首页 2021年人教版高中数学选择性必修第二册(精讲)5.2《导数的运算》(解析版)

2021年人教版高中数学选择性必修第二册(精讲)5.2《导数的运算》(解析版)

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2021年人教版高中数学选择性必修第二册(精讲)5.2《导数的运算》(解析版)5.2导数的运算思维导图常见考法考点一初等函数求导【例1】(2020·林芝市第二高级中学高二期末(文))求下列函数的导函数.(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0(4)SKIPIF1<0(5)SKIPIF1<0(6)SKIPIF1<0【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0(4)SKIPIF1<0(5)SKIPIF1<0(6)SKIPIF1<0【解析】(1)由SKIPI...

2021年人教版高中数学选择性必修第二册(精讲)5.2《导数的运算》(解析版)
5.2导数的运算思维导图常见考法考点一初等函数求导【例1】(2020·林芝市第二高级中学高二期末(文))求下列函数的导函数.(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0(4)SKIPIF1<0(5)SKIPIF1<0(6)SKIPIF1<0【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0(4)SKIPIF1<0(5)SKIPIF1<0(6)SKIPIF1<0【解析】(1)由SKIPIF1<0,则SKIPIF1<0;(2)由SKIPIF1<0,则SKIPIF1<0;(3)由SKIPIF1<0,则SKIPIF1<0;(4)由SKIPIF1<0,则SKIPIF1<0;(5)由SKIPIF1<0,则SKIPIF1<0;(6)由SKIPIF1<0,则SKIPIF1<0.【一隅三反】1.(2020·西藏高二期末(文))求下列函数的导数.(1)SKIPIF1<0;(2)SKIPIF1<0;(3)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【解析】(1)SKIPIF1<0SKIPIF1<0(2)SKIPIF1<0SKIPIF1<0(3)SKIPIF1<0SKIPIF1<02.(2020·通榆县第一中学校高二月考(理))求下列函数的导数:(Ⅰ)SKIPIF1<0;(Ⅱ)SKIPIF1<0.【答案】(Ⅰ)SKIPIF1<0;(Ⅱ)SKIPIF1<0.【解析】(Ⅰ)由导数的 计算公式 六西格玛计算公式下载结构力学静力计算公式下载重复性计算公式下载六西格玛计算公式下载年假计算公式 ,可得SKIPIF1<0.(Ⅱ)由导数的乘法法则,可得SKIPIF1<0.3.(2020·山东师范大学附中高二期中)求下列函数在指定点的导数:(1)SKIPIF1<0,SKIPIF1<0;(2)SKIPIF1<0,SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】(1)SKIPIF1<0,SKIPIF1<0(2)SKIPIF1<0,SKIPIF1<0考点二复合函数求导【例2】.(2020·凤阳县第二中学高二期末(理))求下列函数的导数:(1)SKIPIF1<0;(2)SKIPIF1<0.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0或SKIPIF1<0.【解析】(1)SKIPIF1<0;(2)SKIPIF1<0.或SKIPIF1<0.【一隅三反】1.(2020·陕西碑林·西北工业大学附属中学高二月考(理))求下列函数的导数:(1)SKIPIF1<0;(2)SKIPIF1<0;(3)SKIPIF1<0;(4)SKIPIF1<0.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0;(3)SKIPIF1<0;(4)SKIPIF1<0.【解析】(1)SKIPIF1<0;(2)SKIPIF1<0;(3)∵SKIPIF1<0∴SKIPIF1<0;(4)SKIPIF1<0.2.(2020·横峰中学高二开学考试(文))求下列各函数的导数:(1)SKIPIF1<0;(2)SKIPIF1<0(3)y=SKIPIF1<0【答案】(1)SKIPIF1<0;(2)SKIPIF1<0.(3)SKIPIF1<0【解析】(1)因为SKIPIF1<0令SKIPIF1<0,SKIPIF1<0所以SKIPIF1<0(2)SKIPIF1<0.(3)令SKIPIF1<0,则SKIPIF1<0,所以SKIPIF1<0;考点三求导数值【例3】.(2020·甘肃城关·兰州一中高二期中(理))已知函数SKIPIF1<0的导函数为SKIPIF1<0,且满足SKIPIF1<0,则SKIPIF1<0A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】A【解析】SKIPIF1<0,求导得SKIPIF1<0,则SKIPIF1<0,解得SKIPIF1<0.故选:A.【一隅三反】1.(2020·广东湛江·高二期末(文))已知函数SKIPIF1<0,则SKIPIF1<0()A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】A【解析】SKIPIF1<0,SKIPIF1<0,因此,SKIPIF1<0.故选:A.2.(2020·四川高二期中(理))若函数SKIPIF1<0,则SKIPIF1<0的值为()A.0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】B【解析】因为SKIPIF1<0,所以令SKIPIF1<0,则SKIPIF1<0,所以SKIPIF1<0,则SKIPIF1<0,故选:B.3.