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.
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