If y=x+y+x+y+......∞,then dydx is equal to :

y=x+y+x+y.... y2=x+y-x+y =x+y+y (y2-x)2=2y 2(y2-x) 2y dydx-1=2 dydx dydx 2y3-2xy-1=y2-x dydx=y2-x2y3-2xy-1

If y=x+y+x+y+......∞,then dydx is equal to :

y=x+y+x+y.... y2=x+y-x+y =x+y+y (y2-x)2=2y 2(y2-x) 2y dydx-1=2 dydx dydx 2y3-2xy-1=y2-x dydx=y2-x2y3-2xy-1

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$If y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + . . . . . . \infty,}}}} then \frac{dy}{dx} is equal to :$

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$\frac{y^{2} - x}{2 y^{3} - 2 xy - 1}$

$\frac{y^{2} + x}{2 y^{3} - 2 xy - 1}$

$\frac{y^{2} - x}{2 y^{3} - 4 xy + 1}$

$\frac{y^{2} - x}{y^{3} - 2 xy - 1}$

answer is A.

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Detailed Solution

$y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y . . . .}}}} y^{2} = x + \sqrt{y - \sqrt{x + \sqrt{y}}} = x + \sqrt{y + y} {(y^{2} - x)}^{2} = 2 y 2 (y^{2} - x) (2 y \frac{dy}{dx} - 1) = 2 \frac{dy}{dx} \frac{dy}{dx} (2 y^{3} - 2 xy - 1) = y^{2} - x \frac{dy}{dx} = \frac{y^{2} - x}{2 y^{3} - 2 xy - 1}$