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

Answer: $\inline \frac{dy}{dx}=2$

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

Solution:

Here it is given that,

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

This can be written as:

$\inline y=(\tan x)^{y}$

Taking log on both sides, we get:

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

$\inline \therefore \log y=y\log (\tan x)$                                                                         …(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}

$\therefore y=1$

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

$\frac{dy}{dx}=2$

Hence, it is proved.