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{tanx+\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+....\infty}}}} $then$ $ find $ (2y-1)\frac{dy}{dx}$

Solution 1:

Hint

1. $$ y = x^y $$
2. $$ y^2 = x + y $$

As mentioned in comments, to solve the first one; you might need to take a look at some of the applications of Omega Function.

Solution 2:

$y = x^{x^{x^{x^{{.^{.^{.^{}}}}}}}} = x^{y}$

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 $y$