Question Number 184185 by Mastermind last updated on 03/Jan/23
$$\mathrm{Differentiate},\:\mathrm{y}=\left(\mathrm{log}_{\mathrm{e}} \mathrm{x}\right)^{\mathrm{x}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$
Answered by SEKRET last updated on 03/Jan/23
$$\:\:\:\:\:\boldsymbol{\mathrm{y}}\:\:=\:\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)^{\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{y}}\right)\:\:=\:\:\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{y}}\:'}{\boldsymbol{\mathrm{y}}}\:=\:\boldsymbol{\mathrm{x}}'\:\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)\:+\:\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)\:' \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{y}}\:'\:=\boldsymbol{\mathrm{y}}\centerdot\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)+\frac{\mathrm{1}}{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}\right) \\ $$$$\:\:\:\:\boldsymbol{\mathrm{y}}\:'\:=\:\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)^{\boldsymbol{\mathrm{x}}} \:\centerdot\:\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)+\frac{\mathrm{1}}{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)}\right) \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{ABDULAZIZ}}\:\:\boldsymbol{\mathrm{ABDUVALIYEV}} \\ $$
Commented by Mastermind last updated on 03/Jan/23
$$\mathrm{Good}! \\ $$