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(B) Given, $x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.Applying componendo and dividendo, we get:$\frac{1+x}{1-x} = \frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{

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$\frac{4(x-1)}{(1+x)^3}$

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(B) Given, $x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.Applying componendo and dividendo, we get:$\frac{1+x}{1-x} = \frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y})-(1-\sqrt{y})}$$\frac{1+x}{1-x} = \frac{2}{2\sqrt{y}} = \frac{1}{\sqrt{y}}$Squaring both sides, we get $\sqrt{y} = \frac{1-x}{1+x}$, so $y = \left(\frac{1-x}{1+x}\right)^2$.Now, differentiating with respect to $x$ using the chain rule:$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v-uv'}{v^2}$:$\frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(-1)(1+x) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$Therefore, $\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \left(\frac{-2}{(1+x)^2}\right) = \frac{-4(1-x)}{(1+x)^3} = \frac{4(x-1)}{(1+x)^3}$.

$x = \frac{1-\sqrt{y}}{1+\sqrt{y}} \Rightarrow \frac{dy}{dx}$ is equal to

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