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Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 1 maths textbook solution

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Answer: \frac{d y}{d x}=\frac{1}{2 y-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{x+\sqrt{x+\sqrt{x+\ldots \ldots \ldots+\infty}}}


Here it is given that,

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

This can be written as:


Squaring on both sides, we get:

       y^{2}=x+y                                                                                                    …(1)

Differentiating (1) w.r.t x,

      \begin{aligned} &2 y \frac{d y}{d x}=1+\frac{d y}{d x} \\ &\frac{d y}{d x}(2 y-1)=1 \\ &\therefore \frac{d y}{d x}=\frac{1}{2 y-1} \end{aligned}

Hence, it is proved.


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