1.3.1 函数的单调性与导数（学案）-【成才之路】2020-2021学年高中新课程数学同步学习指导（人教A版选修2-2）

数学 （选修 2 - 2·人教 A 版） ∵ y′ = 2,∴ 1 1 + cosx = 2, ∴ cosx = - 1 2 , 又 - π < x < π,∴ x = ± 2π 3 . 故应选 D. 3. B　 由已知得:3k + 2 = 1,∴ k = - 1 3 ,又 g(x) = xf(x),f ′(3) = - 1 3 , ∴ g′ ( x) = f(x) + x f ′ ( x), ∴ g′ (3 ) = f(3) + 3 f ′ ( 3 ) = 1 + 3 × - 1 3( ) = 0. 4. 3x2 + 6x + 6　 函数的导数为 y′ = 3x2 + 6x + 6. 互动探究·攻重难 　 　 典例试做 1:(1)3　 ∵ f ′(x) = 2ex + (2x + 1)ex = (2x + 3)ex ∴ f ′(0) = 3. (2)①y′ = x′·ex + x·(ex )′ = ex + xex = (1 + x)ex. ②y′ = ( 2x x2 + 1 )′ = (2x)′(x 2 + 1) - 2x(x2 + 1)′ (x2 + 1)2 = 2(x 2 + 1) - 4x2 (x2 + 1)2 = 2 - 2x 2 (x2 + 1)2 . ③y′ = (xsinx)′ - ( 2 cosx )′ = sinx + xcosx - 2sinx cos2 x . ④y = cos2 x 2 = 1 + cosx 2 = 1 2 + 1 2 cosx, ∴ y′ = 1 2 ( - sinx) = - 1 2 sinx. 　 　 跟踪练习 1:(1)y′ = (x·tanx)′ = xsinxcosx( )′ = (xsinx)′cosx - xsinx(cosx)′ cos2 x = (sinx + xcosx)cosx + xsin 2 x cos2 x = sinxcosx + x cos2 x . (2)解法 1:y′ = [(x + 1)(x + 2)(x + 3)]′ = [(x + 1)(x + 2)]′(x + 3) + (x + 1)(x + 2)(x + 3)′ = [(x + 1)′(x + 2) + (x + 1)(x + 2)′] ( x + 3) + ( x + 1) ( x + 2) = (x + 2 + x + 1)(x + 3) + (x + 1)(x + 2) = (2x + 3)(x + 3) + x2 + 3x + 2 = 3x2 + 12x + 11; 解法 2:∵ (x + 1)(x + 2)(x + 3) = (x2 + 3x + 2)(x + 3) = x3 + 6x2 + 11x + 6, ∴ y′ = [(x + 1)(x + 2)(x + 3)]′ = (x3 + 6x2 + 11x + 6)′ = 3x2 + 12x + 11; (3)解法 1:y′ = x - 1x + 1( )′ = (x - 1)′(x + 1) - (x - 1)(x + 1)′ (x + 1)2 = x + 1 - (x - 1) (x + 1)2 = 2 (x + 1)2 ; 解法 2:∵ y = x - 1 x + 1 = x + 1 - 2 x + 1 = 1 - 2 x + 1 , ∴ y′ = 1 - 2x + 1( )′ = - 2 x + 1( )′ = 2 (x + 1)2 . 　 　 典例试做 2:(1)因为 y = xln x = xlnx 1 2 = 1 2 xlnx, 所以 y′ = ( 1 2 xlnx)′ = 1 2 (x)′lnx + 1 2 x(lnx)′ = 1 2 lnx + 1 2 ; (2)因为 y = x 3 - x5 + x7 x = x - x2 + x3 , 所以 y′ = (x - x2 + x3 )′ = 1 - 2x + 3x2 ; (3)因为 y = cos2x sinx - cosx = cos 2 x - sin2 x sinx - cosx = - sinx - cosx, 所以 y′ = ( - sinx - cosx)′ = sinx - cosx. 　 　 跟踪练习 2:(1)因为 y = 1 4 sin2 x 2 = 1 8 (1 - cosx) = 1 8 - 1 8 cosx, 所以 y′ = 1 8 sinx. (2)因为 y = ln2x = lnx·lnx,所以 y′ = (lnx·lnx)′ = 1 x ·lnx + lnx· 1 x =2lnx x . 　 　 典例试做 3:(1)设 y = u2 ,u = 4 - 3x,则 yu′ = 2u,ux′ = - 3,于是 yx′ = yu′