tan ex ln[1 arcsin arctan 12(3).(1x)a (4).ax xln n(6).n1x n

(1

0x 0|x| 2sinxxtan如果limU1,limV則limUVelim(Uf(x)f

1cosx 2f(x)f 直線Lykxb為函數(shù)yf(x)klimf

blim[f(x)

這里的包括和

12常見函數(shù)的導(dǎo)數(shù)12(x)'

()x

)'

(sinx)(n)sin(xn (cosx)(n)cos(xn (sinkx)(n)knsin(xn (coskx)(n)kncos(xn (xn)(n) (ax)(n)(ax)(ln(ex)(n) (1)(n) t (t ( t

(t

[ln(t (t(tanx)2dxtanxx(cotx)2dxcotxx 1arctanxCx2x2

x2lnx2

1ln|x

x2a2 2a xa arcsinxCaa2arcsinxa2x2dx2 x2x2x2a2dx ln|x2

xa2eaxcosbxdx a2esinbxdx a222Swsinwxdx S'3wsinwxdxaaaa

f(x)dx

0[f(x)f(x)]dx

af(x)dx(如果f(x)為偶函數(shù)acoskxdxsinkxdx

(sinkx)2dx

設(shè)klN且klcoskxsinlxdxcoskxcoslxdxsinkxsinlxdx

f(x)dx

f(x)dx 2f2 f(x)dxeu2du eaxdx1(a

feptsinwtdt0

p2

(p0,weptcoswtdt (p0,w0)0sinxdx

p2

02f(sinxdx2f(cosx 特別的

02(sinx)ndx

2(cosx)n

f(sinx)dx

2f(sinx)dx

2f(cosx)0 特別的

(sinx)ndx

2(sinx)ndx

2(cosx)n(cosxndx n為奇數(shù) 22(cosx)n0

（n為偶數(shù)2(sinxndx n為奇數(shù) 42(sinx)n0

（n為偶數(shù)2(cosxndx n為奇數(shù) 42(cosx)n0

（n為偶數(shù) 2(sinx)ndx 2(cosx)n 2(sinx)ndx

n1n3n

n為正奇數(shù) n2n n1n3n5.........1 n2n

n為正偶數(shù)xf(sinx)dx 2

f(sinxlimn!n limxlnxlimxxlimannf(x)

若f'(a0，只能得到結(jié)論：f(x)在a點嚴(yán)格增加。即x(aa)有f(xfx(aa)有f(x)

設(shè)f(x)=|x-a|g(x),其中g(shù)(x)在x=a處連續(xù)，則f(x)在x=a處可導(dǎo)顯然為12 設(shè)f(x)在處n階可導(dǎo)，且f )f )f )LLf(n1)(x 0f(n)(x) (20

n2k且f(n)(x000n2k且f(n)(x000

00 若fxfxLLf(n1)x0，f(nx00 則x0,fx0))設(shè)An1f(An),An 若f(x)在區(qū)間I上單調(diào)遞

A2

注意：若f(x)在區(qū)間I題目中如果出現(xiàn)f''(x0f'(x)ln(x1x2 (x0nmxmo(xn0nmo(xmo(xno(xn當(dāng)0nm 當(dāng)mn0xmo(xno(xmno(xm)o(xn)o(xmn LL) (其中有無窮多個n kn (其中nk1,2,

1n2n

LL n!n! arctanxarctan1 ,0t

t21求A(ba|xb|dx1a 結(jié)論：當(dāng)b 2 (b) (aa ba(xb

10lnxdx11xm(1x)ndx1xn(1 作用：1x(1x)9dx1x9(1

則af(x)dxaf(abbf(x)dx1b[f(x)f(ab 2 f(x)dx f(b bf(x)dx1b[f(x)f(b 2f(x)dx

f(1)dx1[f(x) 2f2

f( f(x)f'(x)dx

f(sinx)f(sinx)dx2f(sinx)dx bxf(sinx)dxnbf(sin n為偶 若f(x)bxf(sinx)dxnbf(sin bxf(cosx)dx bf(cos

f(cosLL

Lf(x,y)ds2Lf(x,a22Lf(x,ya22 I

解：I=(xyx2dsxydsx2ds02L 222a2cos2ad2 I (xy3)ds,L為x2y2L 解：I (xy3 y3ds=0+0= Lx0的半個區(qū)域，則：

LP(x,y)dx2LP(x,2LP(x,y)dx2

R2x2,方向為從左到解:I xy(ydxxdy) x2ydy0 ILx2ydy,其中L為雙紐線的右半支：(x2+y2)2=a2(x2-y2),x0的逆時針方向 由于圖像關(guān)于x軸對稱，則I0

LP(x,y)dy0(上面講到的就是用的這個結(jié)論LP(x,y)dy2L1P(x,例 I

例 I dxdy,其中ABCD是A(1,0)B(0,1)C(-1,0)D(0,-1)ABCD|x||y解：I ABCD|x||y ABCD|x||y |x||y2設(shè)分片光滑的曲面關(guān)于yoz平面對稱，f(x,y,z)在上連續(xù)，是中x02,

f(x,y,z)dsf(x,y,z)ds=2f(x,y, 2例 I(xyz)ds,其中為球面x2y2z2a2上z（0<ha）的部關(guān)于yozxoz面對稱，故Izdsa(a2h2例 I(xyyzzx)ds,其中為z

x2y2被柱面x2y22ax關(guān)于xoz面對稱，故Izxds642 2設(shè)分片光滑的曲面關(guān)于yoz面對稱，函數(shù)p(xyz)在上連續(xù)，是中x02

f(x,y,z)dydzf(x,y,z)dydz=2f(x,y, 2例 I xyzdxdy,其中是球面x2y2z21的外側(cè)在x0,y25解關(guān)于xoy面對稱，故I xyzdxdy2xyzdxdy25 2例 I=x2dydzy2dzdxz2dxdy,其中曲線弧段z=y2(x0,1z解：顯然曲面關(guān)于yozzox面對稱，故Iz2dxdy, f(x,y)dxdyf(y, 例 I(3x2y)dxdy,其中D為xy2與兩坐標(biāo)軸圍D,I(3x2y)dxdy=(3y2x)dxdy5(xy)dxdy5xdxdy 2 例 I x2y2

