一、各類函數(shù)的導(dǎo)數(shù)幾何意kfx0切線yy0fx0xx0法線方程：y (xx f(x)0 f(x00 可導(dǎo)必連續(xù)，但是連續(xù)未必可導(dǎo).可導(dǎo) ：(uv)u (uv)uvuuvuv

熟記導(dǎo)vv vv 微 dyf(

變限積分函數(shù)d

f(t)dtf(

d

f(t)dtf( d(x

(x (x (x

f(t)dtf((x))(f(t)dtf((x))(x)f((x))(Previous 例計(jì)算

arctan x0ln13xsin(2

等價(jià)無窮解

xarctan

xarctan2x0ln13x

60

arctan

等價(jià)無 12

Previous

ttxtt

)dtx2

02t(tsint

等價(jià)無解

x

(tanxx)2tttt

02t(tsint

1

2x(xsin limtanxx

x xxo(x)

xsin

x(x1x3o(x33!1

o(x3lim

13!

o(x3

Previous

lim1xet2dt

并求極x0

t

x

x0 0 lim1xet2

limexx0

1 t

湊形

x

x0

e0

t

1

xet2dt lim1 x0

dt1

0等價(jià)無窮etdt lim

ex2

ex0

ex0

3

ex03x

ePrevious 例計(jì)算

x(arccost)2dt.(11x211x2x(arccost)2dt0limu0

u(arccost)2dt11u2

u(arccost)2dt0u

(arccos1

π2Previous 例f(x連續(xù)f(23，計(jì)算函x x

2f(u)du 2解2

(xx x

2f(u)du

0

x f

(x

2(x f( f 0 Previous 例設(shè)函

Fx

0(xt)f(tx求dx解

(注意區(qū)分積分變量和自變量x(xt)f(t)dtx dxf(t)dtxtf(t)dt x f(t)dtxf(x)xf(xx f(txPrevious 例設(shè)函數(shù)f(x)連續(xù)，x>0， f

f(t)dtx2(1x)1解xf(x2)2x2x(1x)取x 2, 2f(2)2

2(1 2)f(2)12

2Previous 2

x

sinxtant2dt,

gxxsin則當(dāng)x0時(shí)，fx與gx階的比

fxgx

sinxtant2dtxsin

tansinx2cos

sinx

12

x012故當(dāng)x0時(shí)，fx與gx是同階但非等價(jià)Previous例設(shè)函f(x及其反g(x都可微，且成f(xg(t

2(x3

f(x的表 3解等式兩xgfx)]fx2 變上限積分函數(shù)即xfx)2

原函數(shù)和反函數(shù) fx)2 故fx)x2又f(x2x2C-f(x)x2PreviousNext復(fù)合函數(shù)求導(dǎo)法則(鏈?zhǔn)椒▌t)由外到內(nèi)，層層yf ug(x)yf[g(dyf(u)g( dydy 例設(shè)yxeπsinxcos(sinx求dy y(x)eπsinxπcosxcos(sineπsinx(sin(sinx))cos y(0)

y(0)dx

PreviousNext分段函數(shù)的導(dǎo)

sin例證明函fx)

,xx

() x可導(dǎo)，并研fxx0處的連續(xù)性x0時(shí)fx)xcosxsinx0時(shí)，f(0)limfxf xlimsinxxlimcosx

2lim2 x02

PreviousNextxcosxsinx x f(x)

xlimf(x)limxcosxsin

limxsin 20ffxx0點(diǎn)處連續(xù)注fx0fx0PreviousNext例當(dāng)0x f(x)0,f(x)連續(xù)，xtf(tx xg(x)

f(t

抽象 x(1)計(jì)算 (2)證明g(x)在[0,)上單調(diào)增加解(1)g(0)

g(x)x

0tf(t)dtxf(t)dtxx0xxlim xf( limf(x)xf(x)xx0x

f(t)dtxf(

x02f(x)xf( 0PreviousNext因?yàn)間(0存在，則g(x在點(diǎn)x0處右連續(xù)，從而g(x)在[0,)上連續(xù).x0時(shí) g(x)

xf(x)0 f(t)dtf(

tf(t 2x

f(t)dtf(

(xt)f(t 2

f(t)dt故證明g(x)在[0,上單調(diào)增加

PreviousNext例設(shè)函f(x為可導(dǎo)函數(shù)limf(a)f(ax) 2則曲yf(x在點(diǎn)(af(a處切線斜率解利用導(dǎo)數(shù)定義和導(dǎo)數(shù)幾何意義 limf(af(ax)1limf(axf(a) 2

x f(a)

1f(a)1

PreviousNext隱函數(shù)求導(dǎo)法 導(dǎo)例設(shè)函yy(x由方程ex解方程兩x

xy

所確定exy(1y)yxy

y

(yexyxex (1xexy(1y)2exyyyyxyPreviousNext則

(exy(1y)22y)xexyx0時(shí)，代入原方y(tǒng)進(jìn)y(0)

y(0)2ePreviousNext例設(shè)y cosx,求注求導(dǎo)過程中含有冪指函數(shù) yeln

esinxlnxcos yesinxlnx(cosxlnxsinx1)sinxxsinx(cosxlnxsinx)sinxPreviousNext(x1)((x1)(xx5x2

