y=tan(x-y)两边求导
y' = (1-y')/cos²(x-y)
y' = 1/[1+cos²(x-y)]
再次求导
y" = -2cos(x-y)[-sin(x-y)](1-y')/[1+cos²(x-y)]²
= sin(2x-2y)(1-y')/[1+cos²(x-y)]²
= sin(2x-2y){1 - 1/[1+cos²(x-y)]}/[1+cos²(x-y)]²
= sin(2x-2y)cos²(x-y)/[1+cos²(x-y)]³
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