If y = sqrt{tan x + sqrt{tan x + sqrt{tan x + | vedclass

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(C) Given $y = \sqrt{	an x + y}$.Squaring both sides, we get $y^2 = 	an x + y$.Differentiating both sides with respect to $x$, we get $2y \frac{dy}{

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$\frac{4}{5}$

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(C) Given $y = \sqrt{	an x + y}$.Squaring both sides, we get $y^2 = 	an x + y$.Differentiating both sides with respect to $x$, we get $2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$.Rearranging terms, $(2y - 1) \frac{dy}{dx} = \sec^2 x$, so $\frac{dy}{dx} = \frac{\sec^2 x}{2y - 1}$.At $x = \frac{\pi}{4}$, $	an x = 1$. Substituting into $y^2 - y - 	an x = 0$, we get $y^2 - y - 1 = 0$.Solving for $y$ using the quadratic formula, $y = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $y > 0$, $y = \frac{1 + \sqrt{5}}{2}$.Then $2y - 1 = 2(\frac{1 + \sqrt{5}}{2}) - 1 = \sqrt{5}$.At $x = \frac{\pi}{4}$, $\sec^2 x = (\sqrt{2})^2 = 2$.Thus, $\frac{dy}{dx} = \frac{2}{\sqrt{5}}$.Therefore, $(\frac{dy}{dx})^2 = (\frac{2}{\sqrt{5}})^2 = \frac{4}{5}$.

If $y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + \dots \infty}}}$, then $(\frac{dy}{dx})^2$ at $x = \frac{\pi}{4}$ is

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