Explanation

(d)
Since, $y = \sqrt{\frac{1 - x}{1 + x}}$
On differentiating w.r.t. $x,\$we get
$\frac{dy}{dx} = \ \frac{\sqrt{1 + x} \times \frac{( - 1)}{2\sqrt{1 - x}} - \sqrt{1 - x} \times \frac{1}{2\sqrt{1 + x}}}{{(\sqrt{1 + x})}^{2}}$
$\Rightarrow \ \ \ \frac{dy}{dx} = - \frac{1}{(1 + x)\sqrt{1 - x^{2}}} \times \frac{1 - x}{1 - x}$
$\Rightarrow \ \left( 1 - x^{2} \right)\frac{dy}{dx} + y = 0\$