求由方程xcosy=sin(x+y)所确定的应函数导数Y✀x的详细过程及结果

解：两边分别对x求导，将y看为x函数，遇到单独的y时用y‘代替 ，(如果出现形如xy的情形时，则表示为(xy)'=y+xy' )

cosy-(xsiny)y' =cos(x+y)(1+y')
然后解出y' 就行了。即y'=[cosy-cos(x+y)]/[cos(x+y)+xsiny]

xcosy=sinxcosy+cosxsiny
cosy-y'xsiny=cosxcosy-y'sinxsiny -sinxsiny+y'cosxcosy
cosy-y' xsiny=cos(x+y)+y'cos(x+y)
cos(x+y)-cosy= -y'[cos(x+y)+xsiny]
y'= -[cos(x+y)-cosy]/[cos(x+y)+xsiny]

解：
xcosy=sin(x+y)
cosy-x(y')siny=(1+y')cos(x+y)
cosy-x(y')siny=cos(x+y)+(y')cos(x+y)
cosy-cos(x+y)=(y')cos(x+y)+x(y')siny
(y')[cos(x+y)+xsiny]=cosy-cos(x+y)
y'=[cosy-cos(x+y)]/[cos(x+y)+xsiny]

左右各求导，得：cosy-xsiny.(y')=cos(x+y)*(1+y'),合并同类项得：y'=(cosy-cos(x+y))/(cos(x+y)+xsiny)