第一章1.21.2.21(2020天津期末)函数f(x)xex的导数是(D)AexB1C1xex1D1ex解析函数的导数为f (x)1ex,故选D2(2020衡水高二检测)等比数列an中,a12,a84,函数f(x)x(xa1)(xa2)(xa8),则f (0)等于(D)A26B29C215D212解析f(x)x(xa1)(xa2)(xa8)(xa1)(xa2)(xa8)x(xa1)(xa2)(xa8)(xa1)(xa2)(xa8)x,所以f(0)(0a1)(0a2)(0a8)(0a1)(0a2)(0a8)0a1a2a8.因为数列an为等比数列,所以a2a7a3a6a4a5a1a88,所以f(0)84212.3(2020河北区一模)已知函数f(x)xex,f (x)为f(x)的导函数,则f (0)_1_.解析函数f(x)xex,则f (x)exxex(1x)ex,f (0)(10)e01.故答案为1.4若函数f(x)在xc处的导数值与函数的值互为相反数,求c的值解析因为f(x),所以f(c).又因为f (x),所以f (c).依题意知f(c)f (c)0,所以0.所以2c10,得c.