(1) 因为 y=x^(1/x),
两边取对数,得
ln y=(1/x)*ln x.
两边求导,得
(y')/y=(-1/ x^2)*ln x +(1/x)(1/x)
=(1-ln x)/(x^2).
所以 (y')=y(1-ln x)/(x^2)
=x^(1/x -2) *(1-ln x).
(2) 设 g(x)=x^(1/x -2),
两边取对数,得
ln g(x)=(1/x -2)*ln x.
两边求导,得
g'(x)/g(x)=(-1/ x^2)*ln x +(1/x -2)(1/x)
=(1-2x-ln x)/(x^2).
所以 g'(x)=x^(1/x -4) *(1-2x-ln x).
所以 (y'')=[g(x)*(1-ln x)]'
=x^(1/x -4)*(1-2x-ln x)*(1-ln x)
+x^(1/x -2)*(-1/x)
=x^(1/x -4)*[(1-2x-ln x)*(1-ln x)-x]
=x^(1/x -4)*[1-3x-2ln x+ 2x lnx +(ln x)^2].