(y2x2 x2y2

(y2x2)dxdy x2y2

(x2y2dxdyI,故I f(x,y,z)dvf(y,x, 求(xyz)dv,為x0,y0,z0,x2y2z2 (xyz)dv3zdv3 求I(zx2y2)dv,為zx2y2和zhh0）圍成的區(qū)I(zx2y2)dv(zy2x2)dv12zdv 2 f(x,y)ds f(y, 例 I x3ds,L為星形線x3y3aL解：顯然L對x,y22 223y333I 1(x 13y333 2 例 求(x2z)ds,F是圓周x2y2z2R2,xyzF解：F關(guān)于x,y,zx2ds=y2ds= xds=yds=1 1

1

故(xz)ds (xyz)ds (xyz)ds ds 輪換對稱性，f(x,y)在L上連續(xù)，則： f(x,y)dsf(y, 或者f(xy)ds+fyx)ds I ydxxdy,L為xyR上A(R,0)到B(0,R)的一段L解：L對坐標(biāo)xy具有輪換對稱性，故ydxL I y3dxy3dx,L為雙紐線(x2y2)22a2xy位于第一象限部L 解：L關(guān)于xy具有輪換對稱性，則y3dxx3Lf(x,y,z)dsf(y,x, 例 I(x21y21z2)ds,:x2y2z2 解：x2dsy2ds I(x21y21z2)ds(111) 4(111)1(x2y2z2)ds7 43 例 I(axbycz)ds,:x2y2z2R2位于第一掛限部解：xdsyds I(abc)zds1R3(ab f(x,y,z)dydzf(y,x, x2例 I(yz)dydz(zx)dzdx(xy)dxdy,x2(0zh),(y (x (xy)dxdy(yx)dydx 所以I例 Ixydydzyzdzdxzxdxdy,為平面xyz1位于第一掛限的外,xydydzzydydx I3xydydz

f(x)

f則： 使得f'()f(x|x|在x0f(x)f(x)

xx 應(yīng)用：設(shè)f(0)0,則f(x)在x0

f(1ehhlimf(hsinh)存 (D)limf(2h)f(h)存 若 ', '且lim

若 ', '且lim 設(shè)f(x)為()上的連續(xù)，函數(shù)F(x)為f(x)f(x)為奇函數(shù)f(x)任意原函數(shù)F(xxf(x)為偶函數(shù)f(x)的原函數(shù)只有一個是奇函數(shù),即為0ff(x)任意原函數(shù)F(x)為周期函數(shù)f(x)Tf(x)以T為周期的函數(shù)且0f(x)dx0f(x)任意原函數(shù)F(x)以T若lima 則lima1a2Lann 若limaa且an

則limna1a2Lan若limana且a 則 a a

但要注意：若 a且a0，不能推出liman 反例：an2(n為偶數(shù)3(n為奇數(shù)

n

必在函數(shù)yx上不一定都在函數(shù)yx上例如：yx2,階乘不等式在極限證n

nn!e(( 應(yīng)用：證明limn!nn證明:n!e2)

,n

li n證明liman0(a為任意實數(shù)na0,0|an|| e

|a|e|() n |a|e0,(|a|e)n 根據(jù)準(zhǔn)則

an

limnn1設(shè)p2且p為實常數(shù)，則nnp()yf(x)滿足f(a)f在區(qū)間(a,b)內(nèi)至少存在一點使得f'(yf(x)在區(qū)間(a,b)內(nèi)可導(dǎo)

f(x)

f在區(qū)間(a,b)內(nèi)至少存在一點使得f'(yf(x)滿足在區(qū)間(a,b)內(nèi)至少存在一點使得f'(f(x)及F(x)滿足

f(b)fb在區(qū)間(a,b)內(nèi)至少存在一點f(bf(a)F(b)F

f'(1.f(x)存在原函數(shù)，但其不一定可積，例如f(x)1x(0,xf(x)在[ab]上可積，但f(x) P(x是既約真分式，Q(x)在復(fù)數(shù)范圍內(nèi)可以分解為(xa)n(xa)nL(xa)n,

b b L ]

] (xa (xa)n1 (xa (xa (xa)n2 (xa

]

(xa (xa)ni (xa (xa (xa)nr (xa 其中bj(i12,Lr;j1,2,Ln

f(j1)(a 設(shè)fi(x(xa)n(xa)nL(xa)n(xa)n

,且bii

(j (x1)(x

解：令f(x) ,則f(1) (x f(x)

,則

'(1)3,

(1)2 = (x1)(x 4x (x x 2xx(x1)(x

解：f(x) 2x ,f(0) (x1)(x f(x)2x7 x(x

f(1) f(x)2x7 f(3) x(x 2x =7 x(x1)(x 4(x 12(x9x324x2 (x1)(x9x324x2解：f1(x)

(x9x324x2

,f1(1)

ff2(x)

,f2(2)24,f2'(2)12, (x f2'''(2)9x324x248x

(x1)(x (x (x (x )) a 2xln(1ex

ff(x)f 1