<s>[INST] Calculate:$y'$ for $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$ and $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$ (1) If  $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$
(2) If $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$
then find $y'$ in both cases
(3)If $ y= \sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+....\infty}}}}\ $ then find$\ (2y-1)\frac{dy}{dx}$
 [/INST] $y = x^{x^{x^{x^{{.^{.^{.^{}}}}}}}} = x^{y}$
Take the log of both sides.
Then $\ln y = y \ln x$
$ \implies\frac{1}{y} \frac{dy}{dx} = \ln x \frac{dy}{dx}  + \frac{y}{x}$
$ \implies \frac{dy}{dx} =  \frac{y^{2}}{x(1-y \ln x)}$ and then substitute for $y$
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