解兩邊先取絕對(duì)ln|y|1ln|x1|ln|x5|ln|x|1ln(

2兩邊x求導(dǎo)

1y1

2 x5x5

x2y

1

2 (x1)(x5 x2 x x(x1)(x5 x2PreviousNexty例設(shè)函數(shù)y=y xy2

確定0由求 (0,0)解方程兩x求12yycos2(yx)(y則y

cos2(yx)12ycos2(y y(0)

(0,0)

y(0)dx

PreviousNext參數(shù)方程求導(dǎo)法則分別對(duì)t11sin2x例設(shè)曲線方cos22tπ處的切線方程2

求此曲線 dx12sintcostsintcost, dy2costsint, 1sin2t 1sin2t dy2(1sin2t)tπxln 2, y0切線y04x y4x 2.

t22

PreviousNextdyddydxdd2y

d

dydx

ddt

dydtdx

dtdy2(1sin2t) ddy4sintcostdtdxd2 4sintcos

4(1sin2t 1sin2

PreviousNext例1cos在對(duì)應(yīng)點(diǎn)π處切線方程2解π所對(duì)應(yīng)02

(點(diǎn)+斜率曲線的參數(shù)方

x(1cos)cosy(1cos)dy

sin2cos(1cosdx

sin(1cos)sincoscoscos2sinsin切線斜率kdydx

2

1,切線方

yxPreviousNext二、高階導(dǎo)數(shù)常用高階導(dǎo)(ax)(n)axlnn (a (ex)(n)ex(sinkx)(n)knsin(kxn2(coskx)(n)kncos(kxn2(x)(n)(1)(n1)xn(ln(ax))(n)(1)n1(n1)!,(a

a

)(

(a(uv)(n)

C (nCn

v(knCk n

k n(n1)(nkk!(nk k

PreviousNext例設(shè)函數(shù)fx)sin4xcos4x,f(n f(x)(sin2xcos2x)(sin2xcos2cos2 f(n)(x)2ncos(2xn2PreviousNext三、函數(shù)可能極值點(diǎn)：駐2例求fxx1x

的極值解D(2f(x)x

2(x1)x1333

1x1(5x3335x5

2

(x)

當(dāng)x0時(shí)fx不存在5x0時(shí)，fx0;當(dāng)0x2fx5x

2

(x)525故fx)極大值f(0

fx)極小值f(23(23 PreviousNext例設(shè)函數(shù)y=y(例設(shè)函數(shù)y=y(x)

teu20所確定，求函數(shù)的極值

dx

et

dy2te

t

, dy

2tet2

令dy 則t 即x

ed2又

e

d2dd2dx2

2所以函yy(xt0對(duì)應(yīng)x3點(diǎn)

PreviousNext四、函數(shù)單sinx

x例證明

則

f(x)sinxxx2xf(x)cosx12f(x)sinxxfx單調(diào)遞增，fxf(0f(x單調(diào)遞fx)f(0即結(jié)論成立PreviousNext例設(shè)0abπ 證明sinasinb

fx)sinx在x

π]上單2f(x)xcosxsin設(shè)g(x)xcosxsin x

(0,π]2gx)xsinxπ所以g(x(0,2]上單調(diào)遞從 g(x)g(0)PreviousNext于 f(x)xcosxsinxπf(x(0,2]上單調(diào)遞 f(a)f sinasinb PreviousNext例證明：當(dāng)x0，成立1 1 ex 證明：原不等式xexx2ex2設(shè)g(x)xexx2ex 則g(0)g(x)xexex g(0)g(x)xex0g(x)g(0)g(x)g(0)xexx2ex2PreviousNext五、中值設(shè)函f(x[0,1]上連(0,1)內(nèi)可f(0)1f(1)1, fx)0, (0,1), f() f(想辦法構(gòu)造g(xgx)fx)f(g(x) f(x)

PreviousNextf(x)證明構(gòu)造函數(shù)g(x)f(x)則函g(x[0,1]上連(0,1)內(nèi)可導(dǎo)，且g(0)g(1)1由羅爾定理知， g()f( f()f( f( f().注和驗(yàn)證定理的

PreviousNext例函數(shù)f(x)esinx在點(diǎn) 的二階泰 π解f )2

f(x)esin

cos

f(π)2f(x)esinx(cosx)2esin

sin

f(π)2f(x)esinx(cosx)33esinxcosxsinxesinxcosesinx(cosx)3cosx3sinxcosxf(x)ee(xπ)2

xπ21esin(cos)3cos3sincos(xπ)32 Previous五、其ysinxxπ處的曲率2 ysin3K |y