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

Question

If $y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + . . t o \infty}}}$ , prove that $\frac{d y}{d x} = \frac{\sec^{2} x}{2 y - 1}$ ?

Solution

$We have, y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + . . . t o \infty}}}$
$\Rightarrow y = \sqrt{\tan x + y}$
$Squaring both sides, we get,$
$y^{2} = \tan x + y$
$\Rightarrow 2 y \frac{d y}{d x} = \sec^{2} x + \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x} (2 y - 1) = \sec^{2} x$
$\Rightarrow \frac{d y}{d x} = \frac{\sec^{2} x}{2 y - 1}$

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Simple Problems on Applications of Derivatives

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Q 4 Q 3 Q 5

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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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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