y = (tan x)^{(tan x)^{tan x}},then at x = fra | vedclass

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(C) Given $y = (	an x)^{(	an x)^{	an x}}$.Taking $\log$ on both sides,we get $\log y = (	an x)^{	an x} \log(	an x)$.Let $u = (	an x)^{	an x}$.

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$2$

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(C) Given $y = (	an x)^{(	an x)^{	an x}}$.Taking $\log$ on both sides,we get $\log y = (	an x)^{	an x} \log(	an x)$.Let $u = (	an x)^{	an x}$. Then $\log u = 	an x \log(	an x)$.Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = \sec^2 x \log(	an x) + 	an x \cdot \frac{1}{	an x} \cdot \sec^2 x = \sec^2 x (\log(	an x) + 1)$.So,$\frac{du}{dx} = u \sec^2 x (\log(	an x) + 1) = (	an x)^{	an x} \sec^2 x (\log(	an x) + 1)$.Now,$\log y = u \log(	an x)$.Differentiating with respect to $x$,$\frac{1}{y} \frac{dy}{dx} = \frac{du}{dx} \log(	an x) + u \cdot \frac{1}{	an x} \cdot \sec^2 x$.At $x = \frac{\pi}{4}$,$	an x = 1$,so $u = 1^1 = 1$ and $y = 1^1 = 1$.Also,$\log(	an x) = \log 1 = 0$ and $\sec^2 x = \sec^2(\frac{\pi}{4}) = 2$.Substituting these values,$\frac{du}{dx} = 1 \cdot 2 \cdot (0 + 1) = 2$.Then $\frac{1}{1} \frac{dy}{dx} = 2 \cdot 0 + 1 \cdot \frac{1}{1} \cdot 2 = 2$.Therefore,$\frac{dy}{dx} = 2$.

$y = (\tan x)^{(\tan x)^{\tan x}}$,then at $x = \frac{\pi}{4}$,the value of $\frac{dy}{dx} = $

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