Question

微积分,求一阶导数,

Answer 1

设u=x^x v=x^(1/x)

则y=u+v y`=u`+v`

lnu=xlnx

u`/u=lnx+1

u`=x^x(lnx+1)

lnv=lnx/x

v`/v=(1-lnx)/x²

v`=x^(1/x)(1-lnx)/x²

于是y`=u`+v`=x^x(lnx+1)+x^(1/x)(1-lnx)/x²