y = cos³ u(x)u(x) = (1-lnx)/x y' = -3cos² u(x) sin u(x) u'(x)u' = [-1-(1-lnx)]/x² = (lnx -2)/x²y' = 3(2 - lnx)cos² u(x) sinu(x) /x² = 3[(2 - lnx)/ x²]cos² [(1-lnx)/x] sin [(1-lnx)/x] 此结果与答案完全一致。
