Question

Prove the following.
i) If $y=a^{x^{a^{x^{{.}^{{.}^{{.}^{\infty }}}}}}}$, then prove that $\frac{dy}{dx}=\frac{y^2\log y}{x(1-y\log x\log y)}$
ii) If $y=\cos x^{\cos x^{\cos x^{\cos x^{{.}^{{.}^{{.}^{\infty }}}}}}}$, then prove that $\frac{dy}{dx}=\frac{-y^2\tan x}{1-y\log \cos x}$

Answer 1

i) The given series maybe written as $y=a^{x^y}$
$\Rightarrow \log y=x^y\log a$
$\log \left(\log y\right)=y\log x+\log \left(\log a\right)$
$\frac{1}{\log y}\frac{d}{dx}\log y=\frac{dy}{dx}\log x+y\frac{d}{dx}\log x+0$
$\frac{1}{\log y}\frac{1}{y}\frac{dy}{dx}=\frac{dy}{dx}\log x+y\frac{1}{x}$
$\frac{dy}{dx}(\frac{1}{y\log y}-\log x)=\frac{y}{x}$
$\frac{dy}{dx}\left(\frac{1-y\log y\log x}{y\log y}\right)=\frac{y}{x}$
$\frac{dy}{dx}=\frac{y^2\log y}{x(1-y\log y\log x)}$

ii) The given series maybe written as $y=\cos x^y$
$\log y=y\log \cos x$
$\frac{1}{y}\frac{dy}{dx}=y\frac{1}{\cos x}(-\sin x)+\log \cos x\frac{dy}{dx}$
$\frac{1}{y}\frac{dy}{dx}=-y\tan x+\log \cos x\frac{dy}{dx}$
$(\frac{1}{y}-\log \cos x)\frac{dy}{dx}=-y\tan x$
$\left(\frac{1-y\log \cos x}{y}\right)\frac{dy}{dx}=-y\tan x$
$\frac{dy}{dx}=\frac{-y^2\tan x}{1-y\log \cos x}$