Challenging recurrence relation problem I am starting out with the following:
$$
\frac{d^n}{dx^n}[g(x)^{f(x)}] = \sum_{c=0}^n g(x)^{f(x)-c}\lambda_{n,c}(x)
$$
Therefore:
$$
\frac{d^{n+1}}{dx^{n+1}}[g(x)^{f(x)}] = \sum_{c=0}^{n+1}g(x)^{f(x)-c}\lambda_{n+1,c}(x) = \frac{d}{dx}\sum_{c=0}^n g(x)^{f(x)-c}\lambda_{n,c}(x)
$$
$\lambda_{n,c}(x)$ is defined like so:
$$
\lambda_{n,c}(x) = \sum_{k=c}^n \sum_{j=0}^{k-c} {k-c \choose j} \ln(g(x))^{k-c-j} \frac{d^j}{df^j}[f(x)_c] B_{n,k}^{(f \diamond g)^c}(x)
$$
My goal is to find a recurrence relation for $B_{n,k}^{(f \diamond g)^c}(x)$ by setting the two expressions equal to eachother. This is my work so far:
$$
\frac{d}{dx}[g(x)^{f(x)-c} \lambda_{n,c}(x)] = \left((f(x)-c)\frac{g'(x)}{g(x)} + \ln(g(x)) f'(x)\right)g(x)^{f(x)-c} \lambda_{n,c}(x) + g(x)^{f(x)-c} \frac{d}{dx}[\lambda_{n,c}(x)]
$$
Note from now on i will denote $\frac{d^j}{df^j}[f(x)_c] = f_c^{(j)}$
$$
\frac{d}{dx}[\lambda_{n,c}(x)] = \sum_{k=c}^n \sum_{j=0}^{k-c} {k-c \choose j} \left(\frac{g'(x)}{g(x)}(k-c-j) \ln(g(x))^{k-c-j-1} f_c^{(j)} B_{n,k}^{(f \diamond g)^c} + \ln(g(x))^{k-c-j} \frac{d}{dx}[f_c^{(j}B_{n,k}^{(f \diamond g)^c}]\right)
$$
Now, for me to find an expression that will result in a recurrence relation i am going to attempt to group all the $\ln(g(x))^{k-c-j}$ together and set all these terms equal to:
$$
\sum_{c=0}^{n+1} g(x)^{f(x)-c} \lambda_{n+1,c}(x)
$$
By doing this i will have found a way to isolate the $g(x)^{f(x)-c}$ terms as well as the $\ln(g(x))^{k-c-j}$ terms.
To do this i will seperate each individual term and attempt to manipulate it in order to fit these conditions:
$$
A = f'(x) \ln(g(x)) \lambda_{n,c}(x) = f'(x) \sum_{k=c}^n \sum_{j=0}^{k-c} {k-c \choose j} \ln(g(x))^{k-c-j+1} f_c^{(j)} B_{n,k}^{(f \diamond g)^c}(x) = f'(x) \sum_{k=c+1}^{n+1} \sum_{j=0}^{k-c-1} {k-c-1 \choose j} \ln(g(x))^{k-c-j} f_c^{(j)} B_{n,k-1}^{(f \diamond g)^c}(x)
$$
Now,for $B$ i will shift over a step backwards so that instead of $c$ we will be dealing with $c-1$, this is because of the differentiation of the natural log which in turn results in $\frac{g'(x)}{g(x)}$. When we multiply $\frac{g'(x)}{g(x)}$ with $g(x)^{f(x)-c}$ we will get $g'(x) g(x)^{f(x)-c-1}$ therefore by evaluating the expression at $c-1$ we will be evaluating the part of the summation that is dealing with $g(x)^{f(x)-c}$ instead of dealing with the summation that deals with $g(x)^{f(x)-c-1}$. If there is any questions about this please do not hesitate to ask in the comments.
$$
B = g'(x) (f(x)-c+1) \lambda_{n,c-1}(x) = g'(x) (f(x)-c+1) \sum_{k=c-1}^n \sum_{j=0}^{k-c+1} {k-c+1 \choose j} \ln(g(x))^{k-c-j+1} f_{c-1}^j B_{n,k}^{(f \diamond g)^{c-1}}(x) = g'(x) (f(x)-c+1) \sum_{k=c}^{n+1} \sum_{j=0}^{k-c} {k-c \choose j} \ln(g(x))^{k-c-j} f_{c-1}^j B_{n,k-1}^{(f \diamond g)^{c-1}}(x)
$$
Now, for $C$ and $D$ i will split up the two parts in the part where i differentiated the $\lambda_{n,c}(x)$, in variable $C$ we will be using the same logic as i used for "shifting" the $c$ variable to $c-1$.
$$
C = \lambda_{n,c}'(x)_{part \space 1} = g'(x)\sum_{k=c-1}^{n} \sum_{j=0}^{k-c+1} {k-c+1 \choose j} (k-c-j+1) \ln(g(x))^{k-c-j} f_{c-1}^{(j)} B_{n,k}^{(f \diamond g)^{c-1}}(x) = g'(x)\sum_{k=c}^{n+1} \sum_{j=0}^{k-c} {k-c \choose j} (k-c-j) \ln(g(x))^{k-c-j} f_{c-1}^{(j)} B_{n,k}^{(f \diamond g)^{c-1}}(x)
$$
Now for $C$ i did a little bit of trickery, first of all, when $k=c-1$ the term is equal to zero due to the $(k-c+1)$ term and when $j = (k-c+1)$ the term is equal to zero due to the $(k-c-j+1)$ term.
$$
D = \lambda_{n,c}'(x)_{part \space 2} = \sum_{k=c}^n \sum_{j=0}^{k-c} {k-c \choose j} \ln(g(x))^{k-c-j} \frac{d}{dx}[f_c^{(j)} B_{n,k}^{(f \diamond g)^c}(x)]
$$
Now the problem arises when i try to add $A+B+C+D$ and set it equal to $\lambda_{n+1,c}(x)$. i have attempted to do this many times but i have hit some points where it becomes troubling or that the identity does now work at all. If someone can please help me with the issue it would be alot of help to me. Thank you all for reading this if you have gotten this far, i appreciate it a lot.
 A: 
I've checked some parts of your question. Your approach seems feasible and many calculations look quite ok. But I see two possible sources of problems:



