cos(xy)+x^6=y^6
两边同时对x求导得
-sin(xy)·(xy) '+6x^5=6y^5·y'
-sin(xy)·(y+xy ')+6x^5=6y^5·y'
得y '=[6x^5-ysin(xy)]/[6y^5+xsin(xy)]为什么会有-sin(xy)·(y+xy ')+6x^5=6y^5·y'这步出现啊？-sin(xy)·(xy) '+6x^5=6y^5·y'→这步看懂了吗等等...-sin(xy)·(xy) '+6x^5=6y^5·y'-sin(xy)·(y+xy ')+6x^5=6y^5·y'这一步中的y+xy '是上面的xy求导得到的，其他的一样 注意(xy) '=y+xy '嗯看懂了...最后一步呢？最后一步是就是上面一步化简合并同类项，得到y '想了半天，结果不应该是y'=[-sin(xy)·(y+xy ')+6x^5]/[6y^5]吗？怎么算出y '=[6x^5-ysin(xy)]/[6y^5+xsin(xy)]的？-ysin(xy)-xy 'sin(xy)+6x^5=6y^5·y'6y^5·y'+xy 'sin(xy)=6x^5-ysin(xy)[6y^5+xsin(xy)]y '=6x^5-ysin(xy)y'=[6x^5-ysin(xy)]/[6y^5+xsin(xy)] 请选为满意回答把