Find the $n$th derivative of $f(x) =\frac{x^n}{1-x}$ 
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*Question
Find the $n$th derivative of $f(x) =\frac{x^n}{1-x}$



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*What I've managed thus far
First I thought that I might be able to discern a pattern by calculating the first few derivatives of $f(x) =\frac{x^n}{1-x}$ using the multiplication- or quotient rule of derivatives, but that soon turned out to be extremely tedious.
Can anybody give me a hint regarding an approach to this problem? Keep in mind that this should be done with the tools of introductory derivatives.
 A: Note that
$$\frac{(1-y)^n}y=p(y)+\frac1y,$$
where the degree of the polynomial $p$ is $n{-}1$. Setting $y=1{-}x$ we get
$$\frac{x^n}{1-x}=p(1{-}x)+\frac1{1-x}$$
So the $n$-th derivative is
$$\left(\frac1{1-x}\right)^{(n)}=\frac{n!}{(1-x)^{n+1}}$$
as the $n$-th derivative of the polynomial is $0$.
A: First find the $n^{\textrm{th}}$ derivative of $\frac{1-x^n}{1-x}$.
Hint: $(1-x^n) = (1-x)(1+x+\ldots + x^{n-1})$
A: Let $f(x)=h(x)g(x)$ where $h(x)=x^n$, $g(x)=\frac{1}{(1-x)}$.
It's easy to show that:
$$g^{(k)}(x)=\frac{n!}{(n-k)!}x^{n-k}$$
$$h^{(k)}(x)=\frac{k!}{(1-x)^{k+1}}$$
You can prove this for example by discerning the pattern.
Next by induction and chain rule prove that:
$$f^{(k)}(x)=(hg)^{(k)}(x)=\sum_{k=0}^{n}{n \choose k}g^{(k)}(x)h^{(n-k)}(x)$$
So:
$$f^{(k)}(x)=\sum_{k=0}^{n}{n \choose k}\frac{n!}{(n-k)!}x^{n-k}\frac{(n-k)!}{(1-x)^{n-k+1}}=\frac{n!}{1-x}\sum_{k=0}^{n}{n \choose k}\left(\frac{x}{(1-x)}\right)^{n-k}=\\=\frac{n!}{1-x}\left(\frac{x}{1-x}+1\right)^n=\frac{n!}{(1-x)^{n+1}}$$
A: Ignoring issues of convergence, we have 
$$
f(x)=\frac{x^n}{1-x}=x^n\cdot\frac1{1-x}=x^n(1+x+x^2+\cdots)
=x^n+x^{n+1}+x^{n+2}+\cdots
$$
so that
$$f^{(n)}(x)=n!+\frac{(n+1)!}{1!}x+\frac{(n+2)!}{2!}x^2+\cdots=\frac{n!}{(1-x)^{n+1}}$$ by noting that the middle term is the Taylor series of the right side.
A: Maybe you can use a little bit tricky method:
In complex analysis, you have $f^{(n)}(x)= n! \oint f(z)/(z-x)^{n+1} dz$, where the contour is around $z=x$, then, you can consider a contour in infinity, according to residue theorem , the integral equals to the residue at $z=x, z=1$, but the infinity integral is zero, so we must have the integral equal to minus the residue at $z=1$, which is a simple pole so we directly have the answer. 
A: Note: This solution was inspired by the hints provided by Simon S in the comment section of the question.


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*$f(x) = \frac{x^n}{1-x}$, now let $u=x-1$, such that $$f(u) = -\frac{(u+1)^n}{u}$$


*Note that $\frac{df}{dx} = \frac{df}{du} \frac{du}{dx} = \frac{df}{du}\cdot 1 = \frac{df}{du}$


*Expand $f(u) = -\frac{(u+1)^n}{u}$ using the binomial theorem, such that $$f(u) = -\frac{1}{u}\sum\limits_{i=0}^n {n\choose i} u^{n-1}$$
$$= -\frac{1}{u}\Bigg[{n\choose 0}u^{n} + {n\choose 1}u^{n-1} + {n\choose 2}u^{n-2} + \cdot\cdot\cdot + {n\choose n-1}u^{1} + 1\Bigg]$$
$$= -{n\choose 0}u^{n-1} - {n\choose 1}u^{n-2} - {n\choose 2}u^{n-3} - \cdot\cdot\cdot - {n\choose n-1} - \frac{1}{u}$$

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*Now $f(u)$ can be viewed as a polynomial of degree $n-1$ plus $-\frac{1}{u}$.


*The $nth$ derivative of a polynomial of degree $n-1$ is equal to $0$, that leaves us with $$\frac{d^n}{du^n} [f(u)] = \frac{d^n}{du^n}\Big[-\frac{1}{u}\Big] = \frac{d^n}{du^n}\Big[-u^{-1}\Big]$$
$$= \frac{n!}{(1-x)^{n+1}}$$
