xy-sin(x-y)=2的二阶导数？
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xy-sin(x-y) = 2 , 两边对 x 求导， 得
y + xy' - (1-y')cos(x-y) = 0, (1)
y' = [cos(x-y)-y]/[x+cos(x-y)]
式（1）两边对 x 求导， 得
2y' + xy'' + y''cos(x-y) + (1-y')^2 sin(x-y) = 0,
y'' = -[2y'+(1-y')^2 sin(x-y)]/[x+cos(x-y)]
= -【{2[cos(x-y)-y]/[x+cos(x-y)]+{1-[cos(x-y)-y]/[x+cos(x-y)]}^2 sin(x-y)】/[x+cos(x-y)]
= -{2[cos(x-y)-y][x+cos(x-y)]+(x+y)^2 sin(x-y)}/[x+cos(x-y)]^3
y + xy' - (1-y')cos(x-y) = 0, (1)
y' = [cos(x-y)-y]/[x+cos(x-y)]
式（1）两边对 x 求导， 得
2y' + xy'' + y''cos(x-y) + (1-y')^2 sin(x-y) = 0,
y'' = -[2y'+(1-y')^2 sin(x-y)]/[x+cos(x-y)]
= -【{2[cos(x-y)-y]/[x+cos(x-y)]+{1-[cos(x-y)-y]/[x+cos(x-y)]}^2 sin(x-y)】/[x+cos(x-y)]
= -{2[cos(x-y)-y][x+cos(x-y)]+(x+y)^2 sin(x-y)}/[x+cos(x-y)]^3
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f”（x）=sin（x-y）
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