Compact formula for the $n$th derivative of $f\left(\sqrt{x+1}\right)$? I'm interested in a general formula for
$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big].$$
In particular, Fàa di Bruno's formula gives
$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]=\sum_{k=1}^n f^{(k)}\left(\sqrt{x+1}\right)B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big),$$
where $g(x)=\sqrt{x+1}$ and $B_{n,k}$ denote Bell polynomials.
I suspect we can do better than this, using the identities
$$\frac{d^k}{dx^k}\Big[\sqrt{x+1}\Big]=\left(\frac{1}{2}\right)^\underline{k}x^{1/2-k}=-\frac{(2k-3)!!}{(-2)^k}x^{1/2-k}=\frac{\sqrt{\pi}}{2\Gamma(3/2-k)}x^{1/2-k}.$$
In particular, every term in $B_{n,k}$ for fixed $n$ and $k$ will be some numerical factor times $(x+1)^{k/2-n}$ (see notes below), so the result is always of the form
$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]
=\sum_{k=1}^n a_{n,k} \frac{f^{(k)}\left(\sqrt{x+1}\right)}{(x+1)^{n-k/2}}$$
for some constants $a_{n,k}.$ Is there an explicit expression for the $a_{n,k}$ above? For example, it seems likely that there is some expression using Stirling numbers, factorials, binomial coefficients, etc., given the combinatorial nature of the problem.

