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Question
If \[y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + . . to \infty}}}\] , prove that \[\frac{dy}{dx} = \frac{\sec^2 x}{2 y - 1}\] ?
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Solution
\[\text{ We have, y } = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + . . . to \infty}}}\]
\[ \Rightarrow y = \sqrt{\tan x + y}\]
\[\text{ Squaring both sides, we get}, \]
\[ y^2 = \tan x + y\]
\[ \Rightarrow 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx}\left( 2y - 1 \right) = \sec^2 x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\sec^2 x}{2y - 1}\]
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