第二章導數(shù)與微分 一、變速直線運動的瞬時sf(t)t0t0t這段時間內質點的平均速度vsf(t0t)f(t0 t0時刻的0vlimf(t0t)f(t0 0

f(t0

f(t0st0

sf(t0t)f(t0曲線的切線斜 0旋轉而趨向MT，直線0

yf(x) MMN0,NMT

0xxxM(x0,f(x0N((x0xf則割線MN的斜率為 fN曲線,x0

x)f(x0所以切線MT的斜率為：tan f

x)f(x0)vlimf(t0t)f(t0

f(t0t

f(t0t) 瞬時速度 klimf(x0xf(x0

yf(xN 兩個問題的共性

0 xx0limf(x0x)f(x0) yf(x0xf(x0),limyx0 設函數(shù)y=f(x)在點x0的某鄰域內有定義limf(x0x)f(x0)lim

x0存在，則稱函數(shù)f(x)在點x0處可導，并稱此極限為函數(shù)y=f(x)在點x0的導數(shù). 0yxx0

f(x0)

dxx

df(x)

x

tt

f(t0曲線在點(x0,f(x0導數(shù)定義的等價形式若記xx0x 則xxx0當xx0時，x

x

f(x0f(x)limf

x)f(x0 limf(x)f(x0 limf

h)f(x0h

xI若函數(shù)y=f(x)在I內可導，則對于xI f(xxf(x)I上的函數(shù)，f(xy=f(x的導函數(shù)導數(shù)，記作

y f(x) dy df 注意f(x0f

x

df(x0①、yf(xxf(x②、求兩增量的

x

f(xx)f(x)③、ylimyx0例 求函數(shù)f(x)C(C為常數(shù))的導數(shù)解limf(xx)f limCC C

求函數(shù)yx3在x=1處的導數(shù)解

(1x)3

3x3(x)23

例 求函數(shù)f(x)xn(nN)在x=a處的導數(shù)n limf(x)f limn

x x

x

xlim(xn1axn2a2xxnan1yx(為常數(shù)(x)

x x

x (x2x

x 11x (xx

3

xx xx

4 x 例 求函數(shù)f(x)sinx的導數(shù) limf(xh)f limsin(xh)sin

lim2cosxhsin

x

2

cos2 22

2 即

(sinx)cos(cosx)sin(ex)(lnx)x例 求函數(shù)f(x)a(ex)(lnx)x解limf(xhf(x)

axh

h ah

a

ax axln

例 求函數(shù)f(x)logax(a0,a1)的導數(shù) limf(xh)f(x) limloga(xh)loga a x x log1a x x lim lim

設函數(shù)y=f(x)在點x0的左側[x0(x0)有定義，若極限

x,x0]limy

f(x0x)f(x0x0 x0 f(xx0的f(x0)f(x0)x0

f(x0x)f(x0定理1yf(x在點x0處可導的充分必要條件是函數(shù)yf(xx0處的左、右導數(shù)都存在且相等.f(x0)f(x0)說明yf(x(abf(af(b都存在，則稱f(x[ab上可導例 證明函數(shù)f(x)|x|在x=0處不可導f(0h)fh |h|f(0h)fh hlimf(0h)f(0)1f limf(0h)f(0)1f

f(0)f(xx0處不可導四、導數(shù)的幾何意yyf(xx0

Cyf(xTx TxM(x0,y0) f(x0

x0yf(xM(x0,y0處的切線方程yy0f(x0)(xx0M(x0,y0yf(xM處的f(x00

yy0

f(x0

(x

(f

)說明①、f(x00

xyy0②、若f(x0,xxx0例 求等邊雙曲線y1在點1,2處的切線2 2 解y1x2

x2

x2

2y24x 2 4 y21x4

4xy4即2x8y1532 問曲線y3x哪一點有垂直切線？哪一點處的切線與直線y1x1平行？寫出其切線方程.32y(y(x

1x3

33

故在原點(00x3令33

1, x1,y1,3則在點(1,1),(1,1)處與直線y1x1平行的切線方程 y11(x1), y11(x1)3 x3y2五、函數(shù)的可導性與連續(xù)性的關定理 如果函數(shù)y=f(x)在點x0處可導，則它注意該定理的逆命題不成立 y|x|x0處連續(xù)，但不可導.limf(x)limx0 limf(x)lim(x)0 這說明函數(shù)y|x|x0處連續(xù)，7x=0處不可導.

yy|x 例 f(x)xsinx

x xx0處的解sin1xlimxsin10 由limf(x)0f(0f(xx0處連續(xù)x0 (0

0

sin x0時，極限不存在，f(xx0推論yf(xx0x0處不可導例 sinx

xf(x)

xx0處的解

limsinx10f f(xx0處不連續(xù)，x0處不導數(shù)的實質增量比的極限f(x0) :可導必連續(xù),但連續(xù)不一定可導(logx)

(x)(lnx) xln (ax)axlna

(ex)

ex

不連續(xù),一定不可導.作 1、f(x0)f(x0h)f(x0)

f(x

f(x0h)f