$$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{x}^{\sqrt{\mathrm{x}}} } ,\:\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}} \\$$
$${y}={x}^{{e}^{\sqrt{{x}}{ln}\left({x}\right)} } ={e}^{{e}^{\sqrt{{x}}{ln}\left({x}\right)} {ln}\left({x}\right)} \\$$$$\frac{{dy}}{{dx}}=\left({e}^{\sqrt{{x}}{ln}\left({x}\right)} {ln}\left({x}\right)\right)'.{e}^{{e}^{\sqrt{{x}}{ln}\left({x}\right)} {ln}\left({x}\right)} \\$$$$=\left[\left({e}^{\sqrt{{x}}{ln}\left({x}\right)} \right)'{ln}\left({x}\right)+\left({ln}\left({x}\right)\right)'{e}^{\sqrt{{x}}{ln}\left({x}\right)} \right]{x}^{{x}^{\sqrt{{x}}} } \\$$$$\left({e}^{\sqrt{{x}}{ln}\left({x}\right)} \right)'=\left(\sqrt{{x}}{ln}\left({x}\right)\right)'{e}^{\sqrt{{x}}{ln}\left({x}\right)} =\left(\frac{{ln}\left({x}\right)}{\mathrm{2}\sqrt{{x}}}+\frac{\sqrt{{x}}}{{x}}\right){x}^{\sqrt{{x}}} \\$$$${so}\:\frac{{dy}}{{dx}}=\left[\left(\frac{{ln}\left({x}\right)}{\mathrm{2}\sqrt{{x}}}+\frac{\sqrt{{x}}}{{x}}\right){x}^{\sqrt{{x}}} {ln}\left({x}\right)+\frac{{x}^{\sqrt{{x}}} }{{x}}\right]{x}^{{x}^{\sqrt{{x}}} } \\$$$$\frac{{dy}}{{dx}}=\left[\left(\frac{{ln}\left({x}\right)}{\mathrm{2}\sqrt{{x}}}+\frac{\sqrt{{x}}}{{x}}\right){ln}\left({x}\right)+\frac{\mathrm{1}}{{x}}\right){x}^{\sqrt{{x}}+{x}^{\sqrt{{x}}} } \\$$
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\$$