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Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 8 maths textbook solution

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Answer: \frac{d y}{d x}=\frac{-y^{2} \tan x}{1-y \log \cos 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 ( \cos x\right )^{\left ( \cos x \right )^{...\infty }}


Here it is given that,

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

This can be written as:

                                y=\left ( \cos x\right )^{y}

Taking log on both sides, we get:

                             \begin{aligned} &\log y=\log (\cos x)^{y} \\ &\therefore \log y=y \log (\cos x) \end{aligned}                                                                                                  …(1)

Differentiating (1) w.r.t x,

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

Hence, it is proved.


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