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高中数学选修2-2导数--导数的运算(解析版)

高中数学选修2-2导数--导数的运算(解析版)1.若f (x )=sin π3-cos x ,则f ′(α)等于( )A .Sin αB .Cos αC .sin π3+cos αD .cos π3+sin α[答案] A[解析] ∵f (x )=sin π3-cos x ,∴f ′(x )=sin x ,∴f ′(α)=sin α,故选A.2.设函数f (x )=x m +ax 的导数为f ′(x )=2x +1,则数列{1f (n )}(n ∈N *)的前n 项和是( )A.n n +1B .n +2n +1C.nn -1 D .n +1n[答案] A[解析] ∵f (x )=x m +ax 的导数为f ′(x )=2x +1,∴m =2,a =1,∴f (x )=x 2+x , ∴f (n )=n 2+n =n (n +1),∴数列{1f (n )}(n ∈N *)的前n 项和为:S n =11×2+12×3+13×4+…+1n (n +1)=⎝⎛⎭⎫1-12+⎝⎛⎭⎫12-13+…+⎝⎛⎭⎫1n -1n +1 =1-1n +1=nn +1,故选A.3.已知二次函数f (x )的图象如图所示,则其导函数f ′(x )的图象大致形状是( )[答案] B [解析] 依题意可设f (x )=ax 2+c (a <0,且c >0),于是f ′(x )=2ax ,显然f ′(x )的图象为直线,过原点,且斜率2a <0,故选B.4.已知函数f (x )的导函数为f ′(x ),且满足f (x )=2xf ′(e)+ln x ,则f ′(e)=( )A .e -1B .-1C .-e -1D .-e[答案] C [解析] ∵f (x )=2xf ′(e)+ln x ,∴f ′(x )=2f ′(e)+1x ,∴f ′(e)=2f ′(e)+1e ,解得f ′(e)=-1e,故选C.5.曲线y =x sin x 在点⎪⎭⎫⎝⎛22-ππ,处的切线与x 轴、直线x =π所围成的三角形的面积为( ) A.π22B .π2C .2π2D .12(2+π)2 [答案] A [解析] 曲线y =x sin x 在点⎝⎛⎭⎫-π2,π2处的切线方程为y =-x ,所围成的三角形的顶点为O (0,0),A (π,0),C (π,-π),∴三角形面积为π22.6.已知f (x )=log a x (a >1)的导函数是f ′(x ),记A =f ′(a ),B =f (a +1)-f (a ),C =f ′(a +1),则( )A .A >B >C B .A >C >B C .B >A >CD .C >B >A[答案] A [解析] 记M (a ,f (a )),N (a +1,f (a +1)),则由于B =f (a +1)-f (a )=f (a +1)-f (a )(a +1)-a,表示直线MN 的斜率,A =f ′(a )表示函数f (x )=log a x 在点M 处的切线斜率;C =f ′(a +1)表示函数f (x )=log a x 在点N 处的切线斜率.所以,A >B >C .7.设曲线y =ax -ln(x +1)在点(0,0)处的切线方程为y =2x ,则a =( )A .0B .1C .2D .3[答案] D[解析] 本题考查导数的基本运算及导数的几何意义.令f (x )=ax -ln(x +1),∴f ′(x )=a -1x +1.∴f (0)=0,且f ′(0)=2.联立解得a =3,故选D.8.设f 0(x )=sin x ,f 1(x )=f 0′(x ),f 2(x )=f 1′(x ),…,f n +1(x )=f n ′(x ),n ∈N ,则f 2017(x )等于( )A .Sin xB .-sin xC .cos xD .-cos x [答案] C [解析] f 0(x )=sin x ,f 1(x )=f 0′(x )=(sin x )′=cos x ,f 2(x )=f 1′(x )=(cos x )′=-sin x , f 3(x )=f 2′(x )=(-sin x )′=-cos x ,f 4(x )=f 3′(x )=(-cos x )′=sin x ,∴4为最小正周期, ∴f 2017(x )=f 1(x )=cos x .故选C.9.已知f (x )为偶函数,当x ≤0时,f (x )=e -x -1-x ,则曲线y =f (x )在点(1,2)处的切线方程是________________.