Question

If $y=a^{x^{a^{x...\infty }}}$,then show that $\frac{dy}{dx}=\frac{y^2\log y}{x(1-y\log x\log y)}$

Answer 1

$y=a^{x^{a^{x...\infty }}}$

$\Rightarrow y=a^{x^y}$

$\therefore \log y=x^{y}a$

$\Rightarrow \log \log y=y\log x+\log a$

Differentiating both sides w.r.t $x$, we get
$\frac{1}{\log y}\cdot \frac{1}{y}\frac{dy}{dx}=\frac{dy}{dx}.\log x+y.\frac{1}{x}$

$\therefore (\frac{1}{y\log y}-\log x)\frac{dy}{dx}=\frac{y}{x}$

$\therefore \frac{dy}{dx}=\frac{y^2\log y}{x(1-y\log x\log y)}$