If y - -x - -y - -x - -y - - then dy-dx is equal to

∴ y = √(x + √(y + y)) ⇒ (y^2 x) = √(2y) or (y^2 x)^2 = 2y Differentiating both sides w.r.t. x, then 2(y^2 x) ( 2y dy/dx 1 ) = 2 dy/dx ∴dy/dx = (y^2 x) ...

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Byju's Answer

Standard XI

Mathematics

Implicit Differentiation

If y = √x + √...

Question

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

\frac{1}{2 y - 1}

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

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(2 y - 1)

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None of these

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Solution

The correct option is B $\frac{y^{2} - x}{2 y^{3} - 2 x y - 1}$
$∴ y = \sqrt{x + \sqrt{y + y}}$
$\Rightarrow (y^{2} - x) = \sqrt{2 y}$
$o r (y^{2} - x)^{2} = 2 y$
Differentiating both sides w.r.t. x, then
$2 (y^{2} - x) (2 y \frac{d y}{d x} - 1) = 2 \frac{d y}{d x}$
$∴ \frac{d y}{d x} = \frac{(y^{2} - x)}{2 y^{3} - 2 x y - 1}$

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

The correct option is B y2−x2y3−2xy−1∴y=√x+√y+y ⇒(y2−x)=√2y or (y2−x)2=2y Differentiating both sides w.r.t. x, then 2(y2−x)(2ydydx−1)=2dydx ∴dydx=(y2−x)2y3−2xy−1

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