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最佳答案:两边同时对x求导得cos(xy)(y+xy')=1解出y'即得dy/dx=1/xcos(xy)-y/x
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最佳答案:y+sin(xy)=1两边对x求导得:y'+cos(xy)*(xy)'=0即y'+cos(xy)*(y+xy')=0所以y'=-ycos(xy)/(1+xcos
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最佳答案:两边对x求导:y'=cos(X+y)×(1+y')=cos(x+y)+y'cos(x+y){1-cos(x+y)}y'=cos(x+y)y'=cos(x+y)/
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最佳答案:两边同时对x求导得到e^x-e^y*y'+cos(x+y)(1+y')=0解得y=[e^x+cos(x+y)]/[e^y-cos(x+y)]如有其它问题请采纳此
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最佳答案:整理得xcosy-sin(x+y)=0得dy/dx=-Fx/Fy=-[cosy-cos(x+y)]/[-xsiny-cos(x+y)]=[cosy-cos(x+
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最佳答案:前两个错了.第一个人错在:xy对x求导是(y+x×dy/dx).第二个人是白痴不解释.两边对x求导:e^xy(y+xy') y'lnx y/x=0得:y'=(-
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最佳答案:两边对x求导:y'cosx-ysinx-(1+y')cos(x+y)=0y'=[ysinx+cos(x+y)]/[cosx-cos(x+y)]因此dy=[ysi
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最佳答案:y'=cos(x+y)(1+y')y'=cos(x+y)/(1-cos(x+y))
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最佳答案:将原方程两边微分得d[xe^y+sin(xy)]=0→e^ydx+xe^ydy+cos(xy)(ydx+xdy)=0→移项[xe^y+xcos(xy)]dy=-
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最佳答案:sin(xy)+In(y-x)=x两边同时对x求导得cos(xy)·(xy) '+1/(y-x)·(y-x) '=1cos(xy)·(y+xy ')+1/(y-