微积分作业对外经济贸易大学远程教育

1、yx21A,2xB,A,C,A,(14xC,(12x22(1 x )D,(14xx )4x2 2(1 x )2x2cosx4xcosx2x2 sin xB,4xcosx所以A,cosx2xs in x2cos x4xcosx 2x2sin x2cos xD,4 cosx 2x2 s in. 2sin x ,cosx2yuUxyx2 cos x(2x2/cosx)4xcosx 2x2 sin xB,x2，那么cosu2x ,yuUx2 cos x2xcosx2y sin u,2xcosx2。ln(x 1 x2)，那么 yB,1 x2)。C,x2C,2cosx2)。2xD,x22xcosxD,2x

2、 I 1 x21令y=lnu，uxv2，2v=1+x111 1那么yuuv1-v2 ，vx 2xu2所以yxYuuvvx1J x2A,c,3sin(ln x)，贝U y =(cos(ln x)3 (ln x)2 B,3cos(lnx)3 (lnx)xD,。cos(ln x)3332-cos(ln x) (In x) xsinv, v uln x,yv Vu ux33cos(ln x)x2 (ln x)。A,C,3sin2x,那么 y =(sin 2x2cos2x 3 B,2cos2x 3sin 2x ln3 d,y3sin2xcos2x 2。2cos2x In 3sin2x .cos2x 3

3、ln 3In 3今后可约定yxy，省略下标2cos2x 3sin 2x ln3。7,设函数 yearccos2x)，那么 dy 等于 dxA.arccos(2 x)eB .、14x2arccos(2x)2ec .14x2arccos(2 x)2e2x2arccos(2 x)2eCx2解答：arccos(2x)arccos(2x)-y (e) =earccos(2x)arccos(2 x)arccos(2 x)e2e=(2x)=1 4x214x211,12,a,C,8，导数是1的函数是3A，屋3 B，右2解答：丄 3=2x2(x19，函数十的导数是(3x3 B , Cxx(g) =(x 3) =

4、-3xxA,解答:10,设.2 y sin x，那么A, 2 sin2xB, 2设 y . 1 In xA,C,2)14x4-3=x3? x-4sin 2xdxc,sin2xD,sin 2xdx11 ln xe ax sin bx ,e ax(b cosbxeax(bcosbxB,D,a sin bx)a sinbx)12x . 1 ln xln xb,e ax(bcosbxasin bx)d,eax(bcosbxasi n bx)axd (sin bx) sin bx axdee ax cosbxd(bx) sin bx e axd( ax)e ax(bcosbx asinbx)dx。113

5、,(x2)=(),1x1一 1“3一3A,2B,1 2x 2C,1 2xD,1 2x222214,2x(e)=()A,e2xB,e2x C,2x2ed,2e 2xdx15, (log 2 x)=(e b,)1Iog2exC,1 dxxD,1x八2A,log 2x16,(5x)=()A, 5xl n5B,5xC,5xdxD,5I n517,(In 2x)=()A, Lnx B,2C,1D,1x2xx18,设 y=sin7x ,那么y =()A,-7cos7xB, 7cosxC,7cos7xD, cos7x19,设 y = xcos(-x),那么y =()A, cos(-x) - xsin(x )

6、 B, cos(-x)+ xsin(_x)C, cos(-x)+ sin(x)D, cos(-x)- si n(-x)20, (tgx 8)=(A,亠cos xB,12cos xC,D,cosx1cosx一、导数的运算答案1,( D )2, ( C )3,(B )4, ( A )5, ( D )6, ( C )7,(B )8, ( A )9, ( D )10, ( C )11,( B) 12, ( A)13, ( D) 14, ( C )15, (B ) 16, (A )17, ( D ) 18, ( C ) 19, ( B ) 20, ( A )1,2,3,4,5,、函数的微分1dx 2A,

7、B,1dxC,D,de 2xA,d log 2A,d5xA,2e2xB,2xC,2e2xdxD,2e2xdxB,1log 2 edxxC,D,5Xln 5dxB,5Xl n5C,5XdxD,5xd In 2x =(A, Lnx dx B,dxC,6,ds in 7x=(A, 7cosxdx7,dcos(-x)=(A, -sin xdx8, d(tgx 1)=A,9,10,11,dx2xD,1dxxB, 7cosxC, 7cos7xdxD, 7cos7xB, sin(-x)dxC, sin(-x)D, -si n(-x)dx12cos xdxB,12cos xC,dxcosxD,cosxd(2c

8、tgx)=(A,亠sin xd arcs in2x =A,C, 1 4x21 4x2B,B,dx,d arccos(x 1)=(D,12 dxsin xdx,4x2C,12 dxsin xD,2 -2 sin-dxx1 4x2dx,A,.dx,1 (x 1)2B,dx,1)2C,12 dx,D,.1 (x 1)215,16,A,C,12,A,C,13,A,C,darctgx 22x .4dx1 x2x1 x4darcctg 3x =31 9x2 dx,1 9x2B,14,设 y sin2 x,那么a, 2 sin2xb, 2设 y , 1 In x，那么a,C,B,D,D,dy =dy 12x

9、1 In xJxx11 x2dx, xdx,9x2sin 2xdxdy =(B,1dy dx21 In xe ax sin bx，贝那么 dy =e ax(b cosbxe ax(bcosbx17,函数ya, dyD,c,dysin 2xdx d, sin2xdx2x . 1 In x1dy dx/ In xa sin bx)dxa sin bx)B,d,e ax( b cos bxe ax(bcosbxe axd (sin bx) sin bx e ax cosbxd(bx)e axln (5tgx)的微分是(1 dx b , dy tgxcosxa sin bx)dxasin bx)axd

