一阶导求完之后适当化简可以使二阶导的难度降低。
y=cot(x+y)=cos(x+y)/sin(x+y)
y'=[-sin(x+y)·(1+y')·sin(x+y)-cos(x+y)·(1+y')·cos(x+y)]/sin²(x+y)
=-(1+y')/sin²(x+y)
y'[sin²(x+y)+1]=-1→y'=-1/[sin²(x+y)+1]
y''=[sin²(x+y)+1]'/[sin²(x+y)+1]²
=[sin2(x+y)·(1+y')]/[sin²(x+y)+1]²
=[sin2(x+y)·sin²(x+y)/[sin²(x+y)+1])]/[sin²(x+y)+1]²
=[sin2(x+y)·sin²(x+y)]/[sin²(x+y)+1]³
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