Current progress:
Using the first identity above for derivatives of $\sqrt{x+1}$ [where $(\cdot)^\underline{k}$ denotes the $k$th falling factorial], we have
$$B_{n,k}\big(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\big)
=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1}x^{1/2-1},\left(\frac{1}{2}\right)^\underline{2}x^{1/2-2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}x^{1/2-(n-k+1)}\right].$$
The definition of the Bell polynomials guarantees that the power of $x$ in every term of $B_{n,k}(\cdots)$ comes out to $x^{k/2-n}$, with pre-factor given by
$$a_{n,k}=B_{n,k}\left[\left(\frac{1}{2}\right)^\underline{1},\left(\frac{1}{2}\right)^\underline{2},\dots,\left(\frac{1}{2}\right)^\underline{n-k+1}\right],$$
so the remaining task is to simplify the above expression, if possible.
 A: Heuristic
By analyzing the matrix of the first values for $a_{n,m}$ it is possible to
deduce that
$$ \bbox[lightyellow] {  
a_{\,n,\,m}  =\frac{ \left( { - 1} \right)^{n - m} }{2^n} \left( \matrix{  2n - 1 - m \cr   2n - 2m \cr}  \right)
\left( {2n - 2m - 1} \right)!!\quad \left| {\,0 \le m \le n} \right.
} \tag{1}$$
which can be further simplified in terms of Gamma or other, and which can be
of help for a rigorous demonstration.
Demonstration
We put
$$
{{d^{\,n} } \over {dx^{\,n} }}f\left( {\sqrt {x + 1} } \right)
 = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} {{f^{\,\left( k \right)} \left( {\sqrt {x + 1} } \right)}
 \over {\left( {x + 1} \right)^{\,n - k/2} }}} 
$$
The $a_{\,n,\,k} $ are supposed to be constant, independent from $f$.
Then they shall be valid for any infinitely differentiable function, in particular
for analytic functions, and thus in particular for
$$
f(x) = x^{\,2m} \quad \left| {\,0 \le m \in Z} \right.
$$
For this we have
$$
\left\{ \matrix{
  f\left( {\sqrt {x + 1} } \right) = \left( {x + 1} \right)^{\,m}  \hfill \cr 
  {{d^{\,n} } \over {dx^{\,n} }}f\left( {\sqrt {x + 1} } \right)
 = m^{\,\underline {\,n\,} } \left( {x + 1} \right)^{\,m - n}  \hfill \cr 
  f^{\,\left( n \right)} \left( {\sqrt {x + 1} } \right)
 = \left( {2m} \right)^{\,\underline {\,n\,} } \left( {\sqrt {x + 1} } \right)^{\,2m - n}  \hfill \cr}  \right.
$$
and therefore
$$
\eqalign{
  & m^{\,\underline {\,n\,} } \left( {x + 1} \right)^{\,m - n}  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} {{\left( {2m} \right)^{\,\underline {\,k\,} }
 \left( {x + 1} \right)^{\,m - k/2} } \over {\left( {x + 1} \right)^{\,n - k/2} }}}
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \left( {2m} \right)^{\,\underline {\,k\,} }
 \left( {x + 1} \right)^{\,m - n} }   \cr 
  & \quad \quad  \Downarrow   \cr 
  & m^{\,\underline {\,n\,} }
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \left( {2m} \right)^{\,\underline {\,k\,} } }
 \quad  \Leftrightarrow \quad \left( {{m \over 2}} \right)^{\,\underline {\,n\,} }
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} m^{\,\underline {\,k\,} } }  \cr} 
$$
Since
$$
\eqalign{
  & \left( {{m \over 2}} \right)^{\,\underline {\,n\,} }
  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {\left( { - 1} \right)^{\,n - j} \left[ \matrix{  n \cr   j \cr}  \right]\left( {{m \over 2}} \right)^{\,j} }
  = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)}
 {{{\left( { - 1} \right)^{\,n - j} } \over {2^{\,j} }}\left[ \matrix{ n \cr  j \cr} \right]m^{\,j} }   \cr 
  & \sum\limits_{0\, \le \,k} {a_{\,n,\,k} m^{\,\underline {\,k\,} } }
  = \sum\limits_{0\, \le \,k} {a_{\,n,\,k} \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)}
 {\left( { - 1} \right)^{\,k - j} \left[ \matrix{k \cr  j \cr}  \right]m^{\,j} } }  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,j\,} {\left( {\sum\limits_{0\, \le \,k}
 {\left( { - 1} \right)^{\,k - j} a_{\,n,\,k} \left[ \matrix{ k \cr  j \cr}  \right]} } \right)m^{\,j} }  \cr} 
$$
i.e.
$$
{{\left( { - 1} \right)^{\,n - j} } \over {2^{\,j} }}\left[ \matrix{  n \cr  j \cr}  \right]
 = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k - j} a_{\,n,\,k}
 \left[ \matrix{  k \cr   j \cr}  \right]} 
$$
which can be viewed as a matrix relation involving the matrix $A:\; A_{\,n,\,k} = a_{\,n,\,k}$
multiplied by the matrix of the signed Stirling N. of 1st kind, which is well known to be
the inverse of the Stirling N. of 2nd kind, so:
$$ \bbox[lightyellow] {  
a_{\,n,\,m}  = \sum\limits_{0\, \le \,k}
 {{{\left( { - 1} \right)^{\,n - k} } \over {2^{\,k} }}
\left[ \matrix{  n \cr   k \cr}  \right]\left\{ \matrix{  k \cr  m \cr}  \right\}} 
}\tag{2}$$
The computation of the above checks with the heuristic formula:
it remains to demonstrate this fact analytically.
A: Thanks to G Cab's inference of the correct answer and Gary's comment on the recursive formula for the $a_{n,k}$'s, I have now found a proof by induction of the result. I'll post it here for posterity.