(2020·广西桂林·高二期末(文))已知函数SKIPIF1<0,则SKIPIF1<0()A.3B.0C.2D.1【答案】A【解析】由题得SKIPIF1<0.故选:A考点四求切线方程【例4】.(2020·郸城县实验高中高二月考(理))已知曲线SKIPIF1<0(1)求曲线在点SKIPIF1<0处的切线方程;(2)求曲线过点SKIPIF1<0的切线方程【答案】(1)SKIPIF1<0;(2)SKIPIF1<0或SKIPIF1<0.【解析】(1)∵SKIPIF1<0,∴在点SKIPIF1<0处的切线的斜率SKIPIF1<0,∴曲线在点SKIPIF1<0处的切线方程为SKIPIF1<0,即SKIPIF1<0.(2)设曲线SKIPIF1<0与过点SKIPIF1<0的切线相切于点SKIPIF1<0,则切线的斜率SKIPIF1<0,∴切线方程为SKIPIF1<0,即SKIPIF1<0.∵点SKIPIF1<0在该切线上,∴SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0.故所求切线方程为SKIPIF1<0或SKIPIF1<0.【一隅三反】1.(2020·黑龙江大庆实验中学高三月考(文))曲线SKIPIF1<0在点SKIPIF1<0处的切线方程为A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】A【解析】SKIPIF1<0的导数为SKIPIF1<0,可得曲线SKIPIF1<0在点SKIPIF1<0处的切线斜率为SKIPIF1<0,所以曲线SKIPIF1<0在点SKIPIF1<0处的切线方程为SKIPIF1<0,即SKIPIF1<0,故选A.2.(2020·河南高三其他(理))曲线SKIPIF1<0在某点处的切线的斜率为SKIPIF1<0,则该切线的方程为()A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】D【解析】求导得SKIPIF1<0,根据题意得SKIPIF1<0,解得SKIPIF1<0(舍去)或SKIPIF1<0,可得切点的坐标为SKIPIF1<0,所以该切线的方程为SKIPIF1<0,整理得SKIPIF1<0.故选:D.3.(2020·北京高二期末)过点P(0,2)作曲线y=SKIPIF1<0的切线,则切点坐标为()A.(1,1)B.(2,SKIPIF1<0)C.(3,SKIPIF1<0)D.(0,1)【答案】A【解析】设切点SKIPIF1<0,SKIPIF1<0SKIPIF1<0,即切点SKIPIF1<0故选:A4.(2020·吉林洮北·白城一中高二月考(理))已知函数f(x)=x3-4x2+5x-4.(1)求曲线f(x)在点(2,f(2))处的切线方程;(2)求经过点A(2,-2)的曲线f(x)的切线方程.【答案】(1)x-y-4=0(2)x-y-4=0或y+2=0【解析】(1)∵f′(x)=3x2-8x+5,∴f′(2)=1,又f(2)=-2,∴曲线f(x)在点(2,f(2))处的切线方程为y-(-2)=x-2,即x-y-4=0.(2)设切点坐标为(x0,x03-4x02+5x0-4),∵f′(x0)=3x02-8x0+5,∴切线方程为y-(-2)=(3x02-8x0+5)(x-2),又切线过点(x0,x03-4x02+5x0-4),∴x03-4x02+5x0-2=(3x02-8x0+5)(x0-2),整理得(x0-2)2(x0-1)=0,解得x0=2或x0=1,∴经过A(2,-2)的曲线f(x)的切线方程为x-y-4=0或y+2=0.考点五利用切线求参数【例5】.(2020·全国高三其他(理))已知曲线SKIPIF1<0在点SKIPIF1<0处的切线方程为SKIPIF1<0,则SKIPIF1<0()A.SKIPIF1<0B.0C.1D.SKIPIF1<0【答案】D【解析】令SKIPIF1<0,则SKIPIF1<0,所以SKIPIF1<0,因为曲线SKIPIF1<0在点SKIPIF1<0处的切线方程为SKIPIF1<0,所以该切线过原点,所以SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0.故选:D.【一隅三反】1.(2020·岳麓·湖南师大附中月考)已知函数SKIPIF1<0,若曲线SKIPIF1<0在SKIPIF1<0处的切线与直线SKIPIF1<0平行,则SKIPIF1<0______.【答案】SKIPIF1<0【解析】因为函数SKIPIF1<0,所以SKIPIF1<0,又因为曲线SKIPIF1<0在SKIPIF1<0处的切线与直线SKIPIF1<0平行,所以SKIPIF1<0,解得SKIPIF1<0,故答案为:SKIPIF1<02.(2020·安徽庐阳·合肥一中高三月考(文))曲线SKIPIF1<0在点(0,1)处的切线的斜率为2,则a=_____.【答案】1【解析】SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0.故答案为:1.3.(2020·山东莱州一中高二月考)已知直线SKIPIF1<0是曲线SKIPIF1<0的一条切线,则SKIPIF1<0________.【答案】4【解析】设SKIPIF1<0,切点为SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,故切点为SKIPIF1<0,又切点在切线SKIPIF1<0上,故SKIPIF1<0.故答案为:4
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