已知涵数z=y sin(x+y),求dz
答案:3 悬赏:70
解决时间 2021-02-05 05:06
- 提问者网友：萌萌小主
- 2021-02-04 07:27
我什么都不知道
最佳答案
- 二级知识专家网友：我的任性你不懂
- 2021-02-04 08:14
z=ysin(x+y)
dz=dy*sin(x+y)+ysin'(x+y)
=sin(x+y)dy+ycos(x+y)(x+y)'
=sin(x+y)dy+ycos(x+y)(dx+dy)
=ycos(x+y)dx+[sin(x+y)+ycos(x+y)]dy.
dz=dy*sin(x+y)+ysin'(x+y)
=sin(x+y)dy+ycos(x+y)(x+y)'
=sin(x+y)dy+ycos(x+y)(dx+dy)
=ycos(x+y)dx+[sin(x+y)+ycos(x+y)]dy.
全部回答
- 1楼网友：年轻没有失败
- 2021-02-04 09:50
x+y=sin(x·y)+1
y=sin(x·y)-x+1
dy=[ (y+y'x)cos(xy)-1]dx
- 2楼网友：堕落奶泡
- 2021-02-04 08:35
dz=ycos(x+y)dx+sin(x+y)dy+ycos(x+y)dy
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