Question

高数：求函数的偏导数.z = ( 1 + xy ) ^ y ,

Answer 1

lnz=yln(1 + xy )

z=e^{yln(1 + xy )}

dz/dy=e^[yln(1 + xy )]{ln(1+xy)+xy/(1+xy)}

dz/dx=e^[yln(1 + xy )]{y^2/(1+xy)}