新高考數(shù)學(xué)一輪復(fù)習(xí)考點(diǎn)練習(xí)考點(diǎn)43 導(dǎo)數(shù)及幾何意義、導(dǎo)數(shù)的運(yùn)算 (含解析)

考點(diǎn)43導(dǎo)數(shù)及幾何意義、導(dǎo)數(shù)的運(yùn)算考向一導(dǎo)數(shù)的概念及幾何意義1.導(dǎo)數(shù)的概念(1)在點(diǎn)x0處的導(dǎo)數(shù)limΔx→0ΔyΔx=limΔx→0f(x0+Δx)-f(x0)(2)區(qū)間(a,b)上的導(dǎo)數(shù)

當(dāng)x∈(a,b)時(shí),f'(x)=limΔx→0ΔyΔ2.導(dǎo)數(shù)的幾何意義函數(shù)y=f(x)在點(diǎn)x=x0處的導(dǎo)數(shù)f'(x0)就是函數(shù)圖像在該點(diǎn)處切線的斜率.曲線y=f(x)在點(diǎn)(x0,f(x0))處的切線方程是y-f(x0)=f'(x0)(x-x0).1.【2020全國(guó)高三課時(shí)練習(xí)（理）】函數(shù)SKIPIF1<0的圖像在點(diǎn)SKIPIF1<0處的切線方程為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因此，所求切線的方程為SKIPIF1<0，即SKIPIF1<0.故選：B.2.【2020山東高三其他】已知函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線經(jīng)過坐標(biāo)原點(diǎn)，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0，SKIPIF1<0，切點(diǎn)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以，函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，由于該直線過原點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0，故選A.考向二導(dǎo)數(shù)的運(yùn)算1.常用導(dǎo)數(shù)公式（1）C'=0(C為常數(shù))（2）(xn)'=nxn-1(n∈Z)

（3）(sinx)'=cosx(cosx)'=-sinx（4）(ax)'=axlna(a>0,且a≠1)

（5）(logax)'=1xlna(（6）(ex)'=ex（7）(lnx)'=1x,(ln|x|)'=2.導(dǎo)數(shù)的運(yùn)算法則[f(x)±g(x)]'=f'(x)±g'(x)[f(x)·g(x)]'=f'(x)·g(x)+f(x)·g'(x)[f(x復(fù)合函數(shù)y=f[g(x)]的導(dǎo)數(shù)與函數(shù)y=f(u),u=g(x)的導(dǎo)數(shù)之間具有關(guān)系y'x=y'u·u'x這個(gè)關(guān)系用語言表達(dá)就是“y對(duì)x的導(dǎo)數(shù)等于y對(duì)u的導(dǎo)數(shù)與u對(duì)x的導(dǎo)數(shù)的乘積”

