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$\ (2y-1)\frac{dy}{dx}$

*

*As $y=x^y = e^{y \log{x}}$, take derivatives of both sides:


$$y' = \left( \frac{y}{x} + y' \log{x} \right ) x^y$$
Solve for $y'$.


*

*As $y^2 = x+y$, then a similar implicit differentiation as above yields


$$2 y \,y' = 1 + y'$$
Solve for $y'$.