[答案] y =2x [解析] 当x >0时,-x <0,则f (-x )=e x -1+x .又f (x )为偶函数,所以f (x )=f (-x )=e xe+x ,所以当x >0时, f ′(x )=e x -1+1,则曲线y =f (x )在点(1,2)处的切线的斜率为f ′(1)=2,所以切线方程为y -2=2(x -1),即y =2x .10.设函数f (x )=cos(3x +φ)(0<φ<π),若f (x )+f ′(x )是奇函数,则φ=________.[答案] π6[解析] f ′(x )=-3sin(3x +φ),f (x )+f ′(x )=cos(3x +φ)-3sin(3x +φ)=2sin ⎝⎛⎭⎫3x +φ+5π6. 若f (x )+f ′(x )为奇函数,则f (0)+f ′(0)=0,即0=2sin ⎝⎛⎭⎫φ+5π6,∴φ+5π6=k π(k ∈Z ).又∵φ∈(0,π),∴φ=π6. 11.已知直线y =2x -1与曲线y =ln(x +a )相切,则a 的值为________.[答案] 12ln2[解析] ∵y =ln(x +a ),∴y ′=1x +a ,设切点为(x 0,y 0),则y 0=2x 0-1,y 0=ln(x 0+a ),且1x 0+a =2,解之得a =12ln2.12.设曲线y =e x 在点(0,1)处的切线与曲线y =1x(x >0)上点P 处的切线垂直,则P 的坐标为________.[答案] (1,1)[解析] 设f (x )=e x ,则f ′(x )=e x ,所以f ′(0)=1,因此曲线f (x )=e x 在点(0,1)处的切线方程为y -1=1×(x -0),即y =x +1;设g (x )=1x (x >0),则g ′(x )=-1x 2,由题意可得g ′(x P )=-1,解得x P =1,所以P (1,1).故本题正确答案为(1,1).13.等比数列{a n }中,a 1=2,a 8=4,函数f (x )=x (x -a 1)(x -a 2)…(x -a 8),则f ′(0)=________.[答案] 212[解析] f ′(x )=x ′·[(x -a 1)(x -a 2)…(x -a 8)]+[(x -a 1)(x -a 2)…(x -a 8)]′·x =(x -a 1)(x -a 2)…(x -a 8)+[(x -a 1)(x -a 2)…(x -a 8)]′·x ,所以f ′(0)=(0-a 1)(0-a 2)…(0-a 8)+[(0-a 1)(0-a 2)…(0-a 8)]′·0=a 1a 2…a 8. 因为数列{a n }为等比数列,所以a 2a 7=a 3a 6=a 4a 5=a 1a 8=8,所以f ′(0)=84=212. 14.求下列函数的导数:(1)y =x (x 2+1x +1x 3);(2)y =(x +1)(1x -1);(3)y =sin 4x 4+cos 4x4;(4)y =1+x 1-x +1-x 1+x .[解析] (1)∵y =x ⎝⎛⎭⎫x 2+1x +1x 3=x 3+1+1x 2,∴y ′=3x 2-2x 3. (2)∵y =(x +1)⎝⎛⎭⎫1x -1=-x 12+x -12,∴y ′=-12x -12-12x -32=-12x ⎝⎛⎭⎫1+1x .(3)∵y =sin 4x 4+cos 4x 4=⎝⎛⎭⎫sin 2x 4+cos 2x 42-2sin 2x 4cos 2x 4=1-12sin 2x 2=1-12·1-cos x 2=34+14cos x , ∴y ′=-14sin x .(4)∵y =1+x 1-x +1-x 1+x =(1+x )21-x +(1-x )21-x =2+2x 1-x =41-x -2,∴y ′=⎝⎛⎭⎫41-x -2′=-4(1-x )′(1-x )2=4(1-x )2. 15.偶函数f (x )=ax 4+bx 3+cx 2+dx +e 的图象过点P (0,1),且在x =1处的切线方程为y =x -2,求y =f (x )的解析式.[解析] ∵f (x )的图象过点P (0,1),∴e =1.又∵f (x )为偶函数,∴f (-x )=f (x ). 故ax 4+bx 3+cx 2+dx +e =ax 4-bx 3+cx 2-dx +e .∴b =0,d =0.∴f (x )=ax 4+cx 2+1. ∵函数f (x )在x =1处的切线方程为y =x -2,∴切点为(1,-1).