<s>[INST] Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ 
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .

Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{j}(x-1)^j. \end{align*}
Hence \begin{align*} f^{(n)}(1)&=n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2. \end{align*}
Is it correct? How to compute  $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$
 [/INST] An approach using the Lagrange inversion theorem: let $f(z)$ be analytic around $z=z_0$ with $f'(z_0)\neq0$; then $w=f(z)$ has an inverse $z=g(w)$ analytic around $w=f(z_0)$ with $$g^{(n)}\big(f(z_0)\big)=\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}\left(\frac{z-z_0}{f(z)-f(z_0)}\right)^n.\qquad(n>0)$$
We apply this to $f(z)=\dfrac{\sqrt{z}-1}{\sqrt{z}+1}$ and $z_0=1$, thus $g(w)=\left(\dfrac{1+w}{1-w}\right)^2$ and $$\lim_{z\to1}\frac{d^n}{dz^n}(1+\sqrt{z})^{2n+2}=g^{(n+1)}(0)=\color{blue}{4(n+1)!(n+1)}$$ since $g(w)=1+\displaystyle\frac{4w}{(1-w)^2}=1+4w\frac{d}{dw}\frac1{1-w}=1+4\sum_{n=1}^\infty nw^n$ for $|w|<1$.
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