Let y = sqrt{x + sqrt{x + sqrt{x + dots infty | vedclass

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(D) Given $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$.Squaring both sides,we get $y^2 = x + y$.Differentiating both sides with respect to $x$,

Vedclass · Accepted Answer

All of the above

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(D) Given $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$.Squaring both sides,we get $y^2 = x + y$.Differentiating both sides with respect to $x$,we get $2y \frac{dy}{dx} = 1 + \frac{dy}{dx}$.Rearranging the terms,$(2y - 1) \frac{dy}{dx} = 1$,which gives $\frac{dy}{dx} = \frac{1}{2y - 1}$. This matches option $A$.From $y^2 - y = x$,we have $y(y - 1) = x$,so $y - 1 = \frac{x}{y}$.Substituting $2y - 1 = y + (y - 1) = y + \frac{x}{y} = \frac{y^2 + x}{y}$,we get $\frac{dy}{dx} = \frac{1}{(y^2 + x)/y} = \frac{y}{y^2 + x}$. Since $y^2 = x + y$,we have $y^2 + x = (x + y) + x = 2x + y$. Thus,$\frac{dy}{dx} = \frac{y}{2x + y}$. This matches option $B$.Solving $y^2 - y - x = 0$ for $y$ using the quadratic formula,$y = \frac{1 \pm \sqrt{1 + 4x}}{2}$. Since $y > 0$,$y = \frac{1 + \sqrt{1 + 4x}}{2}$.Differentiating $y = \frac{1}{2} + \frac{1}{2}(1 + 4x)^{1/2}$,we get $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{2} (1 + 4x)^{-1/2} \cdot 4 = \frac{1}{\sqrt{1 + 4x}}$. This matches option $C$.Therefore,all options are correct.

Let $y = \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}}$,then $\frac{dy}{dx} =$

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