x=1-y1+y x+xy=1-y y(x+1)=1-x y=1-x1+x y=(1-x)2(1+x)2dydx= (1+x)2×2(1-x) (-1)-2(1-x)2 (1+x)(1+x)4 =2(1-x2) -1+x-1-x1+x4 =2(1-x2) (-2)(1+x)4=-4(1-x)(1+x)3 =4(x-1)(1+x)3

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

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$\frac{4}{{(x + 1)}^{2}}$

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

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

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

answer is B.

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Detailed Solution

$x = \frac{1 - \sqrt{y}}{1 + \sqrt{y}} x + x \sqrt{y} = 1 - \sqrt{y} \sqrt{y} (x + 1) = 1 - x \sqrt{y} = \frac{1 - x}{1 + x} y = \frac{{(1 - x)}^{2}}{{(1 + x)}^{2}}$

$\frac{dy}{dx} = \frac{{(1 + x)}^{2} \times 2 (1 - x) (- 1) - 2 {(1 - x)}^{2} (1 + x)}{{(1 + x)}^{4}} = \frac{2 (1 - x^{2}) [- (1 + x) - (1 - x)]}{{(1 + x)}^{4}} = \frac{2 (1 - x^{2}) (- 2)}{{(1 + x)}^{4}} = \frac{- 4 (1 - x)}{{(1 + x)}^{3}} = \frac{4 (x - 1)}{{(1 + x)}^{3}}$