z=e^usinv=e^(xsiny)sin(xcosy)∂z/∂x=e^(xsiny)[(siny)]sin(xcosy)-e^(xsiny)cos(xcosy)[(cosy)=e^(xsiny)[siny)sin(xcosy)-cos(xcosy)(cosy)]同理可得：∂z/∂y=.大哥求你把同理可得也写一下 好不，我是完全不懂，照抄答案的孩子！大哥拜托了！555555555555555电脑不好打呀，小妹！∂z/∂y=xcosye^(xsiny)sin(xcosy)-e^(xsiny)cos(xcosy)xsiny=e^(xsiny)[xcosysin(xcosy)-xcos(xcosy)siny]