[-2xsin(2-x^2)]／[cos(2-x^2)]^2有过程吗希望我的解答没错，能给你带来帮助：=﹛-[cos(2-x^2)]^-2﹜*[-sin(2-x^2)]*[-2x]=[-2xsin(2-x^2)]／[cos(2-x^2)]^2记不得这叫换元啥的了……抱歉