求隐函数的导数 sin(x+2y)=ln xy 答案要详细的,谢了

问题描述:

求隐函数的导数 sin(x+2y)=ln xy 答案要详细的,谢了

两边对x求导得:
cos(x+2y)*(1+2y')=1/(xy)*(y+xy')
xycos(x+2y)+2xyy'cos(x+2y)=y+xy'
得:y'=[y-xycos(x+2y)]/[2xycos(x+2y)-x]