I need to prove that the following expression is true:
\begin{align*} \frac{d}{dx}[fg]\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j) f^{k-c-j} g^{c-j} \lambda(k,c,x) = \\ (fg')^{n+1}\sum_{c=2}^{n-1} \sum_{j=0}^{min\{c,n-c+1\}} \Phi(n+1,c,j) \left(\frac{f'g}{g'f}\right)^c \left(\frac{1}{fg}\right)^j \\ +\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j)f^{k-c-j-1}g^{c-j-1} \\ \left(-(k-c-j)f'g \lambda(k,c,x) + fgf'\lambda(k-1,c-1,x) \\ -(c-j)g'f\lambda(k,c,x) + fgg'\lambda(k-1,c,x)\right) \end{align*} Where the following is true: $$ \Phi(k,c,j) = {c \choose j}\frac{(k-c)!}{(k-c-j)!} $$ $$ \lambda(k,c,x) = \sum_{l=1}^{n-1} {n \choose l} B_{n-l,c}^f(x) B_{l,k-c}^g(x) $$ Where $B_{n,k}^f(x) = B_{n,k}(f'(x),f''(x),\ldots,f^{(n-k+1)}(x))$ which is the partial Bell Polynomial. More information can be found here. Also note that the following properties are used: $$ B_{n,k}^f(x) = 0 \quad\quad\quad\quad (k >n) $$ $$ B_{n,0}^f(x) = 0 $$
My Attempt
What I tried to do was factor all the $\lambda(k,c,x)$ into like terms: \begin{align} f'\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j)f^{k-c-j}g^{c-j} \lambda(k-1,c-1,x) \\ = f'g\sum_{c=1}^{n-2} \sum_{k=c+2}^n \sum_{j=0}^{min\{c+1,k-c-1\}} \Phi(k,c+1,j)f^{k-c-j-1}g^{c-j} \lambda(k-1,c,x) \\ =f'g\sum_{c=1}^{n-2} \sum_{k=c+1}^{n-1} \sum_{j=0}^{min\{c+1,k-c\}} \Phi(k+1,c+1,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \end{align} Also: \begin{align} g'\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j)f^{k-c-j}g^{c-j} \lambda(k-1,c,x) \\ =g'f \sum_{c=1}^{n-1} \sum_{k=c}^{n-1} \sum_{j=0}^{min\{c,k-c+1\}} \Phi(k+1,c,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \end{align} Now we have: \begin{align*} \sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j) f^{k-c-j} g^{c-j} \lambda(k,c,x) = \\ (fg')^{n+1}\sum_{c=2}^{n-1} \sum_{j=0}^{min\{c,n-c+1\}} \Phi(n+1,c,j) \left(\frac{f'g}{g'f}\right)^c \left(\frac{1}{fg}\right)^j \\ -g'f\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j)(c-j)f^{k-c-j-1}g^{c-j-1}\lambda(k,c,x) \\ -f'g\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j)(k-c-j) f^{k-c-j-1}g^{c-j-1}\lambda(k,c,x) \\ + g'f \sum_{c=1}^{n-1} \sum_{k=c}^{n-1} \sum_{j=0}^{min\{c,k-c+1\}} \Phi(k+1,c,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \\ + f'g\sum_{c=1}^{n-2} \sum_{k=c+1}^{n-1} \sum_{j=0}^{min\{c+1,k-c\}} \Phi(k+1,c+1,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \end{align*} Now i will match up the bounds on the triple summations that have like terms: \begin{align} \sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j) f^{k-c-j} g^{c-j} \lambda(k,c,x) = \\ (fg')^{n+1}\sum_{c=2}^{n-1} \sum_{j=0}^{min\{c,n-c+1\}} \Phi(n+1,c,j) \left(\frac{f'g}{g'f}\right)^c \left(\frac{1}{fg}\right)^j \\ -g'f\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j)(c-j)f^{k-c-j-1}g^{c-j-1}\lambda(k,c,x) \\ -f'g\sum_{c=1}^{n-2} \sum_{k=c+1}^{n-1} \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j)(k-c-j) f^{k-c-j-1}g^{c-j-1}\lambda(k,c,x) \\ + g'f \sum_{c=1}^{n-1} \sum_{k=c+1}^{n-1} \sum_{j=0}^{min\{c,k-c+1\}} \Phi(k+1,c,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \\ + f'g\sum_{c=1}^{n-2} \sum_{k=c+1}^{n-1} \sum_{j=0}^{min\{c+1,k-c\}} \Phi(k+1,c+1,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \\ - f'g \sum_{c=1}^{n-1} \sum_{j=0}^{min\{c,n-c\}} \Phi(n,c,j)(n-c-1)f^{n-c-j-1}g^{c-j-1} \lambda(n,c,x) \\ +g'f \sum_{c=1}^{n-1} \sum_{j=0}^1 \Phi(c+1,c,j)f^{-j}g^{c-j}\lambda(c,c,x) \end{align} Now notice on the terms where $f^{k-c-j-1} g^{c-j-1}$, we can eliminate this problem by shifting the $j$ index: \begin{align} \sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j) f^{k-c-j} g^{c-j} \lambda(k,c,x) = \\ (fg')^{n+1}\sum_{c=2}^{n-1} \sum_{j=0}^{min\{c,n-c+1\}} \Phi(n+1,c,j) \left(\frac{f'g}{g'f}\right)^c \left(\frac{1}{fg}\right)^j \\ -g'f\sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=1}^{min\{c+1,k-c+1\}} \Phi(k,c,j-1)(c-j+1)f^{k-c-j}g^{c-j}\lambda(k,c,x) \\ -f'g\sum_{c=1}^{n-2} \sum_{k=c+1}^{n-1} \sum_{j=1}^{min\{c+1,k-c+1\}} \Phi(k,c,j-1)(k-c-j+1) f^{k-c-j}g^{c-j}\lambda(k,c,x) \\ + g'f \sum_{c=1}^{n-1} \sum_{k=c+1}^{n-1} \sum_{j=0}^{min\{c,k-c+1\}} \Phi(k+1,c,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \\ + f'g\sum_{c=1}^{n-2} \sum_{k=c+1}^{n-1} \sum_{j=0}^{min\{c+1,k-c\}} \Phi(k+1,c+1,j)f^{k-c-j}g^{c-j} \lambda(k,c,x) \\ - f'g \sum_{c=1}^{n-1} \sum_{j=0}^{min\{c,n-c\}} \Phi(n,c,j)(n-c-1)f^{n-c-j-1}g^{c-j-1} \lambda(n,c,x) \\ +g'f \sum_{c=1}^{n-1} \sum_{j=0}^1 \Phi(c+1,c,j)f^{-j}g^{c-j}\lambda(c,c,x) \end{align} Now to collect like terms: \begin{align} \sum_{c=1}^{n-1} \sum_{k=c+1}^n \sum_{j=0}^{min\{c,k-c\}} \Phi(k,c,j) f^{k-c-j} g^{c-j} \lambda(k,c,x) = \\ (fg')^{n+1}\sum_{c=2}^{n-1} \sum_{j=0}^{min\{c,n-c+1\}} \Phi(n+1,c,j) \left(\frac{f'g}{g'f}\right)^c \left(\frac{1}{fg}\right)^j + g'f \sum_{c=1}^{n-1} \sum_{k=c+1}^{n-1} \lambda(k,c,x) \left(\sum_{j=0}^{min\{c,k-c+1\}}f^{k-c-j}g^{c-j}\Phi(k+1,c,j) \\ -\sum_{j=1}^{min\{c+1,k-c+1\}}f^{k-c-j}g^{c-j}\Phi(k,c,j-1)(c-j+1)\right) \\ + f'g\sum_{c=1}^{n-2} \sum_{k=c+1}^{n-1} \lambda(k,c,x) \left(\sum_{j=0}^{min\{c+1,k-c\}}f^{k-c-j}g^{c-j}\Phi(k+1,c+1,j) \\ - \sum_{j=1}^{min\{c+1,k-c+1\}}f^{k-c-j}g^{c-j}\Phi(k,c,j-1)(k-c-j+1)\right) \\ - f'g \sum_{c=1}^{n-1} \sum_{j=0}^{min\{c,n-c\}} \Phi(n,c,j)(n-c-1)f^{n-c-j-1}g^{c-j-1} \lambda(n,c,x) \\ +g'f \sum_{c=1}^{n-1} \sum_{j=0}^1 \Phi(c+1,c,j)f^{-j}g^{c-j}\lambda(c,c,x) \end{align} Once I get to this point I have absolutely no clue what else I can do with the summation bounds in order to simplify more.
What is the expression for?
The expression is simply a part of an expression that calculated the Complete Bell Polynomial of a product. The expression for a complete bell polynomial of a product is: $$ \frac{d^n}{dx^n}[e^{fg}] = e^{fg}B_n^{fg}(x) $$ Where $$ B_n^{fg}(x) = \sum_{k=1}^n B_{n,k}^{fg}(x) $$