∴a +c +1=-1. ∵f ′(x )|x =1=4a +2c ,∴4a +2c =1.∴a =52,c =-92.∴函数y =f (x )的解析式为f (x )=52x 4-92x 2+1.16.已知f (x )=13x 3+bx 2+cx (b ,c ∈R ),f ′(1)=0,x ∈[-1,3]时,曲线y =f (x )的切线斜率的最小值为-1,求b ,c 的值.[解析] f ′(x )=x 2+2bx +c =(x +b )2+c -b 2, 且f ′(1)=1+2b +c =0.①(1)若-b ≤-1,即b ≥1,则f ′(x )在[-1,3]上是增函数,所以f ′(x )min =f ′(-1)=-1, 即1-2b +c =-1.②由①②解得b =14,不满足b ≥1,故舍去.(2)若-1<-b <3,即-3<b <1,则f ′(x )min =f ′(-b )=-1,即b 2-2b 2+c =-1.③ 由①③解得b =-2,c =3或b =0,c =-1.(3)若-b ≥3,即b ≤-3,则f ′(x )在[-1,3]上是减函数, 所以f ′(x )min =f ′(3)=-1,即9+6b +c =-1.④由①④解得b =-94,不满足b ≤-3,故舍去.综上可知,b =-2,c =3或b =0,c =-1.高中数学选修2-2导数--导数的运算1.若f (x )=sin π3-cos x ,则f ′(α)等于( )A .Sin αB .Cos αC .sin π3+cos αD .cos π3+sin α2.设函数f (x )=x m +ax 的导数为f ′(x )=2x +1,则数列{1f (n )}(n ∈N *)的前n 项和( )A.n n +1B .n +2n +1C.nn -1D .n +1n 3.已知二次函数f (x )的图象如图所示,则其导函数f ′(x )的图象大致形状是( )4.已知函数f (x )的导函数为f ′(x ),且满足f (x )=2xf ′(e)+ln x ,则f ′(e)=( )A .e -1B .-1C .-e -1D .-e5.曲线y =x sin x 在点⎪⎭⎫⎝⎛22-ππ,处的切线与x 轴、直线x =π所围成的三角形的面积为() A.π22B .π2C .2π2D .12(2+π)26.已知f (x )=log a x (a >1)的导函数是f ′(x ),记A =f ′(a ),B =f (a +1)-f (a ),C =f ′(a +1),则( )A .A >B >C B .A >C >B C .B >A >CD .C >B >A7.设曲线y =ax -ln(x +1)在点(0,0)处的切线方程为y =2x ,则a =( )A .0B .1C .2D .38.设f 0(x )=sin x ,f 1(x )=f 0′(x ),f 2(x )=f 1′(x ),…,f n +1(x )=f n ′(x ),n ∈N ,则f 2017(x )等于( )A .Sin xB .-sin xC .Cos xD .-cos x9.已知f (x )为偶函数,当x ≤0时,f (x )=e -x -1-x ,则曲线y =f (x )在点(1,2)处的切线方程是________________.10.设函数f (x )=cos(3x +φ)(0<φ<π),若f (x )+f ′(x )是奇函数,则φ=________. 11.已知直线y =2x -1与曲线y =ln(x +a )相切,则a 的值为________.12.设曲线y =e x 在点(0,1)处的切线与曲线y =1x (x >0)上点P 处的切线垂直,则P 的坐标为________.13.等比数列{a n }中,a 1=2,a 8=4,函数f (x )=x (x -a 1)(x -a 2)…(x -a 8),则f ′(0)=______. 14.求下列函数的导数:(1)y =x (x 2+1x +1x 3);(2)y =(x +1)(1x -1);(3)y =sin 4x4+cos 4x4;(4)y =1+x 1-x +1-x1+x.15.偶函数f (x )=ax 4+bx 3+cx 2+dx +e 的图象过点P (0,1),且在x =1处的切线方程为y =x -2,求y =f (x )的解析式.16.已知f (x )=13x 3+bx 2+cx (b ,c ∈R),f ′(1)=0,x ∈[-1,3]时,曲线y =f (x )的切线斜率的最小值为-1,求b ,c 的值.。

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