y'*sinx+ycosx- [-sin(x+y)*(1+y')]=0
y'(sinx+sin(x+y))=y(1-cosx)
y'=[1-cosx]/[sinx+sin(x+y)]
0/0所以需要洛必达
先关于x
y'=[sinx]/[cosx+cos(x+y)(1+y')]
所以y'[cosx+cos(x+y)(1+y')]=sinx
令x->0
y'[1+cosy *(1+y')]=0
令y->pi
y'[1-(1+y')]=0
(y')^2=0,y'=0
已知ysinx-cos(x+y)=0,求在点(0,π)的dy\/dx值
所以y'[cosx+cos(x+y)(1+y')]=sinx 令x->0 y'[1+cosy *(1+y')]=0 令y->pi y'[1-(1+y')]=0 (y')^2=0,y'=0
已知ysinx-cos(x+y)=0,求在点(0,π\/2)的dy\/dx值
ysinx-cos(x+y)=0,两边对x求导,得 y'sinx+ycosx+(1+y')sin(x+y)=0,解得 y'=-[ycosx+sin(x+y)]\/[sinx+sin(x+y)]dy\/dx=y'(0,π\/2)=-(π\/2+1)\/(0+1)=-(π\/2+1)
ysinx-cos(x+y)=0,求 dy\/dx
故 dy\/dx=y′=-（ycosx+sin（x+y））\/（sinx+sin（x+y））.
由方程ysinx-cos(x-y)=0所确定的函数的导数dy\/dx
ysinx-cos(x-y)=0所确定的函数的导数dy\/dx是：y'= dy\/dx = [ycosx + sin(x-y)]\/[sin(x-y) - sinx]计算过程如下：方程两边同时求导，得到下面式子：y'sinx+ycosx+sin(x-y) (1-y') = 0 整理可得 y'[sinx -sin(x-y)] = -ycosx - sin(x-y)所以 y'=[ycosx + sin(x...
设函数y=y(x)由方程y=sinx-sin(x+y)=0所确定,求dy\/dx
两边同时对x求导有dy\/dx=cosx-cos（x+y）*dy\/dx，合并同类项之后有 （1+cos（x+y））dy\/dx=cosx 所以dy\/dx=cosx\/（1+cos（x+y））希望能采纳，谢谢
已知y*sinx-cos(x-y)=0,求dy的值?
因为 y*sinx-cos(x-y)=0 则：d(ysinx)-dcos(x-y)=0 即：sinxdy+dcosxdx+sin(x-y)(dx-dy)=0 整理得：[sin(x-y)-sinx]dy=[ycosx+sin(x-y)]dx 所以：dy=[ycosx+sin(x-y)]dx\/[sin(x-y)-sinx]
已知y*sinx-cos(x-y)=0,求dy的值?
因为 y*sinx-cos(x-y)=0 则：d(ysinx)-dcos(x-y)=0 即：sinxdy+dcosxdx+sin(x-y)(dx-dy)=0 整理得：[sin(x-y)-sinx]dy=[ycosx+sin(x-y)]dx 所以：dy=[ycosx+sin(x-y)]dx\/[sin(x-y)-sinx]
ysinx+cos(x-y)=0,求dy\/dx|(x=π\/2)
两边对x求导：dy\/dxsinx+ycosx-sin(x-y)(1-dy\/dx)=0,将x=π\/2带入已知方程得到y,再把x、y带入上式求得结果
ysinx+cos(x-y)=0,求dy\/dx|(x=π\/2)
两边对x求导：dy\/dxsinx+ycosx-sin(x-y)(1-dy\/dx)=0,将x=π\/2带入已知方程得到y，再把x、y带入上式求得结果
由方程ysinx-cos(x+y)=0确定隐函数y(x),求dy|(0,π\/2)
两边求导：y'sinx+ycosx+sin(x+y)*(1+y')=0 令x=0,y=π\/2：π\/2+1+y'=0 y'=-(π\/2+1)dy=-(π\/2+1)dx
