If x=frac{1-sqrt{y}}{1+sqrt{y}}, then (x+1) f | vedclass

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(B) Given $x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.Multiplying both sides by $(1+\sqrt{y})$, we get $x(1+\sqrt{y}) = 1-\sqrt{y}$.$x + x\sqrt{y} = 1 - \sqrt

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(B) Given $x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.Multiplying both sides by $(1+\sqrt{y})$, we get $x(1+\sqrt{y}) = 1-\sqrt{y}$.$x + x\sqrt{y} = 1 - \sqrt{y}$.Rearranging terms, we have $\sqrt{y}(x+1) = 1-x$, so $\sqrt{y} = \frac{1-x}{1+x}$.Differentiating with respect to $x$:$\frac{1}{2\sqrt{y}} \frac{dy}{dx} = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$.Thus, $\frac{dy}{dx} = \frac{-4\sqrt{y}}{(1+x)^2}$.Substituting $\sqrt{y} = \frac{1-x}{1+x}$, we get $\frac{dy}{dx} = \frac{-4(1-x)}{(1+x)^3} = \frac{4(x-1)}{(x+1)^3}$.Now, differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = 4 \left[ \frac{(x+1)^3(1) - (x-1) \cdot 3(x+1)^2}{(x+1)^6} \right] = 4 \left[ \frac{(x+1) - 3(x-1)}{(x+1)^4} \right] = 4 \left[ \frac{x+1-3x+3}{(x+1)^4} \right] = \frac{4(4-2x)}{(x+1)^4} = \frac{8(2-x)}{(x+1)^4}$.Substituting these into the expression $(x+1) \frac{d^2 y}{d x^2}+\left(\frac{3 \sqrt{y}+1}{\sqrt{y}}\right) \frac{d y}{d x}$:Note that $\frac{3\sqrt{y}+1}{\sqrt{y}} = 3 + \frac{1}{\sqrt{y}} = 3 + \frac{1+x}{1-x} = \frac{3-3x+1+x}{1-x} = \frac{4-2x}{1-x}$.Expression $= (x+1) \frac{8(2-x)}{(x+1)^4} + \left(\frac{4-2x}{1-x}\right) \frac{4(x-1)}{(x+1)^3} = \frac{8(2-x)}{(x+1)^3} - \frac{4(2-x)}{(x+1)^3} \cdot \frac{4(x-1)}{4(x-1)} = \frac{8(2-x)}{(x+1)^3} - \frac{8(2-x)}{(x+1)^3} = 0$.

If $x=\frac{1-\sqrt{y}}{1+\sqrt{y}}$, then $(x+1) \frac{d^2 y}{d x^2}+\left(\frac{3 \sqrt{y}+1}{\sqrt{y}}\right) \frac{d y}{d x}$ equals

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