*

*When calculating the derivative of e.g.
$$
\lambda_{n,c}(x) = \sum_{k=c}^n \sum_{j=0}^{k-c} {k-c \choose j} \ln(g(x))^{k-c-j} \frac{d^j}{df^j}[f(x)_c] B_{n,k}^{(f \diamond g)^c}(x)
$$
and we take boundary values of the indices e.g. $k=c$ and $j=0$, the inner part of the sum gives $$\ln(g(x))^{k-c-j}=\ln(g(x))^{0}=1.$$ So, when doing differentiation the product rule might not be applicable in all cases.

*Your explanation in the case (B) is not easy (for me) to check. Although the switch from $c$ to $c-1$ sounds reasonable its hard to review. Especially since a combination with A,C, and D is not given. In fact providing the complete expression with all sums and settings of the indices is essential for a proper review.

Hint: Select small values of $n,k$ and $c$ which permit a non-trivial (i.e. none of the inner sums is empty) but manageable manual calculation of the formulas in order to check your approach. I suggest $n=3,k=2$ and $c=1$.



Some more aspects: With respect to this related question your main task is to find a recurrence relation for
\begin{align*}
B_{n,k}^{(f \diamond g)^c}(x):=\frac{(a^{(k-c)\diamond}\diamond b^{c\diamond})_n}{(k-c)!c!}\qquad 1\leq k\leq n, 0\leq c \leq k
\end{align*}
$a=(f^\prime,f^{\prime\prime},\ldots),b=(g^\prime,g^{\prime\prime},\ldots)$ with $f,g$ sufficiently often differentiable real-valued functions.

This is an expression involving the convolution identity $\diamond$ of certain partial Bell polynomials.
According to point (6) in this answer we already know an explicit representation in terms of the partial Bell polynomials
\begin{align*}
B_{n,k}^f&:=B_{n,k}(f^\prime,f^{\prime\prime},\ldots,f^{(n-k+1)})\qquad \qquad 1\leq k \leq n\\
B_{n,k}^g&:=B_{n,k}(g^\prime,g^{\prime\prime},\ldots,g^{(n-k+1)})\\
\end{align*}
namely (arguments in the following sometimes omitted)

The following is valid for $1\leq k \leq n$:
\begin{equation*}
B_{n,k}^{(f \diamond g)^c}=
\begin{cases}
B_{n,k}^f\qquad& c=0\\
\sum_{l=1}^{n-1}\binom{n}{l}B_{l,k-c}^fB_{n-l,c}^g\qquad&  n>1, 0 < c < k\tag{1}\\
B_{n,k}^g\qquad& c=k\\
\end{cases}
\end{equation*}
We also know according to (1) and (2) of this answer
\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{2}
\end{align*}
  and
  \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*}
Approach: So, based upon (1) you could also try to start from 
  $$B_{n+1,k}^{(f \diamond g)^c}=\sum_{l=1}^{n-1}\binom{n+1}{l}B_{l,k-c}^fB_{n+1-l,c}^g\qquad n>1, 0<k<c$$
  and then you could apply (2) in order to properly replace certain parts of the expression together with $\binom{n+1}{l}=\binom{n}{l}+\binom{n}{l-1}$ and some index shifting.

