Question Number 225047 by fantastic last updated on 16/Oct/25
$$\int\:{x}^{{x}} \:{dx} \\ $$
Answered by Ghisom_ last updated on 16/Oct/25
$$\mathrm{impossible} \\ $$
Commented by fantastic last updated on 16/Oct/25
$$\frac{{d}}{{dx}}\left({x}^{{x}} \right) \\ $$
Commented by Ghisom_ last updated on 16/Oct/25
$${x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right) \\ $$
Commented by Raphael254 last updated on 17/Oct/25
$$ \\ $$$${y}\:=\:{u}^{{v}} \:\Leftrightarrow\:{log}_{{u}} \:{y}\:=\:{v} \\ $$$$ \\ $$$${Changing}\:{the}\:{base}\:{of}\:{log}: \\ $$$$ \\ $$$${log}_{{u}} \:{y}\:=\:\frac{{ln}\:{y}}{{ln}\:{u}} \\ $$$$ \\ $$$$\frac{{ln}\:{y}}{{ln}\:{u}}\:=\:{v} \\ $$$$ \\ $$$$\frac{\frac{{y}'}{{y}}×{ln}\:{u}\:−\:{ln}\:{y}×\frac{{u}'}{{u}}}{{ln}^{\mathrm{2}} \:{u}}\:=\:{v}' \\ $$$$ \\ $$$$\frac{{ln}\:{u}×{y}'}{{y}}\:−\:\frac{{ln}\:{y}×{u}'}{{u}}\:=\:{v}'×{ln}^{\mathrm{2}} \:{u} \\ $$$$ \\ $$$${ln}\:{u}×{y}'×{u}\:−\:{ln}\:{y}×{u}'×{y}\:=\:{v}'×{ln}^{\mathrm{2}} \:{u}×{uy} \\ $$$$ \\ $$$${y}'\:=\:\frac{{v}'×{ln}^{\mathrm{2}} \:{u}×{uy}\:+\:{ln}\:{y}×{u}'×{y}}{{ln}\:{u}×{u}} \\ $$$$ \\ $$$${y}'\:=\:{v}'×{ln}\:{u}×{y}\:+\:\frac{{ln}\:{y}×{u}'×{y}}{{ln}\:{u}×{u}} \\ $$$$ \\ $$$${y}'\:=\:{v}'×{ln}\:{u}×\left({u}^{{v}} \right)\:+\:\frac{{ln}\:\left({u}^{{v}} \right)×{u}'×\left({u}^{{v}} \right)}{{ln}\:{u}×{u}} \\ $$$$ \\ $$$${y}'\:=\:{v}'×{ln}\:{u}×{u}^{{v}} \:+\:\frac{{v}×\left({ln}\:{u}\right)×{u}'×{u}^{{v}} }{{ln}\:{u}×{u}} \\ $$$$ \\ $$$${y}'\:=\:{v}'×{ln}\:{u}×{u}^{{v}} \:+\:{v}×{u}'×{u}^{{v}−\mathrm{1}} \\ $$$$ \\ $$$${Applying}\:{in}\:{y}\:=\:{x}^{{x}} : \\ $$$$ \\ $$$${y}\:=\:{x}^{{x}} \:\Rightarrow\:{u},\:{v}\:=\:{x} \\ $$$$ \\ $$$${y}'\:=\:\mathrm{1}×{ln}\:{x}×{x}^{{x}} \:+\:{x}×\mathrm{1}×{x}^{{x}−\mathrm{1}} \\ $$$${y}'\:=\:{ln}\:{x}×{x}^{{x}} \:+\:{x}^{{x}} \\ $$$${y}'\:=\:{x}^{{x}} \left({ln}\:{x}\:+\:\mathrm{1}\right) \\ $$
Answered by Tawa11 last updated on 16/Oct/25
$$\mathrm{Done}\:\mathrm{in}\:\:\mathrm{Q111558} \\ $$