高等数学求解第13题
展开全部
z=e^(2x+y)*cos(x-y)
∂z/∂x=e^(2x+y)*2*cos(x-y)+e^(2x+y)*[-sin(x-y)]*1
=e^(2x+y)*[2cos(x-y)-sin(x-y)]
∂z/∂y=e^(2x+y)*1*cos(x-y)+e^(2x+y)*[-sin(x-y)]*(-1)
=e^(2x+y)*[cos(x-y)+sin(x-y)]
所以dz=∂z/∂x*dx+∂z/∂y*dy
=e^(2x+y)*[2cos(x-y)-sin(x-y)]dx+e^(2x+y)*[cos(x-y)+sin(x-y)]dy
∂z/∂x=e^(2x+y)*2*cos(x-y)+e^(2x+y)*[-sin(x-y)]*1
=e^(2x+y)*[2cos(x-y)-sin(x-y)]
∂z/∂y=e^(2x+y)*1*cos(x-y)+e^(2x+y)*[-sin(x-y)]*(-1)
=e^(2x+y)*[cos(x-y)+sin(x-y)]
所以dz=∂z/∂x*dx+∂z/∂y*dy
=e^(2x+y)*[2cos(x-y)-sin(x-y)]dx+e^(2x+y)*[cos(x-y)+sin(x-y)]dy
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
