填1.首先明白y'=dy/dx.其次,y=x^(2/x),两边取对数推出lny=lnx^(2/x)=(2/x)lnx,两边求导推出y’/y=(-2lnx)/x^2+1/x^2推出y'=y(1-2lnx)/x^2=(1-2lnx)[x^(2/x-2)],即dy=(1-2lnx)[x^(2/x-2)]dx.当x=1时:dy=dx....