Manipulation of Bell Polynomials Like lets say the following:
$
g(n) = \sum_{k=1}^{n}B_{n,k}(f'(x),f''(x),.....,f^{n-k+1}(x))
$
What would $\frac{d}{dx} g(n)$ be?
Or perhaps how to expand a Bell polynomial in this form into something more managable?
Im finding it difficult to prove things that have bell polynomials inside of them. Can anyone give me advice on proving statements with bell polynomials? Thank you
 A: 
Intro: In the following we assume that the function $f$ is sufficiently often differentiable. Let
  \begin{align*}
g_n(x) = \sum_{k=1}^{n}B_{n,k}(f'(x),f''(x),\dots,f^{(n-k+1)}(x))\qquad\qquad n\geq 1
\end{align*}
  Since the differential operator $\frac{d}{dx}$ is linear, we get
  \begin{align*}
\frac{d}{dx}g_n(x)&=\frac{d}{dx}\left(\sum_{k=1}^{n}B_{n,k}(f'(x),f''(x),\dots,f^{(n-k+1)}(x))\right)\\
&=\sum_{k=1}^{n}\frac{d}{dx}B_{n,k}(f'(x),f''(x),\dots,f^{(n-k+1)}(x))\\
\end{align*}
and concentrate therefore on:
  \begin{align*}
\frac{d}{dx}B_{n,k}(f'(x),f''(x),\dots,f^{(n-k+1)}(x))\qquad\qquad n,k\geq 1
\end{align*}

In the following we use the abbreviation
$$B_{n,k}^f(x):=B_{n,k}(f'(x),f''(x),\dots,f^{(n-k+1)}(x))\qquad\qquad n,k\geq 1$$
We will show that

The following is valid:
\begin{align*}
\frac{d}{dx}&B_{n,k}^f(x)=B_{n+1,k}^f(x)-f^{\prime}(x)B_{n,k-1}^f(x)\tag{1}\\
\\
B_{n,k}^f(x)&=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}\left(-f(x)\right)^{k-j}\frac{d^n}{dx^n}\left(f(x)\right)^j\tag{2}
\end{align*}

Note:


*

*The expression (2) should be useful for calculations around $B_{n,k}$. We will use it to show (1). 

*The recurrence relation (1) can also be used to build a Pascal-like triangle of $B_{n,k}$ starting with $B_{1,1}^f(x)=f^{\prime}(x)$.

\begin{array}{ccccccc}
&&&B_{1,1}&\\
&&B_{2,1}&&B_{2,2}&\\
&B_{3,1}&&B_{3,2}&&B_{3,3}&\\
B_{4,1}&&B_{4,2}&&B_{4,3}&&B_{4,4}\\
&&&\ldots&&&\\
\end{array}
\begin{align*}
B_{1,1}&=f^{\prime}&\\
B_{2,1}&=\frac{d}{dx}B_{1,1},&B_{2,2}&=f^{\prime}\cdot B_{1,1}\\
B_{3,1}&=\frac{d}{dx}B_{2,1},&B_{3,2}&=f^{\prime}\cdot B_{2,1}+\frac{d}{dx}B_{2,2},&B_{3,3}&=f^{\prime}\cdot B_{2,2}&\\
B_{4,1}&=\frac{d}{dx}B_{3,1},&B_{4,2}&=f^{\prime}\cdot B_{3,1}+\frac{d}{dx}B_{3,2},&B_{4,3}&=f^{\prime}\cdot B_{3,2}+\frac{d}{dx} B_{3,3},&\ldots\\
&\ldots&
\end{align*}



Proof of (2):
  \begin{align*}
B_{n,k}^f(x)&=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}\left(-f(x)\right)^{k-j}\frac{d^n}{dx^n}\left(f(x)\right)^j
\end{align*}

We start with the Taylor series expansion of $y=y(z)$ at a point $x$:
\begin{align*}
y(x+z)=\sum_{n\geq 0}y^{(n)}(x)\frac{z^n}{n!}\tag{3}
\end{align*}

We consider the composition $y=g\circ f$ of sufficiently often differentiable functions $f$ and $g$ and get according to (3)
  \begin{align*}
g\left(f(x+z)\right)=\sum_{k\geq 0}\frac{d^k}{dx^k}g\left(f(x)\right)\frac{z^k}{k!}\tag{4}
\end{align*}

We also get according to (3) for proper differentiable functions $u$ and $v$
\begin{align*}
g(u+v)=\sum_{k\geq 0}\frac{d^k}{du^k}g(u)\frac{v^k}{k!}\tag{5}
\end{align*}

Now letting $u=f(x)$ and $v=f(x+z)-f(x)$ and substituting it in (5) we get
  \begin{align*}
g\left(f(x+z)\right)=\sum_{k\geq 0}\frac{d^k}{df^k}g\left(f(x)\right)\frac{1}{k!}\left(f(x+z)-f(x)\right)^k\tag{6}
\end{align*}

