question_answer

If \[y={{a}^{{{x}^{{{a}^{x...\infty }}}}}},\] prove that \[\frac{dy}{dx}=\frac{{{y}^{2}}\log \,\,y}{x[1-y\log \,x\log \,y]}.\]

Answer:

We have, \[y={{a}^{{{x}^{{{a}^{{{x}^{{{a}^{{{x}^{.....a}}}}}}}}}}}}\]          \[\Rightarrow y={{a}^{{{x}^{y}}}}\] \[\Rightarrow \]   \[\log y={{x}^{y}}\log a\][taking log on both sides] \[\Rightarrow \]   \[\log \,(\log y)=y\log x+\log \,(\log a)\] [taking log on both sides]             \[\Rightarrow \]\[\frac{1}{\log y}\frac{d}{dx}(\log y)=\frac{dy}{dx}\log x+y\frac{d}{dx}(\log x)+0\][differentiating both sides w.r.t. x]             \[\Rightarrow \]   \[\frac{1}{\log y}\cdot \frac{1}{y}\cdot \frac{dy}{dx}\log x+yx\frac{1}{x}\]             \[\Rightarrow \]   \[\frac{dy}{dx}\left\{ \frac{1}{y\log y}-\log x \right\}=\frac{y}{x}\]             \[\Rightarrow \]   \[\frac{dy}{dx}\left\{ \frac{1-y\log y\log x}{y\log y} \right\}=\frac{y}{x}\]             \[\Rightarrow \]   \[\frac{dy}{dx}=\frac{{{y}^{2}}\log y}{x\{1-y\log y\log x\}}\] Hence proved.