Define $a_{n,k}=0$ if $k=0$ or $k>n$, and otherwise let it be defined as in the original post. Then we can write
$$\frac{d^n}{dx^n}\Big[f\left(\sqrt{x+1}\right)\Big]
=\sum_{k=0}^\infty a_{n,k} \frac{f^{(k)}\left(\sqrt{x+1}\right)}{(x+1)^{n-k/2}},$$
and differentiating both sides gives the recurrence relation $a_{n+1,k}=\frac{1}{2}a_{n,k-1}+(k/2-n)a_{n,k}$ for all $n,k≥1.$ This is equivalent to
$$a_{n+1,k+1}=\frac{1}{2}a_{n,k}+\left(\frac{k+1}{2}-n\right)a_{n,k+1}\qquad\Bigg|\qquad n≥1,k≥0.$$
Now define
$$c_{n,k}:=\frac{(-1)^{n-k}}{2^n}\frac{(2n-1-k)!}{(2n-2k)!!(k-1)!}$$
for $n≥1,1≤k≤n$, and $c_{n,k}=0$ otherwise. (In fact, the first expression also works for these cases if we interpret negative factorials as $\infty$ and $1/\infty$ as zero.) It is straightforward to check that this formula for $c_{n,k}$ is equivalent to $(1)$ in G Cab's answer.
Induction Part 1
It is quick to check that $c_{n,k}=0=a_{n,k}$ for $k=0$ or $k>n$, and that $c_{1,1}=1/2=a_{1,1}.$
Induction Part 2
Let $n≥1,k≥0$ be fixed, and assume that $c_{n,k}=a_{n,k}$ and $c_{n,k+1}=a_{n,k+1}$ for these particular values. We will show that $c_{n+1,k+1}=a_{n+1,k+1},$ which proves the desired result by induction.
$\begin{align*}
a_{n+1,k+1}&=\frac{1}{2}a_{n,k}+\left(\frac{k+1}{2}-n\right)a_{n,k+1}
\\\\
&=\frac{1}{2}c_{n,k}+\left(\frac{k+1}{2}-n\right)c_{n,k+1}
\\\\
&=\frac{1}{2}\frac{(-1)^{n-k}}{2^n}\frac{(2n-1-k)!}{(2n-2k)!!(k-1)!}
+\left(\frac{k+1}{2}-n\right)\frac{(-1)^{n-k-1}}{2^n}\frac{(2n-2-k)!}{(2n-2k-2)!!k!}
\\\\
&=\left[
\frac{k}{2n-k}+(2n-k-1)\frac{2n-2k}{(2n-k)(2n-k-1)}
\right]
\frac{(-1)^{n-k}}{2^{n+1}}\frac{(2n-k)!}{(2n-2k)!!k!}
\\\\
&=\left[
\frac{k}{2n-k}+\frac{2n-2k}{2n-k}
\right]
c_{n+1,k+1}
\\\\
&=c_{n+1,k+1}
\end{align*}$
A: A formula to calculate the $n$-th derivative of composite functions is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

*

*Hoppe Form of Generalized Chain Rule
Let $D_g$ represent differentiation with respect to $g$ and $g=g(x)$. Hence $D^n_x f(g)$ is the $n$-th derivative of $f$ with respect to $x$.  The following is valid for $n\geq 1$:
\begin{align*}
D_x^n f(g)=\sum_{k=1}^nD_g^kf(g)\frac{(-1)^k}{k!}\sum_{j=1}^k(-1)^j\binom{k}{j}g^{k-j}D_x^ng^j\tag{1}
\end{align*}

Here we have $g(x)=\sqrt{1+x}$. The right-most part of (1) is
\begin{align*}
g(x)^{k-j}D_x^ng(x)^j&=(1+x)^{\frac{k-j}{2}}D_x^n(1+x)^{\frac{j}{2}}\\
&=(1+x)^{\frac{k-j}{2}}\left(\frac{j}{2}\right)^{\underline{n}}(1+x)^{\frac{j}{2}-n}\\
&=(1+x)^{\frac{k}{2}-n}\left(\frac{j}{2}\right)^{\underline{n}}\tag{2}\\
\end{align*}
We obtain with $g(x)=\sqrt{1+x}$ from (1) and (2):
\begin{align*}
\color{blue}{D_x^n f\left(\sqrt{1+x}\right)
=\sum_{k=1}^n(1+x)^{\frac{k}{2}-n}D_g^kf(g)\frac{(-1)^k}{k!}\sum_{j=1}^k(-1)^j\binom{k}{j}\left(\frac{j}{2}\right)^{\underline{n}}}
\end{align*}

