Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\left(\tan x\right)}^{{\left(\tan x\right)}^{{\left(\tan x\right)}^{...\infty }}} \\ \Rightarrow y={\left(\tan x\right)}^y \end{gathered}$$


Taking log on both sides,
$$\begin{gathered} \log y=\log {\left(\tan x\right)}^y \\ \Rightarrow \log y=y\log \tan x \end{gathered}$$

Differentiating with respect to x using chain rule ,
$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=y\frac{d}{dx}\left\{\log \tan x\right\}+\log \tan \frac{dy}{dx} \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{y}{\tan x}\frac{d}{dx}\left(\tan x\right)+\log \tan \frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(\frac{1}{y}-\log \tan x\right)=\frac{y}{\tan x}{\sec }^{2}x \\ \Rightarrow \frac{dy}{dx}=\frac{y}{\tan x}{\sec }^{2}x\times \left(\frac{y}{1-y\log \tan x}\right) \\ \mathrm{Now},{\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{y{\sec }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}\times \frac{y}{1-y\log \tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{y^2{\left(\sqrt{2}\right)}^2}{1\left(1-y\log \tan 1\right)} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{4}}=\frac{2{\left(1\right)}^2}{\left(1-0\right)}\left[\begin{array}{c}\because {\left(y\right)}_{\frac{\mathrm{\pi }}{4}}={\left(\tan \frac{\mathrm{\pi }}{4}\right)}^{{\left(\tan \frac{\mathrm{\pi }}{4}\right)}^{{\left(\tan \frac{\mathrm{\pi }}{4}\right)}^{...\infty }}}\end{array}=1\right] \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{4}}=2 \end{gathered}$$