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Question Number 26126 by NECx last updated on 20/Dec/17

using 1st principle find the  derivative of          y=x^x

$${using}\:\mathrm{1}{st}\:{principle}\:{find}\:{the} \\ $$$${derivative}\:{of} \\ $$$$\:\:\:\:\:\:\:\:{y}={x}^{{x}} \\ $$

Commented by abdo imad last updated on 20/Dec/17

answer to question  ew have y= e^(xlnx)   −−> (dy/dx)  =  ((d(xln(x)))/dx)  .x^x   =  (lnx  + 1)  x^x   with the condition  x>0  .

$${answer}\:{to}\:{question}\:\:{ew}\:{have}\:{y}=\:{e}^{{xlnx}} \:\:−−>\:\frac{{dy}}{{dx}}\:\:=\:\:\frac{{d}\left({xln}\left({x}\right)\right)}{{dx}}\:\:.{x}^{{x}} \\ $$$$=\:\:\left({lnx}\:\:+\:\mathrm{1}\right)\:\:{x}^{{x}} \:\:{with}\:{the}\:{condition}\:\:{x}>\mathrm{0}\:\:. \\ $$

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