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(1)f(x)

x(x0)

(2)f(x)

1(x0)(3)f(x)3x2(x0)

(4)f(x)logax(x0)1解:(1)f(x)(x2)

1 2x222f(x)(x2)'2f(x)(x3

2x3f(x)(logax)'1logae1

1 xloge

xyx3x2y5xycosxx2y1的经过点(2,0)的切线x解:(1)y5xk

yx3

y3x22x5

x1

x5 y2y2505,且经过(1,2或(5,-50y5x3y5x1y'sin

- xy2

3 斜率为1

(,0)ykk

ykxb过(2,0)

2kb

bykx

yyx

y1k

x

y

(k1,k) ykx2k上代入k1yx1,k)st22t1求在t3时运动的瞬时速度st22t1s2t

当t3时s8vt38mf(xxa

f(a)f(ax)

f(a5h)f(a3h);f(at

f(a)f(a

16解:(1)

f(a)f(ax)

f(a)f(ax)fa(a

f(a5h)f(a3h)

f(a)f(ax)fa(a

1

1f(a)t0f(a)f(a 33t0f(a)f(a f(x可导,求

[f(xx)]2[f(x)]2

[f(xx)]2[f

lim[f(xx)f

[f(xx)ff'

lim[f(xx)f(x)]f(x)2f(x)2f(x)f'f(xx0

f(x32f(x)x0xf'(0

f(x)3x

f(x)在x0点连续,limf(x32x,f(xx0f(0)

f(xx)f(x)

f(x)f(0)

f(x)3f(x)xx2x的不可导点f(x)xx2xx2x0x1x

f(x)(x3x2)3x2 x2x0时,即0x1,f'(x)2x3x2f(0)f(0) f(1) f(1)1f(1)f(1)

x1为f(x) 设

ff(x)

1,x af(x)x0x(1)(2)x2sin

x f(x)

x

,

xasin1

xasin1f(0)0ax(x0)a f

(x)asinf(0)lim 0

x

lim(x)a1sin

a10a 此

f(0)axa1sin1xa2cos

xf(x)

x

fx)再x0则

f(0)

f(0)f(0)0af(xf(0)0F(x)f(x)(1sinx)x0ff(x)(1f(x)(1sinx

F(0x)F(0)

f(x)0f

fx

f(xxf(xx)fx解:f(x)xx0处为0f'(0不存在f(x)(xa)g(x)g(x)xaf'(a)xag(x)f(xxa

f(x)(xa)g(x),f(a)g(x)(x2)limg(x)A,则f'(a)A.limg(a不存在,则f'(a)不存在 f(x)

e1x,x

x则下列结论正确的是(f(x)x0f(xx0f(xx0fxx0f'(xx0 x2解:D. x2

x,x 1

1ex

1e

e

1lim

x)(lim

x)(lim e x4lim

ex0

f(x)f(x)f(x)

f(x)x0连f(x)lim4f(1x) yf(x在点(1,f(1处切线的方程lim4f(1x)14+f(1)=0 y1

11111cos2

11

f(1x)f(1) 2

4f(1(2x))k=2且经过(1,4)y2xf(xx0f(0)f(x)f(x)所以f(xf(x)x=0时,等式变为2f(0)0f(0)(1)y5x43x21x

(2)y(x1)x(3)y

1x2;

y1lnx1x33x1x33xy

xnloga

y y2xsinx (8)y3xx2(9)y

1cos

ytanxxcotxyxarctans (12)yexarccosx(13)y

cos2x sinxcosx

yx2tanxlnxy2xx2log2

y 1x11x11x11x1x

y'20x36x123123x2y[(x

x

(x1)y

5(1x2)

1x2

(1x21(1

(1x2lny

(1lnx) 1ln

lnx)-(1y(xnlogax)nxn1logax

xyy(2xsinxx x

3xln3 y(ex) )

x3x(2xlny

)1cosxxsiny(tanx-xcotx)sec2x-cotxx

1-x11-x

y

ex(arccosx y'

cos2xsinxcosx

)2sin2x(sinxcosx)(cossinx)cos2x1(sinxcos12sin2xsinx2sin2xcosxcosxcos2xsinxcos2x(cosxsins)sin2x(sinxcosx)(sinxcosx)21sin2x(sinxcosx)

(sinxcosx)2sinx(sinxcos

y(x2tanx1nx)2xtanxlnxx2sec2xlnxxtanxx(2tanxlnxxsec2xlnxtany(2xx2

5ln2

log1x1x11x11x1x21 111

.1

1111

1x1 xx111(arctanx)

1

解:1)(arctanx)

