Let's do some calculus! (35)

To shorten the presentation, let $\cos^{(n)} = \underbrace{\cos(\cos(\cos ...\cos x))}_{n \times \cos}$ . Then, we have:

$\begin{aligned} f(x) & = \cos^{(8)} x \\ f'(x) & = \left(-\sin \left(\cos^{(7)}x\right)\right)\left(- \sin \left(\cos^{(6)}x\right) \right)\left(-\sin \left(\cos^{(5)}x\right)\right)...\left(-\sin \left(\cos x\right)\right)(-\sin x) \end{aligned}$

Now, we are given $\cos j = j$ , $\implies \cos (\cos j) = \cos j = j$ , $\implies \cos^{(n)} j = j$ , $\implies \sin (\cos^{(n)}j) = \sin j$ . Therefore, we have:

$\begin{aligned} f'(x) & = \sin \left(\cos^{(7)}x\right)\sin \left(\cos^{(6)}x\right)\sin \left(\cos^{(5)}x\right)...\sin \left(\cos x\right) \sin x \\ & = \sin^8 j \\ & = \left(\sqrt{1-\cos^2 j}\right)^7\sqrt{1-\cos^2 j} \\ & = \left(\sqrt{1-\cos^2 j}\right)^8 \\ & = \left(1-j^2 \right)^4 \quad \quad \small \color{#3D99F6}{\text{By binomial expansion}} \\ & = 1 - 4j^2 + 6j^4 - 4j^6 + j^8 \end{aligned}$

$\implies |a|+|b|+|c|+|d|+|e| = |1|+|-4|+|6|+|-4|+|1| = \boxed{16}$

We can get $f'(j)=\sin^8(j),$ using the formulas $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ we obtain the following result $f'(j)=\left(\frac{1}{2^8}\right)\left(2\cos(8j)-16\cos(6j)+56\cos(4j)-112\cos(2j)+70\right),$ from our knowledge $\cos(nx)=T_n(\cos(x)),$ where $T_n$ are the Chebsyhev polynomials; using this fact and doing some work $f'(j)=j^8-4j^6+6 j^4-4j^2+1,$ so $|a|+|b|+|c|+|d|+|e|=16$