ye^x+sin(xy)=100,求dy/dx,跪求过程.

ye^x+sin(xy)=100,求dy/dx,跪求过程.
答案是 -[(e^x)y+ycos(xy)]/[e^x+xcos(xy)]
谢谢! 浪子南枫 1年前他留下的回答 已收到1个回答

ICTFU 网友

该名网友总共回答了24个问题，此问答他的回答如下：采纳率：83.3%

ye^x+sin(xy)=100
y'e^x+ye^x+cos(xy)(y+xy')=0
y'e^x+ycos(xy)+xy'cos(xy)=-(e^x)y
y'(e^x+xcos(xy))=-(e^x)y+ycos(xy)
y'=-(e^x)y+ycos(xy)/(e^x+xcos(xy))
dx/dy=-(e^x)y+ycos(xy)/(e^x+xcos(xy))
d(ye^x+sin(xy))/dx=d(100)/dx
d(ye^x)dx+d(sin(xy))/dx=0
(e^x)dy/dx+yd(e^x)/dx+cosx(xy)d(xy)/dx=0
(e^x)dy/dx+y(e^x)+cosx(xy)(y+xdy/dx)=0
dx/dy=-(e^x)y+ycos(xy)/(e^x+xcos(xy))

1年前他留下的回答

1

　　以上就是小编为大家介绍的ye^x+sin(xy)=100,求dy/dx,跪求过程. 的全部内容，如果大家还对相关的内容感兴趣，请持续关注上海建站网！