11

yarctanxxx(tany)

cos2

y1cos2y

1tan2

112)(arccosx)1

11x(cosy)sin

y1

11111cos2

yf(x)ax3bx1g(x)yg(x)在点(2,1)y1x4,求常数 解:yf(x)ax3bx1g(x)yg(x)在点(2,1)y1x4f(x)过点(1,2),yg(x)在点 y1x4,所以a=1,b= f(xfxf(0),f(1

arctanx,0 (2)f(x) xx

x

tan

-x0.

f(x)

1,1x2)f(x)

1

2,f(0)11.f(1)11

x,2

x

cos2

1 (1)y(3x5)3(5x

(2)y(3

14x2yln(3x)3lnx

y 1y(sinnx)(cos1

yln1x1

ylntanx2

ycos5x2yxsec2xcsc2x (10)yx2cot1x2x21(11)1

(12)y

a2ln(x

x2a2y

2x;1x2

ee

ax

y

ex12exe12;y;y

1xx2x21x2

lnx

ylogasinexyu(x)]bax(a,b常数u(x),(x可导y

xxx

ylnarcsinxy

(1)cotx

y(sinx)cosxxx23x3xx23x3

(28)

1313313(32解:1)y[(3x5)3(51)53*3*(3x5)2(5x1)55*5*(5x1)4*(3x(120x134)(3x5)2(5xy[(32x3)14x21 (3 312x(63x11

2x)2*1y[ln(3x)3lnx]

1*33

x3

221(lnx)3*

1(1 3ln23ln2111y1

]

21211(1x2y[(sinnx)(cosnncosxcosnxsinnx(sinx)*ncosn1n(cosnxcosnxsinnxsinxcosn11y

x] 1111x12 2(1x 1

*1

11

csc[1ntan]

tan2

cos2 2

xcos

y[cos5x]5cos4x(sinx)*15cos3x.sin y[sec2xcsc2x] )2(sec2xtanxcsc2 cos2x

sin2x

cot y

12xcot1

1*12xcot1csc211y1

cotx

xarcsinx]

sin21 1121

1 1

1x2x2yx2

a2ln(x

x2a2x22xx22x2

112x2xx22x2a2y

2x

2(1x2)2x

[arctan1-x2

2x2*1(1x2)

(1x2

1x2111*arccosx*111*arccosx*211

1(1(1x21

y

ax2

a

*

*lna*(2x)

aea

exex

(exex)2(exex

y[exex]

(exex (exex)21-1y1-1121121

]

121112111

*(1x1x)(1x)2y

[logasin

x2]

*

*x22x2

xex2cotex2y[arc

x21

1nxxx22x22x2x2x21x2xlnxln(x2

x2

y[[u(x)]bav(x)]b[u(x)]b1u(x)xxxx2x2xx

*(1 2x2x124x2xx2x8x2xxxx

x]

y

1arcsin11arcsin11212xx2arcsin

1ln()cot cotxln x

1

)ln

x*cotx

(1)cotx

xlnx

cotx);y[(sinx)cosx[ecosxlnsinx(sinx)cosx*[sinxlnsinxcosx*cossin(sinx)cosx(cotxcosx-sinxlnsinx12yxx,即lnyln x12所以(lny)ln

x)1*yy

lnxxxyx

(1lnx)x2xy

x x23x23x3x23x3

(x24xx32

3x

xy[1x3(3x)2]

1x3(9x2)133 (32xy(11)133 (32x (lny)ln(11)x],1*yxln(11 所以y(11)x[ln(11) 2xf(x)(1)yf(ex)ef(x) (2)yf(sin2x)sin[f(x)]2(3)y

f

1)x

yarctan[f解:1)y[f(ex)efxf(ex)*ex*ef(x)f(ex)*ef(x)*fef(x)[exf'(ex)f(x)f(ex2)y(f(sin2x)sin[f(x)]2f(sin2x)sin2x2f(x)f(x)cosf2 1

1

f(arcsinxxx21y[f(arcsin)]f1

*(x2) y(arctan[f(x)])

f'(x);1f2(x)1f(x2x求1

1x2111111211111

)]f

)*( )

f(xln(x2a2

x1x1可导,求a,bx解:因为f(x)在x1处可导,所以f(x)在x1处一定连续.

