《(全國(guó)通用版)2018-2019版高中數(shù)學(xué) 第一章 導(dǎo)數(shù)及其應(yīng)用 1.2 導(dǎo)數(shù)的計(jì)算 第2課時(shí) 導(dǎo)數(shù)的運(yùn)算法則學(xué)案 新人教A版選修2-2》由會(huì)員分享,可在線閱讀,更多相關(guān)《(全國(guó)通用版)2018-2019版高中數(shù)學(xué) 第一章 導(dǎo)數(shù)及其應(yīng)用 1.2 導(dǎo)數(shù)的計(jì)算 第2課時(shí) 導(dǎo)數(shù)的運(yùn)算法則學(xué)案 新人教A版選修2-2(15頁(yè)珍藏版)》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、第2課時(shí)導(dǎo)數(shù)的運(yùn)算法則學(xué)習(xí)目標(biāo)1.理解函數(shù)的和、差、積、商的求導(dǎo)法則.2.理解求導(dǎo)法則的證明過(guò)程,能夠綜合運(yùn)用導(dǎo)數(shù)公式和導(dǎo)數(shù)運(yùn)算法則求函數(shù)的導(dǎo)數(shù)知識(shí)點(diǎn)一和、差的導(dǎo)數(shù)已知f(x)x,g(x).Q(x)f(x)g(x),H(x)f(x)g(x)思考1f(x),g(x)的導(dǎo)數(shù)分別是什么?答案f(x)1,g(x).思考2試求yQ(x),yH(x)的導(dǎo)數(shù)并觀察Q(x),H(x)與f(x),g(x)的關(guān)系答案y(xx)x,1.Q(x) 1.同理,H(x)1.Q(x)的導(dǎo)數(shù)等于f(x),g(x)的導(dǎo)數(shù)的和H(x)的導(dǎo)數(shù)等于f(x),g(x)的導(dǎo)數(shù)的差梳理和、差的導(dǎo)數(shù)f(x)g(x)f(x)g(x)知識(shí)點(diǎn)二積
2、、商的導(dǎo)數(shù)(1)積的導(dǎo)數(shù)f(x)g(x)f(x)g(x)f(x)g(x)cf(x)cf(x)(2)商的導(dǎo)數(shù)(g(x)0)(3)注意f(x)g(x)f(x)g(x),.1若f(x)2x,則f(x)x2.()2函數(shù)f(x)xex的導(dǎo)數(shù)是f(x)ex(x1)()3當(dāng)g(x)0時(shí),.()類型一利用導(dǎo)數(shù)的運(yùn)算法則求導(dǎo)例1求下列函數(shù)的導(dǎo)數(shù)(1)y3x2xcos x;(2)ylg x;(3)y(x23)(exln x);(4)yx2tan x;(5)y.考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)的運(yùn)算法則解(1)y6xcos xx(cos x)6xcos xxsin x.(2)y(lg x)(x2).(3)y(x23)(e
3、xln x)(x23)(exln x)2x(exln x)(x23)ex(x22x3)2xln xx.(4)因?yàn)閥x2,所以y(x2)2x2x.(5)y.反思與感悟(1)先區(qū)分函數(shù)的運(yùn)算特點(diǎn),即函數(shù)的和、差、積、商,再根據(jù)導(dǎo)數(shù)的運(yùn)算法則求導(dǎo)數(shù)(2)對(duì)于三個(gè)以上函數(shù)的積、商的導(dǎo)數(shù),依次轉(zhuǎn)化為“兩個(gè)”函數(shù)的積、商的導(dǎo)數(shù)計(jì)算跟蹤訓(xùn)練1求下列函數(shù)的導(dǎo)數(shù)(1)y;(2)y;(3)y(x1)(x3)(x5)考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)的運(yùn)算法則解(1)y23x1,y3x2.(2)方法一y.方法二y1,y.(3)方法一y(x1)(x3)(x5)(x1)(x3)(x5)(x1)(x3)(x1)(x3)(x5)(
4、x1)(x3)(2x4)(x5)(x1)(x3)3x218x23.方法二y(x1)(x3)(x5)(x24x3)(x5)x39x223x15,y(x39x223x15)3x218x23.類型二導(dǎo)數(shù)公式及運(yùn)算法則的綜合應(yīng)用例2(1)已知函數(shù)f(x)2xf(1),試比較f(e)與f(1)的大小關(guān)系;(2)設(shè)f(x)(axb)sin x(cxd)cos x,試確定常數(shù)a,b,c,d,使得f(x)xcos x.考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用解(1)由題意得f(x)2f(1),令x1,得f(1)2f(1),即f(1)1.f(x)2x.f(e)2e2e,f(1)2,由f(e)f(1)2e20
5、,得f(e)0)在xx0處的導(dǎo)數(shù)為0,那么x0等于()Aa BaCa Da2考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)的運(yùn)算法則答案B解析y,由xa20,得x0a.3若函數(shù)f(x)exsin x,則此函數(shù)圖象在點(diǎn)(4,f(4)處的切線的傾斜角為()A. B0 C鈍角 D銳角考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案C解析f(x)exsin xexcos x,f(4)e4(sin 4cos 4)4,sin 40,cos 40,f(4)0的解集為()A(0,) B(1,0)(2,)C(2,) D(1,0)考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)的運(yùn)算法則答案C解析f(x)x22x4ln x,f(x)2x20,整理得0,解
6、得1x2.又x0,x2.5函數(shù)f(x)xcos xsin x的導(dǎo)函數(shù)是()A奇函數(shù)B偶函數(shù)C既是奇函數(shù)又是偶函數(shù)D既不是奇函數(shù),又不是偶函數(shù)考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案B解析f(x)(xcos x)(sin x)cos xxsin xcos xxsin x.令F(x)xsin x,xR,則F(x)xsin(x)xsin xF(x),f(x)是偶函數(shù)6設(shè)曲線y在點(diǎn)(3,2)處的切線與直線axy10垂直,則a等于()A2 B. C D2考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案D解析y1,y,.a1,即a2.7在下面的四個(gè)圖象中,其中一個(gè)圖象是函數(shù)f(x)x3ax2(a2
7、1)x1(aR)的導(dǎo)函數(shù)yf(x)的圖象,則f(1)等于()A. BC. D或考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案B解析f(x)x22ax(a21),導(dǎo)函數(shù)f(x)的圖象開(kāi)口向上,故其圖象必為.由圖象特征知f(0)0,且對(duì)稱軸a0,a1,則f(1)11,故選B.二、填空題8設(shè)f(5)5,f(5)3,g(5)4,g(5)1,若h(x),則h(5)_.考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)的運(yùn)算法則答案解析由題意知f(5)5,f(5)3,g(5)4,g(5)1,h(x),h(5).9已知某運(yùn)動(dòng)著的物體的運(yùn)動(dòng)方程為s(t)2t2(位移單位:m,時(shí)間單位:s),則t1 s時(shí)物體的瞬時(shí)速度為_(kāi) m/s.
