Explanation

(d)
Let$\ u = \tan^{- 1}{\left( \frac{2x}{1 - x^{2}} \right)and\ v = \cos^{- 1}\sqrt{1 - x^{2}}}$
On differentiating w.r.t. $x,\$respectively, we get
$\frac{du}{dx} = \frac{1}{1 + \left( \frac{2x}{1 - x^{2}} \right)^{2}}.\left\lbrack \frac{\left( 1 - x^{2} \right)2 - 2x( - 2x)}{\left( 1 - x^{2} \right)^{2}} \right\rbrack$
$= \frac{2 + 2x^{2}}{\left( 1 + x^{2} \right)^{2}} = \frac{2}{1 + x^{2}}$
and $\frac{dv}{dx\ } = - \frac{1}{\sqrt{1 - (1 - x^{2})}}.\left\lbrack \frac{( - 2x)}{2\sqrt{1 - x^{2}}} \right\rbrack$
$= \frac{1}{\sqrt{x^{2}}}\left\lbrack \frac{x}{\sqrt{1 - x^{2}}} \right\rbrack = \frac{1}{\sqrt{1 - x^{2}}}$
$\therefore\ \frac{du}{dv} = \frac{\frac{2}{1 + x^{2}}}{\frac{1}{\sqrt{1 - x^{2}}}} = \frac{2\sqrt{1 - x^{2}}}{1 + x^{2}}$
Alternate
Let $u = \tan^{- 1}{\left( \frac{2x}{1 - x^{2}} \right) = 2\tan^{- 1}x}$
$\Rightarrow$ $\frac{du}{dx} = \frac{2}{1 + x^{2}}$
and let $v = \cos^{- 1}{\sqrt{1 - x^{2}} = \sin^{- 1}x}$
$\Rightarrow \ \ \ \ \ \ \ \frac{dv}{dx} = \frac{1}{\sqrt{1 - x^{2}}}$
Now, $\frac{du}{dv} = \frac{2}{1 + x^{2}}/\frac{1}{\sqrt{1 - x^{2}}} = \frac{2\sqrt{1 - x^{2}}}{1 + x^{2}}$