y = arccos(1/x)y' = -1/√[1-(1/x²)] * (-1/x²)= 1/√[(x²-1)/x²] * (1/x²)= x/√(x²-1) * (1/x²)= 1/[x√(x²-1)]y = arccos(1/x)= arcsecxy' = 1/[x√(x²-1)],这是...