Derivative of \(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\) w.r.t. \(\cos ^{-1}\left(4 x^3-3 x\right)\) is

Let \(y=\tan ^{-1} \sqrt{\frac{1-x}{1+x}}, \mathrm{z}=\cos ^{-1}\left(4 x^3-3 x\right)\)
for \(y\), substitute \(x=\cos 2 \theta_1\) and for z , substitute \(x=\cos \theta\)
\(\therefore y=\tan ^{-1} \sqrt{\frac{1-\cos 2 \theta_1}{1+\cos 2 \theta_1}}=\tan ^{-1}\left(\tan \theta_1\right)=\theta_1 \text { and } \)
\( \mathrm{z}=\cos ^{-1}\left(4 \cos ^3 \theta-3 \cos \theta\right) \)
\( =\cos ^{-1}(\cos 3 \theta)=3 \theta \)
\(\therefore y =\frac{\cos ^{-1} x}{2} \text { and } \mathrm{z}=3 \cos ^{-1} x \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x} =\frac{-1}{2 \sqrt{1-x^2}} \text { and } \frac{\mathrm{d} z}{\mathrm{~d} x}=\frac{-3}{\sqrt{1-x^2}} \)
\( \therefore\frac{\mathrm{~d} y}{\mathrm{dz}} =\frac{\frac{\mathrm{d} y}{\mathrm{~d} x}}{\frac{\mathrm{~d} z}{\mathrm{~d} x}}=\frac{1}{6}\)