Menu Close

Question-105929




Question Number 105929 by Study last updated on 01/Aug/20
Answered by Dwaipayan Shikari last updated on 01/Aug/20
lim_(x→2) (((d/dx)(x^x^x  )+x^x (logx+1)+4x^3 +2x+1)/(2x))    =((x^x^x  ((x^x /x)+x^x (logx+1)logx)+x^x (logx+1)+4x^3 +2x+1)/(2x))  =((2^2^2  (2+4(log2+1)log2)+4(log2+1)+32+4+1)/4)  =4(2+4(log2)^2 +4log2)+log2+((41)/4)  x^x^x  =y  x^x logx=logy  x^x .(1/x)+x^x (logx+1)logx=(1/y).(dy/dx)  x^x^x  ((x^x /x)+x^x (logx+1)logx)=(dy/dx)        {(d/dx)(x^x )=x^x (1+logx)
$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left({x}^{{x}^{{x}} } \right)+{x}^{{x}} \left({logx}+\mathrm{1}\right)+\mathrm{4}{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}} \\ $$$$ \\ $$$$=\frac{{x}^{{x}^{{x}} } \left(\frac{{x}^{{x}} }{{x}}+{x}^{{x}} \left({logx}+\mathrm{1}\right){logx}\right)+{x}^{{x}} \left({logx}+\mathrm{1}\right)+\mathrm{4}{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}^{\mathrm{2}} } \left(\mathrm{2}+\mathrm{4}\left({log}\mathrm{2}+\mathrm{1}\right){log}\mathrm{2}\right)+\mathrm{4}\left({log}\mathrm{2}+\mathrm{1}\right)+\mathrm{32}+\mathrm{4}+\mathrm{1}}{\mathrm{4}} \\ $$$$=\mathrm{4}\left(\mathrm{2}+\mathrm{4}\left({log}\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}{log}\mathrm{2}\right)+{log}\mathrm{2}+\frac{\mathrm{41}}{\mathrm{4}} \\ $$$${x}^{{x}^{{x}} } ={y} \\ $$$${x}^{{x}} {logx}={logy} \\ $$$${x}^{{x}} .\frac{\mathrm{1}}{{x}}+{x}^{{x}} \left({logx}+\mathrm{1}\right){logx}=\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dx}} \\ $$$${x}^{{x}^{{x}} } \left(\frac{{x}^{{x}} }{{x}}+{x}^{{x}} \left({logx}+\mathrm{1}\right){logx}\right)=\frac{{dy}}{{dx}} \\ $$$$\:\:\:\:\:\:\left\{\frac{{d}}{{dx}}\left({x}^{{x}} \right)={x}^{{x}} \left(\mathrm{1}+{logx}\right)\right. \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *