Question Number 135536 by islamo last updated on 13/Mar/21
$$\left(\boldsymbol{{u}}^{\boldsymbol{{v}}} \right)'=? \\ $$$${u}\:,{v}\:\:\:{tow}\:{fonctions} \\ $$
Answered by Olaf last updated on 13/Mar/21
$${u}^{{v}} \:=\:{e}^{\mathrm{ln}\left({u}^{{v}} \right)} \:=\:{e}^{{v}\mathrm{ln}{u}} \\ $$$$\left({u}^{{v}} \right)'\:=\:\left({v}\mathrm{ln}{u}\right)'{e}^{{v}\mathrm{ln}{u}} \:=\:\left({v}'\mathrm{ln}{u}+\frac{{vu}'}{{u}}\right){u}^{{v}} \\ $$$$\left({u}^{{v}} \right)'\:=\:{v}'{u}^{{v}} \mathrm{ln}{u}+{u}'{vu}^{{v}−\mathrm{1}} \\ $$