Question

设隐函数y=y(x)由方程ycesx+sin(x-y)=0所确定，求dy

Answer 1

ycosx+sin(x-y)=0
d(ycosx+sin(x-y))=0
ydcosx + cosxdy + cos(x-y)d(x-y) = 0
y(-sinx dx) + cosxdy + cos(x-y)(dx-dy) = 0
[cosx -cos(x-y)]dy = [ysinx-cos(x-y)] dx
dy = { [ysinx-cos(x-y)]/[cosx -cos(x-y)] } dx

Answer 2

cesx？？应该是cosx吧……
ycosx+sin(x-y)=0，两边对x求导得到：
y'cosx+y·(-sinx)+cos(x-y)·(1-y')=0
==> y'cosx-ysinx+cos(x-y)-cos(x-y)·y'=0
==> [cosx-cos(x-y)]·y'=ysinx-cos(x-y)
==> y'=[ysinx-cos(x-y)]/[cosx-cos(x-y)]
则，dy=[……]dx