Explain this derivative identity: $ \frac{1}{2^n} \frac{d^n}{dy^n} \frac{(1+y)^{2n+3}(1-y)}{((1+y)^2 -2yx)^2} \bigg|_{y=0} = (n+1)! x^n $ I have the following result that I believe to be true:
$$
\frac{1}{2^n} \frac{d^n}{dy^n} \frac{(1+y)^{2n+3}(1-y)}{((1+y)^2 -2yx)^2} \bigg|_{y=0} = (n+1)! x^n
$$
The LHS is something that arose in physics research. The RHS has been inferred by checking with Mathematica for n from 0 to 100. However, proving this result has evaded me.
It's not surprising that the RHS is a polynomial in x with coefficient $(n+1)!$ (the only way to get an $x^n$ term is to operate the derivative on the denominator repeatedly). What is surprising is that the coefficients vanish for all but the leading term. This looks like it might be amenable to induction or recursion, but I was unable to make any meaningful headway with those techniques.
Good luck!
 A: We have 
$$\frac1{(1-t)^2}=\sum_{m=0}^\infty (m+1)t^{m},\qquad (|t|<1)$$
Put $t=\left(\dfrac{2y}{(1+y)^2}\right)x$, where $0<|y|<1$,
$$\frac1{\left(1-\left(\frac{2y}{(1+y)^2}\right)x\right)^2}=\sum_{m=0}^\infty (m+1)\left(\dfrac{2y}{(1+y)^2}\right)^mx^{m}.\qquad (|x|<\frac{(1+y)^2}{2|y|})$$
Multiply $(1+y)^{2n-1}(1-y)$ on both sides,
$$\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}=\sum_{m=0}^\infty(m+1)\big((1+y)^{2n-2m-1}(y^m-y^{m+1})\big)(2x)^m.\qquad (|x|<\frac{(1+y)^2}{2|y|})\tag{*}$$

Consider the coefficient of $y^n$ in $(*)$,
  
  
*
  
*when $m<n$, the coefficient is given by
  $$\binom{2n-2m-1}{n-m}y^{n-m}\cdot y^m-\binom{2n-2m-1}{n-m-1}y^{n-m-1}\cdot y^{m+1}=0;$$
  
*when $m=n$, 
  \begin{align*}
(1+y)^{-1}(y^n-y^{n+1})=\left(\sum_{k=0}^\infty (-y)^k\right)(y^n-y^{n+1})=y^n+\underbrace{\text{something}}_{\text{“deg of $y$” $>n$}},
\end{align*}
  so the coefficient is $(n+1) (2x)^n$;
  
*when $m>n$, $2n-2m-1<0$,
  \begin{align*}
(1+y)^{2n-2m-1}(y^m-y^{m+1})=\left(\sum_{k=0}^\infty (-y)^k\right)^{1+2m-2n}(y^m-y^{m+1})
\end{align*}
  does not contain $y^n$.

Therefore,
$$\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}=\color{blue}{\underbrace{\text{something}}_{\text{“deg of $y$” $<n$}}}+\big((n+1) (2x)^n\big)y^n+\color{orange}{\underbrace{\text{something}}_{\text{“deg of $y$” $>n$}}},\qquad (|x|<\frac{(1+y)^2}{2|y|})$$
Take the $n$-th derivative with respect to $y$ and set $y\to 0$,
\begin{align*}
\left.\frac{\partial^n}{\partial y^n}\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}\right|_{y=0}&=\color{blue}{0}+\big((n+1) (2x)^n\big)(y^n)^{(n)}+\color{orange}{0},\qquad (|x|<+\infty)\\
\left.\frac{\partial^n}{\partial y^n}\frac{(1+y)^{2n+3}(1-y)}{((1+y)^2-2yx)^2}\right|_{y=0}&=(n+1)! (2x)^n.\qquad (x\in \Bbb R)
\end{align*}