10、esin bx e axd( ax)(bcosbx asinbx)dx。52c, dy dx d , dydxtgxsin (2x)解答：dy d ln(5tgx) =dln( tgx) In 5 = d ln(tgx) d ln 5d In(tgx)=dtgx =1dx =tgx tgx cos x1sin xcosxdx =2sin (2x)dx18,设 f (x)-ln(1 2x)，那么f(x)(A)-x1 2x(B)1 2x(C)2x(D)1 2x19,设函数y arccos(2x)，贝Udydx等于20, 9.设yln、 x2,那么 yo (-1 x).33C.1AB.-222A.-

11、1 B . -2 C . -3 D . -4二、函数的微分答案1,( D ) 2, ( C ) 3,( B)4, ( A )5, ( D ) 6, ( C ) 7,( B)8, ( A )9, ( D ) 10, ( C ) 11, ( B ) 12, ( A )13, ( D ) 14, ( C )15,(B ) 16,( A)17, ( D ) 18, ( C )19 (B ) 20,( A)D.三、隐函数的导数1,y=f(x)由方程 ysin y 0 决定，那么 yx= oA,2xC,yx1 cosy2xB,yxyx1 cosyD,2x将二元方程两边对x求导,yx2xcos yyx由此解

12、得yx2xo那么由此方程决定的隐函数yf x的导数是1 cosyA,dy 1 x y dxx yB,dydx1 x yx yC,dy 1 x yD,dy *x ydxx ydxx y对方程两边取微分，d(x22xyy2)d(2x),即d(x2)d(2xy) d(y2) 2dx,亦即2xdx2xdy2ydx2ydy 2dx或(2x 2y)dy(2 2x2y)dx ,于是ydy 1x yodxx y3,y arctg (x y)，那么y等于)A.1B ,2(x y)2sin (x y)C,1D1cos (x y)1 (xy)22， x2 2xy y2 2x ,等式两边合并dy=(x y) 2dx,

13、故：y2=dy/dx= (x y)24，x +y2=1，那么由此方程决定的隐函数f (x)的导数是()。xA. By兰C ,1 D , 1yx x5,设方程xyeyIn2确定y是x的函数,(D)(A) Jx e6I,0.设xln y yln x确定函数 y y(x),那么y |x).B.1 eC.D.解：两边取微分：d(x ln y)=d(yl nx)然后按微分的乘法公式:Iny dx+ xd(l ny )=ln xdy+ y d( Inx) Iny dx+ x/ydy =ln xdy+ y/x dx x/ydy- In xdy = y/x dx- Inydx (x/y- Inx)dy =(

14、y/x- Iny)dxdy / dx =( y/x - In y)/ (x/y- Inx) 把x=1,y=1代入即可：dy / dx =1四、高阶导数1求y二a二X的2阶导数，A.y1ax b , y(1)xaC ,ya 2x d , y(1)xa 22求y二=sinx的2阶导数。A.ycosx B , ycosxC ,ysin x D , ysin x3,函数y In(2x 1)2的二阶导数为：()A.y24(2x 1) B。2y8(2x 1)C.y4(2x 1) 2 d。y 8(2x 1) 2解答：2 2y ln(2x 1) =2ln(2x 1) =(2x 1)(2x 1)4=4(2x1)

15、 1(2x1)y 4(2x 1) 1=4(2x1) 2(2x1) = 8( 2x1)24,设(u)具有二阶导数，y x(x)，那么y()(A)x (x)2 (x)(B) x(x)(x)(C)x (x) x (x)(D) yx (x)2x 15,函数y e的二阶导数为:()A.2x 1y 2eb. y(2x2x 11)e2x 12x 1c.y 2(2x1)ed. y4e求函数的极限五、1,设 f (x)x2-2xA,xim2f(x)那么有()6xim2f (x)C,lim f (x)2解答:他 f(X)2x2x_(x24)(x2x 6)=limx2x2 2x 14=4/552,sin (1x)2

16、 x112si n(1x)(D.1B.-1 C.limx 12解答：xm1 x2si n(1x)lim 2x 1 (x21)limx 1cos(12xx)cosO2.xsin3, lim 3 =x 0 x1B. 3 C.D. 1解利用公式limx 0 xlim f (x)x 2!1 cosx4, limx 01A.-2x2B.-1 C.-12 D. 05,求 Ixm12xx2:1 =(3x 2C.-1-D. - 22显然为0型问题0(x2 1)= lim 2x 1 (x2 3x 2)2x Timx 1 2x 3=-2 o6,lim ln xx1A.-2C.1 D.B.-8,1A. 601In

17、x1lim 1lir?01C. -1 D.(x=limolimx 0 0 1e2 xo-2 x sin x lim x 0 x sinB.-1 C.-=(D.cosx原式=lim2x 0 2xsin x x cosx 0cosxlimx 0 6cosx 4xsin x x cosx 2xsin x11,A.A.13A.9, limx62x24x2-=(14B.-亦 C. -1 D.m224XX2X e-3m 1 - 6 H XQ -415lim 2xx 24lim(4x21)15x 2 /39=16 D.415解 因 lim (x29) lim x2x 3x 3后，再用这个定理就可以了，由于X-2Xm3| Klim 2x 1x 4 .x16 B.解所以12,limxX/V3X-Xm3H X2x45x499 90,故不能应用商的极限定理,x 3但x 3，故有1 - 61 X/Vm3H X但对函数做适当变化3=(D.-3x3x231=()B.limx2x3x2x2x