If $y^{y^y}$ $={\log }_e\left(x+{\log }_e(x+\dots )\right)\text{ then }\frac{dy}{dx}\text{ at }\left(x=e^2-2,y=\sqrt{2}\right)$ is

$\text{ Let }{y}^{y^y}={\log }_e\left(x+{\log }_e(x+\dots )\right)=v$

$\therefore y^v={\log }_e(x+v)=v\Rightarrow y=v^{1/v}\text{ and }x=e^v-v\text{ find }$

$\begin{array}{r}\Rightarrow \frac{dy}{dv}=v^{1/v}\left(\frac{d}{dx}\frac{1}{v}\log v\right)\text{ and }\frac{dx}{dv}=\left(e^v-1\right)\\ \text{ i.e }\frac{dy}{dv}=v^{1/v}\left(\frac{1}{v}\cdot \frac{1}{v}-\frac{1}{v^2}\log v\right)={v^{1/v}}^{-2}(1-\log v)\\ \text{ and }\frac{dx}{dv}=\left(e^v-1\right)\therefore \frac{dy}{dx}=\frac{{v^{1/v}}^{-2}(1-\log v)}{e^v-1}\end{array}$

$\text{ at }\mathrm{x}={\mathrm{e}}^2-2,\mathrm{y}=\sqrt{2},v^{1/v}=\sqrt{2},{\mathrm{e}}^{\mathrm{v}}-\mathrm{v}={\mathrm{e}}^2-2$

$\Rightarrow \mathrm{v}=2\text{ or }4\text{ which is not valid }$

$\therefore \mathrm{v}=2,{\mathrm{e}}^{\mathrm{v}}-\mathrm{v}={\mathrm{e}}^2-2=\mathrm{x}\text{ given so true }$

${\left(\frac{dy}{dx}\right)}_{e^2-2,\sqrt{2}}={\left(\frac{dy}{dx}\right)}_{v=2}=\frac{1-\log 2}{2\sqrt{2}\left(e^2-1\right)}$