Expression of $\cos^{-1}\left(4x^3-3x\right)$ The Original question : (translated from French)
Let $f$ be the function
$$
f\left(x\right)=\cos^{-1}\left(4x^3-3x\right)
$$

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*Find the definition domain of $f$.



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*Compute $f'\left(x\right)$. Find an expression of $f\left(x\right)$ in terms of $\cos^{-1}\left(x\right)$.



My attempt :
$f$ is defined on $\left[-1;1\right]$ and
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{1-\left(4x^3-3x\right)^2}}
$$
To answer the question I thought i could express it as
$$
f'\left(x\right)=\frac{3-12x^2}{\sqrt{-1+3x-4x^3}\sqrt{1-3x+4x^3}}
$$
and try a kind of decomposition to find it as
$$
f'\left(x\right)=a\frac{3-12x^2}{\sqrt{-1+3x-4x^3}}+b\frac{3-12x^2}{\sqrt{1-3x+4x^3}}
$$
What am I missing ?

Since the domain is $[-1,1]$, you can let $t=\cos^{-1} x \implies x=\cos t$. Why? Because $4\cos^3 t -3\cos t = \cos 3t $. $$f(t) = \cos^{-1} (\cos 3t) $$
The range of $\cos^{-1} x$ is $[0,\pi] $, and so you need to be careful and divide $f(t)$ into suitable pieces as follows:
$$f(t) =\begin{cases} 3t, & 0\le t\le \pi/3 \\ 2\pi-3t, & \pi/3 \le t \le 2\pi/3 \\ 3t -2\pi, & 2\pi/3 \le t \le \pi \end{cases} $$
Now just replace $t$ while noting that $0\le \cos^{-1} x \le \pi/3 \iff \frac 12 \le x\le 1$ and $\pi/3 \le \cos^{-1} x \le 2\pi/3 \iff -\frac 12 \le x \le \frac 12$ and $2\pi/3 \le \cos^{-1} x \le \pi \iff -1\le x\le -\frac 12$.
$$f(x)  =\begin{cases} 3\cos^{-1} x, & \frac 12 \le x \le 1 \\ 2\pi-3\cos^{-1} x, & -\frac 12  \le x \le \frac 12 \\ 3\cos^{-1} x -2\pi, & -1 \le x \le  -\frac 12 \end{cases} $$
