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Provide solution RD Sharma maths class 12 chapter 10 differentiation exercise 10.6 question 6 maths textbook solution

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Answer: \frac{dy}{dx}=2

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=(\tan x)^{(\tan x)^{(\tan x) \cdot \cdots}}


Here it is given that,

                           y=(\tan x)^{(\tan x)^{(\tan x) \cdot \cdots}}

This can be written as:

                           y=(\tan x)^{y}

Taking log on both sides, we get:

                        \log y=\log (\tan x)^{y}                                                                                

                   \therefore \log y=y\log (\tan x)                                                                         …(1)

Differentiating (1) w.r.t x,

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

\text { If } \mathrm{x}=\frac{\pi}{4}, \text { becomes: }

                                          \begin{aligned} \left(\frac{d y}{d x}\right)_{x=\left(\frac{\pi}{4}\right)} &=\frac{y^{2}}{1-y \log \tan \left(\frac{\pi}{4}\right)} \cdot \frac{\sec ^{2}\left(\frac{\pi}{4}\right)}{\tan \left(\frac{\pi}{4}\right)} \\ &=\frac{y^{2}(\sqrt{2})^{2}}{1-y \log 1} \\ &=\frac{2 y^{2}}{1-y(0)} \end{aligned}

\begin{aligned} \text { Since, }(y) \frac{\pi}{4} &=\tan \left(\frac{\pi}{4}\right)^{\tan \left(\frac{\pi}{4}\right)^{\cdots-\infty}} \\ &=(1)^{\infty} \\ &=1 \end{aligned}

\therefore y=1

\frac{dy}{dx}=\frac{2\left ( 1 \right )^{2}}{1-0}


Hence, it is proved.

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