If y = √(x + √(x + √(x + ...\ ...\ ...))), then Option 1: (2y+1)dy/dx + 3√x = 0 Option 2: 3dy/dx + 3y^2 = 0 Option 3: 3xdy/dx + (2y+1) = 0 Option 4: (2y-1)dy/dx - 1 = 0

Answer: (2y-1)dy/dx - 1 = 0. Solution: Given y = √(x + √(x + √(x + ...))) Since the nested radical repeats infinitely with the same pattern, the expression inside the first square root equals y itself. y = √(x + y) y = √(x + y) y^2 = x + y Taking the derivative with respect to x: d/dx(y^2) = d/dx(x + y) 2ydy/dx = 1 + dy/dx 2ydy/dx = 1 + dy/dx 2ydy/dx - dy/dx = 1 (2y - 1)dy/dx = 1 (2y - 1)dy/dx - 1 = 0 Therefore, (2y-1)dy/dx - 1 = 0

Calculus

Continuity & Differentiability

Medium

applied

If $y = \sqrt{x + \sqrt{x + \sqrt{x + ...\ ...\ ...}}}$ , then

$(2y+1)\frac{dy}{dx} + 3\sqrt{x} = 0$

$3\frac{dy}{dx} + 3y^2 = 0$

$3x\frac{dy}{dx} + (2y+1) = 0$

$(2y-1)\frac{dy}{dx} - 1 = 0$

✅ Correct Option: 4

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

Since the nested radical repeats infinitely with the same pattern, the expression inside the first square root equals $y$ itself.

$y = \sqrt{x + y}$

$y^2 = x + y$

Taking the derivative with respect to $x$ :

$\frac{d}{dx}(y^2) = \frac{d}{dx}(x + y)$

$2y\frac{dy}{dx} = 1 + \frac{dy}{dx}$

$2y\frac{dy}{dx} - \frac{dy}{dx} = 1$

$(2y - 1)\frac{dy}{dx} = 1$

$(2y - 1)\frac{dy}{dx} - 1 = 0$

Therefore, $(2y-1)\frac{dy}{dx} - 1 = 0$

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If $y = \sqrt{x + \sqrt{x + \sqrt{x + ...\ ...\ ...}}}$ , then