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



$y=(\tan x)^{(\tan x)^{(\tan x)}-^{\infty}}$

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

By taking log on both sides,

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

$\log y=y(\log \tan x)$

Differentiating both sides with respect to $x$ using the product rule and chain rule,

$\frac{1}{y} \frac{d y}{d x}=y \frac{d(\log \tan x)}{d x}+\log \tan x \frac{d y}{d x}$

$\frac{1}{y} \frac{d y}{d x}=\frac{y}{\tan x} \frac{d(\tan x)}{d x}+\log \tan x \frac{d y}{d x}$

$\left(\frac{1}{y}-\log \tan x\right) \frac{d y}{d x}=\frac{y}{\tan x}\left(\sec ^{2} x\right)$

$\left(\frac{1-y \log \tan x}{y}\right) \frac{d y}{d x}=\frac{y \sec ^{2} x}{\tan x}$

$\frac{d y}{d x}=\frac{y^{2} \sec ^{2} x}{\tan x(1-y \log \tan x)}$

$\frac{d y}{d x}\left(x=\frac{\pi}{4}\right)=\frac{y^{2} \sec ^{2}\left(\frac{\pi}{4}\right)}{\tan \left(\frac{\pi}{4}\right)\left(1-y \log \tan \left(\frac{\pi}{4}\right)\right)}$

$\frac{d y}{d x}\left(x=\frac{\pi}{4}\right)=\frac{2 y^{2}}{1(1-y \log 1)}$

Since $\left\{(y)_{\frac{\pi}{4}}=\left(\tan \frac{\pi}{4}\right)^{\left(\tan \frac{\pi}{4}\right)^{\left(\tan \frac{\pi}{4}\right) \ldots}}=(1)^{\infty}=1\right\}$



Hence proved.

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