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If Y = √ Tan X + √ Tan X + √ Tan X + . . T O ∞ , Prove that D Y D X = Sec 2 X 2 Y − 1 ? - Mathematics

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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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Chapter 11: Differentiation - Exercise 11.06 [Page 98]

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RD Sharma Mathematics [English] Class 12
Chapter 11 Differentiation
Exercise 11.06 | Q 4 | Page 98

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