（e^y）cos x = 6 + sin(xy) 这个求导怎么求啊...要详细过程RT在线等

对x求导是吧,
(e^y)cos x = 6 + sin(xy),
所以对等式两边求导得到,
(e^y)' *cosx +(e^y) *(cosx)' = [sin(xy)]'
显然(e^y)'= (e^y) *y',(cosx)'= -sinx,[sin(xy)]'= cos(xy) *(xy)'=cos(xy) *(y+xy')
于是
(e^y) *y' *cosx - (e^y) *sinx= cos(xy) *(y+xy'),
那么化简得到
y'= [y* cos(xy) +(e^y) *sinx] / [ (e^y) *cos x - x *cos(xy)]