow, very interesting result! can you show a step by step derivation please? – johnny israeli Jun 24 '13 at 7:34
@johnnyisraeli: I'll try to do it later. – Mhenni Benghorbal Jun 24 '13 at 7:39
@johnnyisraeli: I think you can prove it by induction. – Mhenni Benghorbal Jun 24 '13 at 7:46
Whoever voted this answer to be good should undo it, it makes others think this actually answers my question but it doesn't really get us closer to the answer as far as I can tell.. – johnny israeli Jun 24 '13 at 7:54
Note that the denominator in the question is $(1-x)^2$, and not $1-x^2$. – Christian Blatter Jun 24 '13 at 8:27

You could write $$\frac{\partial^n}{\partial x^n}x^n(1-x)^{-2}=\sum_{k=0}^n\binom{n}{k}\left(\frac{\partial^k}{\partial x^k}x^n\right)\left(\frac{\partial^{n-k}}{\partial x^{n-k}}(1-x)^{-2}\right)$$ and proceed from there. You'd have it as a finite sum.

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I need a closed form expression like I'd get with that same method for $\frac{x^{n}}{1-x}$, any ideas in that direction? – johnny israeli Jun 24 '13 at 7:14
You say you are looking for a closed form expression for this derivative. Just curious - why the specification that $0<x<1$? Surely a closed form expression or even a finite sum will be valid on almost all $x$. It's the kind of specification one would have if the expected answer were an infinite series. – alex.jordan Jun 24 '13 at 7:27
In my specific application with protein concentrations that just happens to be the case so I mentioned it in case infinite series can help in the same way that yields $\frac{\partial^n }{\partial x^n} \frac{x^{n}}{1-x}=n!(1-x)^{-(n+1)}$ – johnny israeli Jun 24 '13 at 7:33

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