8、考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案5解析因?yàn)閟(t)2t22t22t2,所以s(t)24t,所以s(1)1245,即物體在t1 s時(shí)的瞬時(shí)速度為5 m/s.10已知函數(shù)f(x)fcos xsin x,則f的值為_(kāi)考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案1解析f(x)fsin xcos x,ff,得f1.f(x)(1)cos xsin x,f1.11已知函數(shù)f(x)xln x,若直線l過(guò)點(diǎn)(0,1),并且與曲線yf(x)相切,則直線l的方程為_(kāi)考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案xy10解析點(diǎn)(0,1)不在曲線f(x)xln x上,設(shè)切點(diǎn)坐標(biāo)為(x0,y0)又f
9、(x)1ln x,解得x01,y00.切點(diǎn)坐標(biāo)為(1,0),f(1)1ln 11.直線l的方程為yx1,即xy10.12已知曲線yxln x在點(diǎn)(1,1)處的切線與曲線yax2(a2)x1相切,則a_.考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案8解析由yxln x,得y1,得曲線在點(diǎn)(1,1)處的切線的斜率為k2,所以切線方程為y12(x1),即y2x1.此切線與曲線yax2(a2)x1相切,消去y,得ax2ax20,所以a0且a28a0,解得a8.三、解答題13偶函數(shù)f(x)ax4bx3cx2dxe的圖象過(guò)點(diǎn)P(0,1),且在x1處的切線方程為yx2,求f(x)的解析式考點(diǎn)導(dǎo)數(shù)的運(yùn)算法
10、則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用解f(x)的圖象過(guò)點(diǎn)P(0,1),e1.又f(x)為偶函數(shù),f(x)f(x)故ax4bx3cx2dxeax4bx3cx2dxe.b0,d0.f(x)ax4cx21.函數(shù)f(x)在x1處的切線方程為yx2,切點(diǎn)坐標(biāo)為(1,1)ac11.f(x)|x14a2c,4a2c1.a,c.函數(shù)f(x)的解析式為f(x)x4x21.四、探究與拓展14在等比數(shù)列an中,a12,a84,函數(shù)f(x)x(xa1)(xa2)(xa8),則f(0)等于()A26 B29C215 D212考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用答案D解析f(x)x(xa1)(xa2)(xa8)x(xa1
11、)(xa2)(xa8)x(xa1)(xa2)(xa8)(xa1)(xa2)(xa8)x(xa2)(xa8)x(xa1)(xa2)(xa7),f(0)a1a2a8(a1a8)484212.15設(shè)函數(shù)f(x)ax,曲線yf(x)在點(diǎn)(2,f(2)處的切線方程為7x4y120.(1)求f(x)的解析式;(2)證明:曲線yf(x)上任一點(diǎn)處的切線與直線x0和直線yx所圍成的三角形的面積為定值,并求此定值考點(diǎn)導(dǎo)數(shù)的運(yùn)算法則題點(diǎn)導(dǎo)數(shù)運(yùn)算法則的綜合應(yīng)用解(1)由7x4y120,得yx3.當(dāng)x2時(shí),y,f(2),又f(x)a,f(2),由得解得故f(x)x.(2)設(shè)P(x0,y0)為曲線上任一點(diǎn),由y1知,曲線在點(diǎn)P(x0,y0)處的切線方程為yy0(xx0),即y(xx0)令x0,得y,從而得切線與直線x0的交點(diǎn)坐標(biāo)為.令yx,得yx2x0,從而切線與直線yx的交點(diǎn)坐標(biāo)為(2x0,2x0)所以點(diǎn)P(x0,y0)處的切線與直線x0,yx所圍成的三角形面積為|2x0|6.故曲線yf(x)上任一點(diǎn)處的切線與直線x0,yx所圍成的三角形的面積為定值,此定值為6. 15