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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