$y=x^{x^{x}}=>lny=x^xlnx$.Đặt $y_1=x^x$,ta có $lny_1=xlnx$ =>$\frac{y'_1}{y_1}=lnx+1$ Từ đó $y'_1=x^x.(lnx+1)$.Ta có: $lny=y_1lnx$ nên: $\frac{y'}{y}=y'_1lnx+\frac{1}{x}y_1=x^x.(ln^2x+lnx+\frac{1}{x})$ =>$y'=x^{x^{x}}.x^x.(ln^2x+lnx+\frac{1}{x})$ $y'(1)=1$
