The second order derivative of \(\cos^{-1}(4x^3 - 3x)\) with respect to \(\cos^{-1}(2x^2 - 1)\), where \(\dfrac{1}{2} < x < 1\) is

Let \(u = \cos^{-1}(4x^3 - 3x)\) and \(v = \cos^{-1}(2x^2 - 1)\).

Substitute \(x = \cos \theta\). Since \(\dfrac{1}{2} < x < 1\), we have \(0 < \theta < \dfrac{\pi}{3}\).

For \(u\):
\(u = \cos^{-1}(4\cos^3 \theta - 3\cos \theta) = \cos^{-1}(\cos 3\theta)\)
Since \(0 < \theta < \dfrac{\pi}{3}\), we get \(0 < 3\theta < \pi\). In this interval, \(\cos^{-1}(\cos 3\theta) = 3\theta\).
Thus, \(u = 3\theta\).

For \(v\):
\(v = \cos^{-1}(2\cos^2 \theta - 1) = \cos^{-1}(\cos 2\theta)\)
Since \(0 < \theta < \dfrac{\pi}{3}\), we get \(0 < 2\theta < \dfrac{2\pi}{3}\). In this interval, \(\cos^{-1}(\cos 2\theta) = 2\theta\).
Thus, \(v = 2\theta\).

We can express \(u\) in terms of \(v\) as:
\(u = \dfrac{3}{2} v\)

Differentiating \(u\) with respect to \(v\), we get the first order derivative:
\(\dfrac{du}{dv} = \dfrac{3}{2}\)

Differentiating again with respect to \(v\), we get the second order derivative:
\(\dfrac{d^2u}{dv^2} = \dfrac{d}{dv}\left(\dfrac{3}{2}\right) = 0\)

Answer: \(0\)