Note that the leading power of \(\lambda\) in each term cannot be taken out in this state, due to it being concentated into \(f_{n+1}\). However, this option may be feasible within sums of \(\lambda^{n-k}\). To derive bounds for each power, we can:
\[ f_1^{'} \left(\frac12\right) = \lambda\]
\[ f_2^{'} \left(\frac12\right) = \lambda^2 (4-\lambda) = 4\lambda^2-\lambda^3\]
\[ f_3^{'} \left(\frac12\right) =\lambda(4\lambda^2-\lambda^3)(16-4\lambda^2+\lambda^3)= 64\lambda^3-16\lambda^5+4\lambda^6-16\lambda^4+4\lambda^6-\lambda^7\]\[= 64\lambda^3-16\lambda^4-16\lambda^5+8\lambda^6-\lambda^7\]
The lowest order \(k\) of \(\lambda^k\) is \(n\), as a constant times \(\lambda\) times \(\lambda^{n-1}\), and the highest order is \(2^n-1\), as \(\left(\lambda^{2^{n-1}-1}\right)^2 \) times \(\lambda\) produces \(\lambda^{2(2^{n-1}-1)+1}=\lambda^{2^n-1}\). Given the coefficients \(c_{n,k}\), we can express \(f_n^{'}(x)\) as
\[\sum_{k=0}^{2^n-n-1} c_{n,k} x^{k+n}\]
More iterations of \(n\) will be needed to figure out a general formula for these coefficients.

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