e^(x+y)-xy=1,求y''(0)
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e^(x+y)-xy=1,求y''(0)
e^(x+y)-xy=1,求y''(0)
e^(x+y)-xy=1,求y''(0)
原式可得x=0,y=0 两边同时对x求导,得:e^(x+y)(1+y')-y'-xy'=0(1) 把x=0,y=0代入(1)得到y'(0)=-1 把(1)继续对x求导,可得:e^(x+y)(x+y)^2+e^(x+y)y''-y'-y'-xy''=0(2) 把x=0.y=0,y'=-1代入(2)得到y''(0)=-2
根据e^(x+y)-xy=1 得y'=[e^(x+y)-y]/[-e^(x+y)+x] so y'(0)=-1 对y'=[e^(x+y)-y]/[-e^(x+y)+x]求导 y''(0)=-1
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