早教吧 育儿知识 作业答案 考试题库 百科 知识分享

用微分法求由方程1+sin(x+y)=e^-xy确定的函数y=y(x)的微分与导数.

题目详情
用微分法求由方程1+sin(x+y)=e^-xy确定的函数y=y(x)的微分与导数.
▼优质解答
答案和解析
∵1+sin(x+y)=e^(-xy),
∴[1+sin(x+y)]e^(xy)=1,
∴[1+sin(x+y)]′e^(xy)+[1+sin(x+y)][e^(xy)]′=0,
∴[0+(x+y)′cos(x+y)]e^(xy)+[1+sin(x+y)][e^(xy)](xy)′=0,
∴(1+y′)cos(x+y)+[1+sin(x+y)](x′y+xy′)=0,
∴(1+y′)cos(x+y)+[1+sin(x+y)](y+xy′)=0,
∴cos(x+y)+y′cos(x+y)+y[1+sin(x+y)]+xy′[1+sin(x+y)]=0,
∴{cos(x+y)+x[1+sin(x+y)]}y′=-cos(x+y)-y[1+sin(x+y)],
∴dy=-{cos(x+y)+y[1+sin(x+y)]}dx/{cos(x+y)+x[1+sin(x+y)]}.
∴y′=-{cos(x+y)+y[1+sin(x+y)]}/{cos(x+y)+x[1+sin(x+y)]}.