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Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 5 maths textbook solution

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Answer: \frac{d y}{d x}=\frac{y^{2} \cot x}{1-y \log \sin x}

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=\left ( \sin x \right )^{\left ( \sin \right )^{..... \infty }}


Here it is given that,

                           y=\left ( \sin x \right )^{\left ( \sin \right )^{..... \infty }}

This can be written as:

                          \log y=\log \left ( \sin x \right )^{y}                          

Taking log on both sides, we get:

                        \log y=\log \left ( \sin x \right )^{y}                                                                                             

                    \therefore \log y=y\log \left ( \sin x \right )                                                                               …(1)

Differentiating (1) w.r.t x,

                        \begin{aligned} &\frac{1}{y} \frac{d y}{d x}=y \frac{d}{d x}(\log \sin x)+\log \sin x \cdot \frac{d}{d x}(y) \\ &\frac{1}{y} \frac{d y}{d x}=y \frac{1}{\sin x}(\cos x)+\log \sin x \frac{d y}{d x} \\ &\frac{d y}{d x}\left(\frac{1}{y}-\log \sin x\right)=y \cot x \\ &\frac{d y}{d x}\left(\frac{1-y \log \sin x}{y}\right)=y \cot x \\ &\therefore \frac{d y}{d x}=\frac{y^{2} \cot x}{1-y \log \sin x} \end{aligned}

Hence, it is proved.

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