If \(\tan u=\sqrt{\frac{1-x}{1+x}}, \cos v=4 x^{3}-3 x\), then \(\frac{d u}{d v}=\)

Given \(\tan u=\sqrt{\frac{1-x}{1+x}}\)
Put \(x=\cos \theta \Rightarrow \theta=\cos ^{-1} x\)
Now, \(\tan u=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}}=\tan \frac{\theta}{2}\)
\(\therefore \mathrm{u}=\frac{1}{2} \theta \Rightarrow \mathrm{u}=\frac{1}{2} \cos ^{-1} \mathrm{x}\)
\(\therefore \frac{\mathrm{du}}{\mathrm{dx}}=\frac{-1}{2 \sqrt{1-\mathrm{x}^{2}}}\)
We have \(\cos v=4 x^{3}-3 x\)
Put \(x=\cos \theta\)
\(\cos v=4 \cos ^{3} \theta-3 \cos \theta\)
\(\cos \mathrm{v}=\cos 3 \theta \Rightarrow \mathrm{v}=3 \theta\)
\(\therefore v=3 \cos ^{-1} x\)
\(\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-3}{\sqrt{1-\mathrm{x}^{2}}}\)
\(\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\left(\frac{\mathrm{du}}{\mathrm{dx}}\right)}{\left(\frac{\mathrm{dv}}{\mathrm{dx}}\right)}=\frac{-1}{2 \sqrt{1-\mathrm{x}^{2}}} \times \frac{\sqrt{1-\mathrm{x}^{2}}}{-3}=\frac{1}{6}\)