高数上复习题2导数与微分

一、各类函数的导数几何意kfx0切线yy0fx0xx0法线方程：y (xx f(x)0 f(x00 可导必连续，但是连续未必可导.可导 ：(uv)u (uv)uvuuvuv

熟记导vv vv 微 dyf(

变限积分函数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 例计算

arctan x0ln13xsin(2

等价无穷解

xarctan

xarctan2x0ln13x

60

arctan

等价无 12

Previous

ttxtt

)dtx2

02t(tsint

等价无解

x

(tanxx)2tttt

02t(tsint

1

2x(xsin limtanxx

x xxo(x)

xsin

x(x1x3o(x33!1

o(x3lim

13!

o(x3

Previous

lim1xet2dt

并求极x0

t

x

x0 0 lim1xet2

limexx0

1 t

凑形

x

x0

e0

t

1

xet2dt lim1 x0

dt1

0等价无穷etdt lim

ex2

ex0

ex0

3

ex03x

ePrevious 例计算

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

u(arccost)2dt11u2

u(arccost)2dt0u

(arccos1

π2Previous 例f(x连续f(23，计算函x x

2f(u)du 2解2

(xx x

2f(u)du

0

x f

(x

2(x f( f 0 Previous 例设函

Fx

0(xt)f(tx求dx解

(注意区分积分变量和自变量x(xt)f(t)dtx dxf(t)dtxtf(t)dt x f(t)dtxf(x)xf(xx f(txPrevious 例设函数f(x)连续，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则当x0时，fx与gx阶的比

fxgx

sinxtant2dtxsin

tansinx2cos

sinx

12

x012故当x0时，fx与gx是同阶但非等价Previous例设函f(x及其反g(x都可微，且成f(xg(t

2(x3

f(x的表 3解等式两xgfx)]fx2 变上限积分函数即xfx)2

原函数和反函数 fx)2 故fx)x2又f(x2x2C-f(x)x2PreviousNext复合函数求导法则(链式法则)由外到内，层层yf ug(x)yf[g(dyf(u)g( dydy 例设yxeπsinxcos(sinx求dy y(x)eπsinxπcosxcos(sineπsinx(sin(sinx))cos y(0)

y(0)dx

PreviousNext分段函数的导

sin例证明函fx)

,xx

() x可导，并研fxx0处的连续性x0时fx)xcosxsinx0时，f(0)limfxf xlimsinxxlimcosx

2lim2 x02

PreviousNextxcosxsinx x f(x)

xlimf(x)limxcosxsin

limxsin 20ffxx0点处连续注fx0fx0PreviousNext例当0x f(x)0,f(x)连续，xtf(tx xg(x)

f(t

抽象 x(1)计算 (2)证明g(x)在[0,)上单调增加解(1)g(0)

g(x)x

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

f(t)dtxf(

x02f(x)xf( 0PreviousNext因为g(0存在，则g(x在点x0处右连续，从而g(x)在[0,)上连续.x0时 g(x)

xf(x)0 f(t)dtf(

tf(t 2x

f(t)dtf(

(xt)f(t 2

f(t)dt故证明g(x)在[0,上单调增加

PreviousNext例设函f(x为可导函数limf(a)f(ax) 2则曲yf(x在点(af(a处切线斜率解利用导数定义和导数几何意义 limf(af(ax)1limf(axf(a) 2

x f(a)

1f(a)1

PreviousNext隐函数求导法 导例设函yy(x由方程ex解方程两x

xy

所确定exy(1y)yxy

y

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

(exy(1y)22y)xexyx0时，代入原方y进y(0)

y(0)2ePreviousNext例设y cosx,求注求导过程中含有幂指函数 yeln

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

解两边先取绝对ln|y|1ln|x1|ln|x5|ln|x|1ln(

2两边x求导

1y1

2 x5x5

x2y

1

2 (x1)(x5 x2 x x(x1)(x5 x2PreviousNexty例设函数y=y xy2

确定0由求 (0,0)解方程两x求12yycos2(yx)(y则y

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

(0,0)

y(0)dx

PreviousNext参数方程求导法则分别对t11sin2x例设曲线方cos22tπ处的切线方程2

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

t22

PreviousNextdyddydxdd2y

d

dydx

ddt

dydtdx

dtdy2(1sin2t) ddy4sintcostdtdxd2 4sintcos

4(1sin2t 1sin2

PreviousNext例1cos在对应点π处切线方程2解π所对应02

(点+斜率曲线的参数方

x(1cos)cosy(1cos)dy

sin2cos(1cosdx

sin(1cos)sincoscoscos2sinsin切线斜率kdydx

2

1,切线方

yxPreviousNext二、高阶导数常用高阶导(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例设函数fx)sin4xcos4x,f(n f(x)(sin2xcos2x)(sin2xcos2cos2 f(n)(x)2ncos(2xn2PreviousNext三、函数可能极值点：驻2例求fxx1x

的极值解D(2f(x)x

2(x1)x1333

1x1(5x3335x5

2

(x)

当x0时fx不存在5x0时，fx0;当0x2fx5x

2

(x)525故fx)极大值f(0

fx)极小值f(23(23 PreviousNext例设函数y=y(例设函数y=y(x)

teu20所确定，求函数的极值

dx

et

dy2te

t

, dy

2tet2

令dy 则t 即x

ed2又

e

d2dd2dx2

2所以函yy(xt0对应x3点

PreviousNext四、函数单sinx

x例证明

则

f(x)sinxxx2xf(x)cosx12f(x)sinxxfx单调递增，fxf(0f(x单调递fx)f(0即结论成立PreviousNext例设0abπ 证明sinasinb

fx)sinx在x

π]上单2f(x)xcosxsin设g(x)xcosxsin x

(0,π]2gx)xsinxπ所以g(x(0,2]上单调递从 g(x)g(0)PreviousNext于 f(x)xcosxsinxπf(x(0,2]上单调递 f(a)f sinasinb PreviousNext例证明：当x0，成立1 1 ex 证明：原不等式xexx2ex2设g(x)xexx2ex 则g(0)g(x)xexex g(0)g(x)xex0g(x)g(0)g(x)g(0)xexx2ex2PreviousNext五、中值设函f(x[0,1]上连(0,1)内可f(0)1f(1)1, fx)0, (0,1), f() f(想办法构造g(xgx)fx)f(g(x) f(x)

PreviousNextf(x)证明构造函数g(x)f(x)则函g(x[0,1]上连(0,1)内可导，且g(0)g(1)1由罗尔定理知， g()f( f()f( f( f().注和验证定理的

PreviousNext例函数f(x)esinx在点 的二阶泰 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处的曲率2 ysin3K |y