Question

If \[y = \left( \tan x \right)^{\left( \tan x \right)^{\left( \tan x \right)^{. . . \infty}}}\], prove that \[\frac{dy}{dx} = 2\ at\ x = \frac{\pi}{4}\] ?

 

Answer 1

\[\text{ We have, y } = \left( \tan x \right)^{\left( \tan x \right)^{\left( \tan x \right)^{. . . \infty}}} \]
\[ \Rightarrow y = \left( \tan x \right)^y\]

Taking log on both sides,\[\log y = \log \left( \tan x \right)^y \]
\[ \Rightarrow \log y = y \log \tan x\]

Differentiating with respect to x using chain rule ,

\[\frac{1}{y}\frac{dy}{dx} = y\frac{d}{dx}\left\{ \log \tan x \right\} + \log \tan\frac{dy}{dx}\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{y}{\tan x}\frac{d}{dx}\left( \tan x \right) + \log \tan\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{1}{y} - \log \tan x \right) = \frac{y}{\tan x} \sec^2 x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{\tan x} \sec^2 x \times \left( \frac{y}{1 - y\log \tan x} \right)\]
\[\text{Now}, \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{4}} = \frac{y \sec^2 \left( \frac{\pi}{4} \right)}{\tan\left( \frac{\pi}{4} \right)} \times \frac{y}{1 - y \log \tan\left( \frac{\pi}{4} \right)}\]
\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{4}} = \frac{y^2 \left( \sqrt{2} \right)^2}{1\left( 1 - y \log \tan 1 \right)}\]
\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{4}} = \frac{2 \left( 1 \right)^2}{\left( 1 - 0 \right)} \left[ \begin{array}\because \left( y \right)_\frac{\pi}{4} = \left( \tan\frac{\pi}{4} \right)^{\left( \tan\frac{\pi}{4} \right)^{\left( \tan\frac{\pi}{4} \right)^{. . . \infty}}}\end{array} = 1 \right] \]
\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = \frac{\pi}{4}} = 2\]