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 $x=\cos \theta$
$\cos v=4 \cos ^{3} \theta-3 \cos \theta$
$\cos \mathrm{v}=\cos 3 \theta \Rightarrow \mathrm{v}=3 \theta$
$\therefore v=3 \cos ^{-1} x$
$\therefore \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-3}{\sqrt{1-\mathrm{x}^{2}}}$
$\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\left(\frac{\mathrm{du}}{\mathrm{dx}}\right)}{\left(\frac{\mathrm{dv}}{\mathrm{dx}}\right)}=\frac{-1}{2 \sqrt{1-\mathrm{x}^{2}}} \times \frac{\sqrt{1-\mathrm{x}^{2}}}{-3}=\frac{1}{6}$