,即

1

,所以a0,b

f(x)

f(x)f(1)

0ln(1yax2ylnx相切,求ae解:因为曲线yax2与ylnx在x 处相切,所以y(x)y(x),即2ax1,所e 11x ,此时y1.所以1 ,所以a111 f(x)f'(x也是周期函数f(x)T为f(x)f(x)f(xT),f(x)f(xT所以f(xxya2xy轴围成的三角形的面积为一常数

y0解:xya2y

,y'

,,任取一点(x,y),则有 ,则切线方程为

0 k

xxy 2xxx

x轴的交点为(2ax0,0y轴的交点为

s

12ax02

,yxarcsinx

yxex21111

yarctan1xlnxx1(5)y (6)ylnf(xf(u有二1解:1)y

[xarcsinx]

x

2(1x2)32)y

x2)(ex2

*2x)2x(3

2x2)ex213)y1

)

*1) (1x21(1x2134xx(1xx

x)(arctan1xx

x)

1 (1x2y(xx)[xx(lnx1)]xx(lnx1)2xx*1xx[(1lnx)21

f(x)f(x)[f

(ln[f

*f(x)]f

f2求下列函数的n(1)1n(1x)

1sin21xex

(4) 解:1)[1n(1x)]n

(1x)n

(n2)[sin2x]n(sin2x)n2n1sin(2xn12[xex]n(exxex)n1(exxex)n1(nx)ex11

)n

11

2]n,令f(x)

11

22

,f'(x)

1*522

fn(x)(-1)*

(2nn(1x)

(1x)即( 1x

*

(2n1)!n(1x)f(x)的n2f(n2(x)

,f(n解:因为f(x)的n2所以f(n)(x)2lnx

f(n-2)(x)

fn1(x)lnx设函数f(x在(,上满足2f(1xf(1x)ex,

f(x).2f(1xf(1x)ex1xt2f(tf(2tet-1式11xt2f(2-t)f(t)e1-

(2)。(1)*2-(23f(t2et-1et1.所以f(t1(2et1e1t3即f(x)1(2ex1e1x3f(x)x23x1x2x依次等于0.1,0.01,0.001时的改变量与微分的差,解:(fdf)x2x)2x依次等于101,102,103时,(fdf)x2依次等于102,104,106所以x愈小则(fdf)x221(1)1

(2)y

1yexcosx

y

1x3xy xycosx1

yexsin2(2x)y5lntan2解:1)dyy2

2xdx 1dy1

1x2x(2x)(1x2)2

dx

(1x2)2dyydx[excosxex(sinx)]dxex(cosxsindy

1121

dx

2(1x3);dy

xx2x2xdyex[2sin4xsin2(x21)sinx2xcosdy

(1x2

dy(5lntanx*ln5

*sec2x)dx

2ln55lntanxdx.sin2xx(x0x的高阶无穷小(1)ex1xtanxx

(2)n1x1xnln(1x)解:1)

exx

limex1

(法则).所以1

xlimx1nf(0f(0)dx1x(x1nx01x(微分求近似解limtanx

1cos2x1x0 1limln(1x)lim1x x0(1)50.95

(2)解:1)50.9551

15

5*(0.05)10.01

221 20.1厘米,如已知用材每米3的重量公斤,求此球壳重的精确值和近似值.解m4R20.001

m [4R34(R0.001)3]约4[23(23

m的精确值4[12.1036.1061093m4R20.001 25cm,底半径为200.05cm求圆柱体体积和侧面积的绝对误差和相解:1)V[(RR)2R2h2520.052202)V157cm3

S2[(RR)R]h500.052.5S7.85cm2

2已知某商品的成本函数为C(Q)100028

解:20

1P

1000 8

(总成本Q4C(Q)2QQ48

Q12030(边际成本QPP1500.01Q(元解WPQ(1500.01Q)Q150Q0.01Q2;W150WQ10014900元(总收益 WQ100148

P60

解:1)W(P20)Q(40

W40

2)当P10元时Q5000总收益QPQ500000P1%Q3990D(P)75P2,P4时的需求价格弹性和收益价格弹性,并解Q

当P时

QPdP2P75P275

RPQ75P

RPdRP753P2

dP

75P

RP270.46S(P)23PP3时的供给价格弹性解:

3P

当p3时

P3931%110.82%(1)Q100(2

p)

p

abQ(a,b解:1)

dq 50p2

100(22(2100(22(2p 22 222(2p所以当2(2p

1时是低弹性,即0p16。当qp 2(22(2P

1性,即16P4.92)qap2,

dq*p2p*

2

.所以当

2p21时是低弹性,a/aa/a

dp aa/a/

a

a0P

,当qpap2

p

(因为padq,b0,q设某商品的需求函数为Q1005p,如需求弹性小于-1p的取值范围解:QpdQP5 dP 100 20P20

1P因为Q1005P P2010P(1)设u(x),(x)在点x0处均可异,且 v(10(,则limu(x)v(x)2(D A.- 3函数f(x) 在点x0处(C3

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