<s>[INST] 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 ?
 [/INST] The problem with your approach is that $f$ is not differentiable at all points of $[-1,1]$.
Since domain is $[-1,1]$, let $x=\cos \theta$, where $\theta \in [0,\pi]$ so that $\cos $ function becomes a bijection and hence $\cos^{-1}x=\theta$. Note that $0\le 3\theta \le 3\pi$.
Now using trig. identities, it follows that:
$f(x)=\cos^{-1} (\cos 3\theta)$
Case 1: $0\le 3\theta \le \pi\implies 0\le \theta\le \pi/3$ 
$f(x)=3\theta =3 \cos^{-1}x$ for $\frac 12\le x \le 1$ 
Case 2: $\pi\lt 3\theta \le 2\pi \implies \pi/3\lt \theta \le 2\pi/3$ 
$\cos 3\theta=\cos (2\pi-(2\pi-3\theta))=\cos(-3\theta+2\pi)\implies f(x)=\cos^{-1}\cos 3\theta=2\pi-3\theta=2\pi-3\cos^{-1}x$ for $-\frac 12\le x\lt \frac 12$. 
Case 3: $2\pi\lt 3\theta \le 3\pi\implies 2\pi/3\lt \theta \le\pi$ 
$f(x)=\cos^{-1}\cos 3\theta=\cos^{-1}\cos (3\theta-2\pi)=3\theta -2\pi=3\cos^{-1}x-2\pi$ for $-1\le x\lt -1/2$.
Can you find the derivative now?
</s>