y=x+y+x+y+........∝ then dydx=_____

y=x+y+yy2=x+2yy2−x2=2y2y4−2xy2+x2−2y=0dydx=−−2y2+2x4y3−4xy−2=−x+y22y3−2xy−1

y=x+y+x+y+........∝ then dydx=_____

y=x+y+yy2=x+2yy2−x2=2y2y4−2xy2+x2−2y=0dydx=−−2y2+2x4y3−4xy−2=−x+y22y3−2xy−1

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$y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + ........ \propto}}}}$ then $\frac{d y}{d x} =_____$

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

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

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

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

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Detailed Solution

$y = \sqrt{x + \sqrt{y + y}}$

$y^{2} = x + \sqrt{2 y}$

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

$y^{4} - 2 x y^{2} + x^{2} - 2 y = 0$

$\frac{d y}{d x} = \frac{- (- 2 y^{2} + 2 x)}{4 y^{3} - 4 x y - 2} = \frac{- x + y^{2}}{2 y^{3} - 2 x y - 1}$

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y=x+y+x+y+........∝ then dydx=_____

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$y = \sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + ........ \propto}}}}$ then $\frac{d y}{d x} =_____$