Solve the following

Let $y=\sec ^{-1}\left(\frac{1}{4 x^{3}-3 x}\right)$

Put $x=\cos \theta \quad \therefore \theta=\cos ^{-1} x$

$\Rightarrow y=\sec ^{-1}\left(\frac{1}{4 \cos ^{3} \theta-3 \cos \theta}\right)$

$y=\sec ^{-1}\left(\frac{1}{\cos 3 \theta}\right) \quad\left[\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right]$

$y=\sec ^{-1}(\sec 3 \theta) \Rightarrow y=3 \theta$

$y=3 \cos ^{-1} x$

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=3 \cdot \frac{d}{d x} \cos ^{-1} x=3\left(\frac{-1}{\sqrt{1-x^{2}}}\right)=\frac{-3}{\sqrt{1-x^{2}}}$

Thus, $\frac{d y}{d x}=\frac{-3}{\sqrt{1-x^{2}}}$.