z=3x²+4e∧xy+cos(x²y)求dz
2个回答
展开全部
z=3x^2+4e^(xy)+cos(x^2.y)
dz
=d[3x^2+4e^(xy)+cos(x^2.y)]
=6x dx + 4e^(xy). d(xy) - sin(x^2.y) . d(x^2.y)
=6x dx + 4e^(xy). (xdy +ydx) - sin(x^2.y) . (2xy dx + x^2.dy)
=[6x +4ye^(xy)-2xysin(x^2.y)] dx + [ 4xe^(xy) -x^2.sin(x^2.y) ] dy
dz
=d[3x^2+4e^(xy)+cos(x^2.y)]
=6x dx + 4e^(xy). d(xy) - sin(x^2.y) . d(x^2.y)
=6x dx + 4e^(xy). (xdy +ydx) - sin(x^2.y) . (2xy dx + x^2.dy)
=[6x +4ye^(xy)-2xysin(x^2.y)] dx + [ 4xe^(xy) -x^2.sin(x^2.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