Question Number 134297 by mathlove last updated on 02/Mar/21
Answered by Ñï= last updated on 02/Mar/21
![(d/dx)ln(x^x +2^x^x )=(1/(x^x +2^x^x ))[x^x (lnx+1)+(ln2)2^x^x ln2^x 2^x^x ] =(1/(x^x +2^x^x ))[(lnx+1)x^x +(ln2)^2 (2^x^x )^2 x] (d/dx)(lnx^(ln(x^x +2^x^x )) ) =(d/dx)[(lnx)ln(x^x +2^x^x )] =(1/x)ln(x^x +2^x^x )+((lnx)/(x^x +2^x^x ))[(lnx+1)x^x +(ln2)^2 (2^x^x )^2 x]](https://www.tinkutara.com/question/Q134317.png)
$$\frac{{d}}{{dx}}{ln}\left({x}^{{x}} +\mathrm{2}^{{x}^{{x}} } \right)=\frac{\mathrm{1}}{{x}^{{x}} +\mathrm{2}^{{x}^{{x}} } }\left[{x}^{{x}} \left({lnx}+\mathrm{1}\right)+\left({ln}\mathrm{2}\right)\mathrm{2}^{{x}^{{x}} } {ln}\mathrm{2}^{{x}} \mathrm{2}^{{x}^{{x}} } \right] \\ $$$$=\frac{\mathrm{1}}{{x}^{{x}} +\mathrm{2}^{{x}^{{x}} } }\left[\left({lnx}+\mathrm{1}\right){x}^{{x}} +\left({ln}\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{2}^{{x}^{{x}} } \right)^{\mathrm{2}} {x}\right] \\ $$$$\frac{{d}}{{dx}}\left({lnx}^{{ln}\left({x}^{{x}} +\mathrm{2}^{{x}^{{x}} } \right)} \right) \\ $$$$=\frac{{d}}{{dx}}\left[\left({lnx}\right){ln}\left({x}^{{x}} +\mathrm{2}^{{x}^{{x}} } \right)\right] \\ $$$$=\frac{\mathrm{1}}{{x}}{ln}\left({x}^{{x}} +\mathrm{2}^{{x}^{{x}} } \right)+\frac{{lnx}}{{x}^{{x}} +\mathrm{2}^{{x}^{{x}} } }\left[\left({lnx}+\mathrm{1}\right){x}^{{x}} +\left({ln}\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{2}^{{x}^{{x}} } \right)^{\mathrm{2}} {x}\right] \\ $$