设e^(x+y)+cos(xy)=0确定y是x的函数求dy

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f(x,y)=e^(x+y)+cos(xy)=0 //: 利用隐函数存在定理：
f 'x(x,y)=e^(x+y) - ysin(xy)
f 'y(x,y)=e^(x+y) - xsin(xy)
dy/dx = - f 'x/f 'y = -[e^(x+y) - ysin(xy)]/[e^(x+y) - xsin(xy)]
dy = -[e^(x+y) - ysin(xy)]/[e^(x+y) - xsin(xy)]dx
2. (1+y')e^(x+y) - (y+xy')sin(xy)=0 //: 直接对x求导：
e^(x+y)+y'e^(x+y)-ysin(xy)-xy'sin(xy)=0
y' [e^(x+y)-xsin(xy)]=-e^(x+y)+ysin(xy)
y'(x) = -[e^(x+y) - ysin(xy)]/[e^(x+y) - xsin(xy)]
dy = -[e^(x+y) - ysin(xy)]/[e^(x+y) - xsin(xy)]dx
3. 两种方法结果相同.

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