Question

怎么求z=(1+xy)^y对y的偏导数?

Answer 1

lnz=y(1+xy)

(1/z)(dz/dy)=ln(1+xy)+(yx)/(1+xy)

dz/dy=[ln(1+xy)+(yx)/(1+xy)]z

dz/dy=[ln(1+xy)+(yx)/(1+xy)](1+xy)^y