Explanation

(d)
Let $u = \sin^{- 1}{\left( \frac{1}{1 - 2x^{2}} \right),\ \ v = \sin^{- 1}{(3x - 4x^{3})}}$
Put $x = \sin{\theta,\ \ \ we\ get}$
$u = \sec^{- 1}{(\sec{2\theta),\ \ }}v = \sin^{- 1}{(\sin{3\theta)}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ u = \ 2\theta,\ \ \ \ \ \ \ \ v = 3\theta\ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ u = 2\sin^{- 1}{x,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = 3{\sin^{- 1}\ x}{\ \ }}\ \ \ \ \ \$
$\Rightarrow \ \ \frac{u}{v} = \frac{2}{3}\ \ \ \Rightarrow$ $u = \frac{2}{3}v$
$\Rightarrow \ \frac{du}{dv} = \frac{2}{3}$