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Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 4 maths textbook solution

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Answer: \frac{dy}{dx}=\frac{\sec ^{2}x}{2y-1}

Hint: The value of y is given as infinite series. If a term is deleted from an infinite series, it remains the same in this case.

Given: y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \ldots \ldots .+\infty}}}


Here it is given that,

             y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \ldots \ldots .+\infty}}}

This can be written as:

             y=\sqrt{\tan x+y}

Squaring on both sides, we get:

           y^{2}=\left ( \tan x+y \right )                                                                             …(1)

Differentiating (1) w.r.t x,

                              2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}

                                \frac{d y}{d x}\left ( 2y-1 \right )=\sec ^{2} x

                            \therefore \frac{d y}{d x}=\frac{\sec ^{2} x}{ 2y-1}

Hence, it is proved.

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