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Question: If y = \(\sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + ......\infty}}}}\), then \(\frac{dy}{dx}\)is equal ...

If y = x+y+x+y+......∞\sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + ......\infty}}}}, then dydx\frac{dy}{dx}is equal to-

A

y+xy2−2x\frac{y + x}{y^{2} - 2x}

B

y3−x2y2−2xy−1\frac{y^{3} - x}{2y^{2} - 2xy - 1}

C

y3+x2y2−x\frac{y^{3} + x}{2y^{2} - x}

D

None

Answer

None

Explanation

Solution

y = x+y+x+y+.............∞\sqrt{x + \sqrt{y + \sqrt{x + \sqrt{y + ....}}..}}.......\infty y = x+y+y\sqrt{x + \sqrt{y + y}}Ž y2

= x +2y\sqrt{2y} [2y\sqrt{2y}= y2 –x]

Ž 2y. dydx\frac{dy}{dx}= 1 + 22y\frac{2}{\sqrt{2y}}×dydx\frac{dy}{dx}

Ž dydx\frac { d y } { d x } [2y−12y]\left\lbrack 2y - \frac{1}{\sqrt{2y}} \right\rbrack= 1 Ždydx\frac{dy}{dx}= 2y2y2y−1\frac{\sqrt{2y}}{2y\sqrt{2y} - 1}

= y2−x2y3−2xy−1\frac{y^{2} - x}{2y^{3} - 2xy - 1}