Question Number 193486 by Mingma last updated on 15/Jun/23
Answered by Subhi last updated on 15/Jun/23
$${x}^{{x}} .{ln}\left({x}\right)={ln}\left({y}\right) \\ $$$${ln}\left({x}^{{x}} .{ln}\left({x}\right)\right)={ln}\left({ln}\left({y}\right)\right)\Rrightarrow\:{xln}\left({x}\right)+{ln}\left({ln}\left({x}\right)\right)={ln}\left({ln}\left({y}\right)\right) \\ $$$${ln}\left({x}\right)+{x}.\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}.{ln}\left({x}\right)}=\frac{\mathrm{1}}{{y}.{ln}\left({y}\right)}.\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}={x}^{{x}} .{x}^{{x}^{{x}} } .{ln}\left({x}\right)\left(\frac{{xln}^{\mathrm{2}} \left({x}\right)+{xln}\left({x}\right)+\mathrm{1}}{{xln}\left({x}\right)}\right)={x}^{{x}^{{x}} +{x}−\mathrm{1}} \left({xln}^{\mathrm{2}} \left({x}\right)+{xln}\left({x}\right)+\mathrm{1}\right) \\ $$$$ \\ $$
Commented by Mingma last updated on 15/Jun/23
Perfect ��