Question Number 184187 by Mastermind last updated on 03/Jan/23
$$\mathrm{Differentiate},\:\mathrm{y}=\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{x}^{\mathrm{x}} \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$
Answered by SEKRET last updated on 03/Jan/23
$$\:\:\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{u}}=\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:\:\:\:\:\boldsymbol{\mathrm{u}}'=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\:\boldsymbol{\mathrm{t}}\:=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:\:\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{t}}\right)\:=\:\boldsymbol{\mathrm{x}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\:\:\:\:\frac{\boldsymbol{\mathrm{t}}\:'}{\boldsymbol{\mathrm{t}}}\:\:=\:\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:+\mathrm{1} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{t}}\:'\:=\:\boldsymbol{\mathrm{t}}\centerdot\left(\mathrm{1}+\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)=\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \centerdot\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)+\mathrm{1}\right) \\ $$$$\:\:\:\boldsymbol{\mathrm{y}}\:'\:=\boldsymbol{\mathrm{u}}\:'\:+\:\:\boldsymbol{\mathrm{t}}\:' \\ $$$$\:\:\:\:\boldsymbol{\mathrm{y}}\:\:'\:\:=\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:+\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \centerdot\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:+\:\mathrm{1}\right) \\ $$
Commented by Mastermind last updated on 03/Jan/23
$$\mathrm{Good}! \\ $$