Explanation

(b)
Given, $\frac{x}{1} = \frac{1 - \sqrt{y}}{1 + \sqrt{y}}$
Applying componendo and dividendo, we get
$\frac{1 + x}{1 - x} = \frac{\left( 1 + \sqrt{y} \right) + (1 - \sqrt{y})}{\left( 1 + \sqrt{y}\ \right)–(1 - \sqrt{y})}$
$\Rightarrow \frac{1 + x}{1 - x} = \frac{2}{2\sqrt{y}}$
$\Rightarrow \ y = \left( \frac{1 - x}{1 + x} \right)^{2}$
On differentiating w.r.t. $x,\$we get
$\frac{dy}{dx} = \frac{- 2(1 + x)^{2}(1 - x) - (1 - x)^{2}.2(1 + x)}{(1 + x)^{4}}$
$= \ \frac{(1 - x)(1 + x)( - 2 - 2x - 2 + 2x)}{(1 + x)^{4}}$
$= \frac{4(x - 1)}{(x + 1)^{3}}$