1.【2020全國(guó)高三課時(shí)練習(xí)（理）】已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且滿足SKIPIF1<0，則SKIPIF1<0等于（）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，故選B.2.【2020河南高三其他（理）】已知函數(shù)SKIPIF1<0，則SKIPIF1<0______.【答案】1【解析】由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.故答案為：1.題組一（真題在線）1.【2020年高考全國(guó)Ⅰ卷理數(shù)】函數(shù)SKIPIF1<0的圖像在點(diǎn)SKIPIF1<0處的切線方程為A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02.【2020年高考全國(guó)III卷理數(shù)】若直線l與曲線y=SKIPIF1<0和x2+y2=SKIPIF1<0都相切，則l的方程為A．y=2x+1 B．y=2x+SKIPIF1<0 C．y=SKIPIF1<0x+1 D．y=SKIPIF1<0x+SKIPIF1<03.【2020年高考北京】為滿足人民對(duì)美好生活的向往，環(huán)保部門要求相關(guān)企業(yè)加強(qiáng)污水治理，排放未達(dá)標(biāo)的企業(yè)要限期整改、設(shè)企業(yè)的污水摔放量W與時(shí)間t的關(guān)系為SKIPIF1<0，用SKIPIF1<0的大小評(píng)價(jià)在SKIPIF1<0這段時(shí)間內(nèi)企業(yè)污水治理能力的強(qiáng)弱，已知整改期內(nèi)，甲、乙兩企業(yè)的污水排放量與時(shí)間的關(guān)系如下圖所示.給出下列四個(gè)結(jié)論：①在SKIPIF1<0這段時(shí)間內(nèi)，甲企業(yè)的污水治理能力比乙企業(yè)強(qiáng)；②在SKIPIF1<0時(shí)刻，甲企業(yè)的污水治理能力比乙企業(yè)強(qiáng)；③在SKIPIF1<0時(shí)刻，甲、乙兩企業(yè)的污水排放都已達(dá)標(biāo)；④甲企業(yè)在SKIPIF1<0這三段時(shí)間中，在SKIPIF1<0的污水治理能力最強(qiáng)．其中所有正確結(jié)論的序號(hào)是____________________．4.【2019年高考全國(guó)Ⅲ卷理數(shù)】已知曲線SKIPIF1<0在點(diǎn)（1，ae）處的切線方程為y=2x+b，則A．SKIPIF1<0 B．a(chǎn)=e，b=1C．SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<05.【2019年高考全國(guó)Ⅰ卷理數(shù)】曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為___________．6.【2020年高考北京】已知函數(shù)SKIPIF1<0．（Ⅰ）求曲線SKIPIF1<0的斜率等于SKIPIF1<0的切線方程；（Ⅱ）設(shè)曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與坐標(biāo)軸圍成的三角形的面積為SKIPIF1<0，求SKIPIF1<0的最小值．題組二1.【2020陜西西安高三二模（理）】已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，則（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02.【2020全國(guó)高三課時(shí)練習(xí)（理）】若曲線SKIPIF1<0與曲線SKIPIF1<0在交點(diǎn)SKIPIF1<0處有公切線，則SKIPIF1<0（）A．SKIPIF1<0 B．0C．2 D．13.【2020邢臺(tái)市第二中學(xué)高二期末】已知函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04.【2020陜西西安高三三?！亢瘮?shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的傾斜角為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05.【2020山東師范大學(xué)附中高二月考】已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且滿足SKIPIF1<0，則SKIPIF1<0等于（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．16.【2020全國(guó)高三課時(shí)練習(xí)（理）】函數(shù)SKIPIF1<0在點(diǎn)(0，f(0))處的切線方程是___________.7.【2020福建高三其他（理）】設(shè)曲線SKIPIF1<0在SKIPIF1<0處的切線與直線SKIPIF1<0平行，則實(shí)數(shù)a的值為_______.8.【2020山東萊陽一中高三月考】已知SKIPIF1<0，則SKIPIF1<0_____．9.【2020河南開封高三二模（理）】已知函數(shù)SKIPIF1<0．則函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為___________．10.【2019山東聊城】如圖，SKIPIF1<0是可導(dǎo)函數(shù),直線l是曲線SKIPIF1<0在SKIPIF1<0處的切線，令SKIPIF1<0，則SKIPIF1<0___________.題組一1.B【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因此，所求切線的方程為SKIPIF1<0，即SKIPIF1<0.故選：B．2.D【解析】設(shè)直線SKIPIF1<0在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，函數(shù)SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，則直線SKIPIF1<0的斜率SKIPIF1<0，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，由于直線SKIPIF1<0與圓SKIPIF1<0相切，則SKIPIF1<0，兩邊平方并整理得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0（舍），則直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.故選：D．3.①②③【解析】SKIPIF1<0表示區(qū)間端點(diǎn)連線斜率的負(fù)數(shù)，在SKIPIF1<0這段時(shí)間內(nèi)，甲的斜率比乙的小，所以甲的斜率的相反數(shù)比乙的大，因此甲企業(yè)的污水治理能力比乙企業(yè)強(qiáng)；①正確；甲企業(yè)在SKIPIF1<0這三段時(shí)間中，甲企業(yè)在SKIPIF1<0這段時(shí)間內(nèi)，甲的斜率最小，其相反數(shù)最大，即在SKIPIF1<0的污水治理能力最強(qiáng)．④錯(cuò)誤；在SKIPIF1<0時(shí)刻，甲切線的斜率比乙的小，所以甲切線的斜率的相反數(shù)比乙的大，甲企業(yè)的污水治理能力比乙企業(yè)強(qiáng)；②正確；在SKIPIF1<0時(shí)刻，甲、乙兩企業(yè)的污水排放量都在污水打標(biāo)排放量以下，所以都已達(dá)標(biāo)；③正確；故答案為：①②③4.D【解析】∵SKIPIF1<0∴切線的斜率SKIPIF1<0，SKIPIF1<0，將SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0.故選D．5.SKIPIF1<0【解析】SKIPIF1<0所以切線的斜率SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0．6.見解析【解析】（Ⅰ）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以切點(diǎn)為SKIPIF1<0，由點(diǎn)斜式可得切線方程：SKIPIF1<0，即SKIPIF1<0.（Ⅱ）顯然SKIPIF1<0，因?yàn)镾KIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為：SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，不妨設(shè)SKIPIF1<0SKIPIF1<0時(shí)，結(jié)果一樣SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，所以SKIPIF1<0時(shí)，SKIPIF1<0取得極小值，也是最小值為SKIPIF1<0.題組二1.D【解析】SKIPIF1<0SKIPIF1<0，SKIPIF1<0將SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0，故選D．2.D【解析】由曲線SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，由曲線SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，因?yàn)榍€SKIPIF1<0與曲線SKIPIF1<0在交點(diǎn)SKIPIF1<0出有公切線，所以SKIPIF1<0，解得SKIPIF1<0，又由SKIPIF1<0，即交點(diǎn)為SKIPIF1<0，將SKIPIF1<0代入曲線SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，故選D．3.D【解析】切點(diǎn)SKIPIF1<0在切線SKIPIF1<0上，∴SKIPIF1<0，得SKIPIF1<0，又切線斜率SKIPIF1<0，∴SKIPIF1<0.故選：D4.B【解析】SKIPIF1<0，則SKIPIF1<0，則傾斜角為SKIPIF1<0．故選：B．5.B【解析】SKIPIF1<0,SKIPIF1<0SKIPIF1<0,令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故選：B6.SKIP