Attempting to derive a formula for logistic map (part 1)

The logistic map comes from concatenations of the function $f(x) = \lambda x(1-x)$, where $x=\frac12$ and$\lambda$ is an argument within the domain $[0,4]$. We'll begin by defining a function $f_n (x)$ as $n$ instances of $f(x)$ concatenated together, observing the fractional form of each incrementation of $n$: $f_1 \left(\frac12\right) = \lambda \frac12 \left(1-\frac12\right) = \frac{\lambda}4$ $f_2 \left(\frac12\right) = \lambda \frac{\lambda}4 \left(1-\frac{\lambda}4\right) = \frac{\lambda^2 (4-\lambda)}{16}$ $f_3 \left(\frac12\right) = \lambda \frac{\lambda^2 (4-\lambda)}{16} \left(1-\frac{\lambda^2 (4-\lambda)}{16} \right) = \frac{\lambda^3(4-\lambda)(16-\lambda^2(4-\lambda))}{256}$ The denominator of each iteration is the square of the previous denominator. Using the fact that $4$ is $2^{2^1}$, we can rewrite $f_n \left(\frac12\right) = \frac{f_n^{'}\left(\frac12\right)}{2^{2^n}}$ (here $f_n^{'}(x)$ just denotes the remaining sections of $f_n (x)$.)
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.