We now consider (6) as series in $z$ and compare the coefficients of $z^n$ in (6) with the coefficients of $z^n$ in (4). Using the  coefficient of operator $[z^n]$ to denote the coefficient of $z^n$, we observe:
\begin{align*}
\frac{d^n}{dx^n}g\left(f(x)\right)&=n![z^n]\sum_{k\geq 0}\frac{d^k}{df^k}g\left(f(x)\right)\frac{1}{k!}\left(f(x+z)-f(x)\right)^k\tag{7}\\
&=\sum_{k=0}^{n}\frac{g^{(k)}\left(f(x)\right)}{k!}
\left.\left(\frac{d^n}{dz^n}\left(f(x+z)-f(x)\right)^k\right)\right|_{z=0}\tag{8}\\
&=\sum_{k=0}^{n}\frac{g^{(k)}\left(f(x)\right)}{k!}
\left.\left(\frac{d^n}{dz^n}\sum_{j=0}^{k}\binom{k}{j}\left(f(x+z)\right)^j\left(-f(x)\right)^{k-j}\right)\right|_{z=0}\tag{9}\\
&=\sum_{k=0}^{n}\frac{g^{(k)}\left(f(x)\right)}{k!}
\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)\right)^{k-j}\left.\left(\frac{d^n}{dz^n}\left(f(x+z)\right)^j\right)\right|_{z=0}\tag{10}\\
&=\sum_{k=0}^{n}\frac{g^{(k)}\left(f(x)\right)}{k!}
\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)\right)^{k-j}\left(\frac{d^n}{dx^n}\left(f(x)\right)^j\right)\tag{11}\\
\end{align*}
Comment:


*

*(7) Comparison of the coefficients of $z^n$ in (6) and (4)

*(8) Note that $n!$ times the coefficient of $z^n$ can be written as the $n$-th derivative. Observe also that since the Taylor series expansion of $f(x+z)-f(x)$ at the point $x$ starts with $z^1$, we can restrict the index range to $0\leq k \leq n$.

*(9) Application of the binomial theorem

*(10) Linearity of $\frac{d^n}{dz^n}$ operator

*(11) Observe, that taking $n$ derivatives of $f(z+x)^j$ with respect to $z$ and then setting $z=0$ in (10) is equivalently, by the chain rule to take $n$ derivatives of $\left(f(z+x)\right)^j$ with respect to $z+x$ and then setting $z=0$. And this is the same as taking $n$ derivatives of $\left(f(x)\right)^j$ with respect to $x$.

Next we consider the well-known formula of Faà di Bruno of the $n$-th derivative of composites of functions:
\begin{align*}
\frac{d^n}{dx^n}g\left( f(x)\right)=\sum_{k=0}^ng^{(k)}\left(f(x)\right)B_{n,k}(f^{\prime}(x),\ldots,f^{(n-k+1)}(x))\tag{12}
\end{align*}
Comparing (11) with (12) gives
\begin{align*}
B_{n,k}^f(x)&=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}\left(-f(x)\right)^{k-j}\frac{d^n}{dx^n}\left(f(x)\right)^j
\end{align*}
  and (2) follows.



Proof of (1):
  \begin{align*}
\frac{d}{dx}&B_{n,k}^f(x)=B_{n+1,k}^f(x)-f^{\prime}(x)B_{n,k-1}^f(x)\end{align*}

We use the identity (2) and we observe:
\begin{align*}
\frac{d}{dx}B_{n,k}\left(f(x)\right)&=\frac{1}{k!}\frac{d}{dx}\sum_{j=0}^k\binom{k}{j}\left(-f(x)\right)^{k-j}\frac{d^n}{dx^n}\left(f(x)\right)^j\\
&=\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}(k-j)\left(-f(x)\right)^{k-j-1}\left(-f^{\prime}(x)\right)\frac{d^n}{dx^n}\left(f(x)\right)^j\\
&\qquad+\frac{1}{k!}\sum_{j=0}^k\binom{k}{j}\left(-f(x)\right)^{k-j}\frac{d^{n+1}}{dx^{n+1}}\left(f(x)\right)^j\\
&=-f^{\prime}(x)\frac{1}{k!}\sum_{j=0}^{k-1}\binom{k-1}{j-1}k\left(-f(x)\right)^{(k-1)-j}\frac{d^n}{dx^n}\left(f(x)\right)^j\\
&\qquad+B_{n+1,k}^f(x)\\
&=-f^{\prime}(x)B_{n,k-1}^f(x)+B_{n+1,k}^f(x)
\end{align*}
and (1) follows.
Note: This proof was derived from the nice paper The Curious History of Faà di Bruno's Formula by Warren P. Johnson.
