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d-dx-x-x-please-give-the-answer-with-proof-




Question Number 11547 by Nayon last updated on 28/Mar/17
          (d/dx)(x^x )=?[please give the answer with proof]
$$\:\:\:\:\:\:\:\:\:\:\frac{{d}}{{dx}}\left({x}^{{x}} \right)=?\left[{please}\:{give}\:{the}\:{answer}\:{with}\:{proof}\right] \\ $$
Answered by sma3l2996 last updated on 28/Mar/17
x^x =e^(xln(x))   so  ((d(x^x ))/dx)=((d(e^(xln(x)) ))/dx)=(d/dx)(xln(x))e^(ln(x^x ))   =(ln(x)+1)x^x
$${x}^{{x}} ={e}^{{xln}\left({x}\right)} \\ $$$${so}\:\:\frac{{d}\left({x}^{{x}} \right)}{{dx}}=\frac{{d}\left({e}^{{xln}\left({x}\right)} \right)}{{dx}}=\frac{{d}}{{dx}}\left({xln}\left({x}\right)\right){e}^{{ln}\left({x}^{{x}} \right)} \\ $$$$=\left({ln}\left({x}\right)+\mathrm{1}\right){x}^{{x}} \\ $$
Answered by ajfour last updated on 28/Mar/17
y= x^x   ln y = xln x  (1/y)(dy/dx) = ln x +1  (dy/dx) = y (1+ln x )  (dy/dx) = x^x (1+ln x) .
$${y}=\:{x}^{{x}} \\ $$$$\mathrm{ln}\:{y}\:=\:{x}\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}\:=\:\mathrm{ln}\:{x}\:+\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}\:=\:{y}\:\left(\mathrm{1}+\mathrm{ln}\:{x}\:\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:{x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right)\:. \\ $$

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