高數(shù)學(xué)研教育導(dǎo)數(shù)與微分

導(dǎo)數(shù)與微返關(guān)dyydyydxydy系limx0

dy導(dǎo)數(shù)的定設(shè)函數(shù)yfx)在點x0的,當(dāng)自變量x在x0處取得增量x(點x0x仍在該鄰域內(nèi))時,相應(yīng)地函數(shù)y取得增量yf(x0x)f(x0如果y與x之比當(dāng)x0時的極限存在,則稱數(shù)yfx)在點x0處可導(dǎo),并稱這個極0yfx)在點x處的導(dǎo)數(shù),記為0

x

,

xx0或df

limylimf(

x)f(x0x單側(cè)導(dǎo)

x0

左導(dǎo)數(shù)

x

fxfx0x

xlimf(

x)f(x00右導(dǎo)數(shù)

fxlimfxfx x xlimf(

x)f(x0xfx0都.

0與(C)(sinx)cosx(tanx)sec2x(secx)secxtgx(ax)axln

（常數(shù)和基本初等函數(shù)的導(dǎo) (x)(cosx)sinx(cotx)csc2x(cscx)cscxctgx(ex)ex x) (lnx) xln x) x)求導(dǎo)法

11x1x

(arccosx)(arccotx)

11x211x函數(shù)的和,差,積,商的求導(dǎo) uvu cu

uvuv

uv v

uv反函數(shù)的求導(dǎo)如果函數(shù)x(y)的反函數(shù)為y f(x),則f(x)

復(fù)合函數(shù)的求導(dǎo)設(shè)yf(u),而u(x),則對復(fù)合函數(shù)y fdydydu

或yx)f(u對數(shù)求多個函數(shù)相乘和冪指函數(shù)u(x)v(x)的情形隱函數(shù)求導(dǎo)y若參數(shù)方程(t)確定ydy

(t高階導(dǎo)數(shù)(二階和二階以上的導(dǎo)數(shù)統(tǒng)稱為高階導(dǎo)數(shù)二階

f(x)limf(xx)f(x),也可記作y

d2 d2f( 一般,函數(shù)fx)的n1fx)的n階導(dǎo)數(shù),記f(n

(

y(n

dn dnf( 微分設(shè)函數(shù)yfx)在某區(qū)間內(nèi)有定義x0及x0屬于此區(qū)間,且yf(x0x)f(x0)Axo(x)(其中A是與x無關(guān)的常數(shù)),則稱函數(shù)y f(在點x0可微,并且稱Ax為函數(shù)y f(x)在點相應(yīng)于自變量增量x的微分,記作 或df(x0x 0dyx

Axfx0 (微分的實質(zhì)0定理:函數(shù)f(x)在點x0可微的充要條f(x)在點x0處可導(dǎo),且A f(x0微分的計dyf( 微分無論x是自變量還是中間變量,微分形式dydyf(例1.設(shè)fxxx1x2)x100),求f解:f(0limfxf

xlim(x1)(x2)(x或fxx1x100x,其中x由x1x2,x100的積及和構(gòu)成,且00.從而可得:f0100!例 設(shè)f'(x0)存在求 f(x

x)f(x0x)

(解lim[f

x)f(x0f(x0)f(x0x)]lim[f(

x)f(x0f(x0x)f(x0)f'

)f'(x0g(x)sin x例 設(shè)f(x)

x

且g(0g0)求f解f(0)

f(x)fx

g(x)sin x

0 0又g(0)limgxg(0)limgx) 所以f'(0)limg(x)sin 1e x例 已知f(x)

，求fxx

x解xx

f'(x)2x x f'(x)exf(0)f'(0) f(0x)f(0)

(x)2 f'(0)

f(0x)f(0)

x0

1e x0

x0 limxx0 f(0)

e x2 x例 確定a，使曲線yax與曲線ylnx相切 設(shè)切點為(x0,y0曲線yax的斜率k1a曲線ylnx的斜率 1x2x a，又切點必須x01x0 ay0 y 0 lny 0例 已知y f((x))，求y解設(shè)ux)，則y

f'(u)'(xyyy

f''(u)['(x)]2f'(u)''(x f'''(u)['(x)]32f''(u)'(x)''(xf''(u)'(x)''(x) f'(u)'''(x f'''((x))['(x)]3f''((x))'(x)''(xf'((x))'''(x例 已知yemarcsinx[cos(marcsinx)sin(marcsinx)]解設(shè)umarcsinx則yeu(cosusinudyeu(cosusinu)eu(sinucosu2eucosu2emarcsinxcos(m xdu 1xdydydu 2m

em2

xcos(marcsinx例 設(shè)y解u

11x1x2,

14

1x11x1xy1arctanu1lnu u1arctanu1ln(u1)ln(u2y

1

) 2(1u2 u u1x1x21x

1x

2x21x1x

) (xy (x例

x2t

,

t0y5t24t 解:當(dāng)t時t不可導(dǎo),故t0時不能 求導(dǎo)

dxdy不存在 lim

5(t)24t

lim5t4t02sgn(t

t

2tlimt[54sgn(t)]故

t

設(shè)函數(shù)yf(x由方程 yyx(x0,yd2所確定,

解兩邊取對數(shù)

lnyx

lnx,即ylnyxlny兩邊對x

y'lnyy1y'lnxx 即(1lnyylnx

ylnx1lny

1(lny1)(lnx1)

(1lnny(lny1)2x(lnxn 設(shè)f(x)xx(x2),求f(解:先去掉絕對值,x2(x2),xf(x)

x2(x2),0x當(dāng)x0時

x2(x

2),xf(0)limf(x)f(0)

x2(x0x0

x

x0 f(0)limf(x)f

x2(x x0f(0)

x

x0 當(dāng)x2或x0時

f(x)3x24當(dāng)0x2時當(dāng)x2時

f(x)3x24f(2)limf(x)f(2)

x2(x2)

x

x f(2)limf(x)f

x2(x2) x2

x

x f(2f(2fx)在x2處不可導(dǎo)3x24xx2,或xf(x)0,x 4x2

( 設(shè)y

x2

,求 4x2

4x2434 解:y

x21

x2

x243 x

)(n

2x x(1)n (x

1x

(1)n)(n)(xy(n

3(1)n (x (x練習(xí) 設(shè)yx(sinx)cosx,求 設(shè)yf(sin2xf(cos2x),求 設(shè)y (x),(x)0且(x)1，(x)(xx),x)可導(dǎo)，求3 設(shè)y x2

x0),求 設(shè)yx(sinx)cosx,求解利用對數(shù)求導(dǎo)法ln|y|lnxcosxlnsin1y(lnxcosxlnsinx)1y

sinxlnsinx

cos2 sinyy1

cos2x sinxlnsinx1

sinx

cos2x(sinx)cosxx

sinxlnsinx

sin設(shè)yf(sin2xf(cos2x),求解

f'(sin2x)sin2xf'(cos2x)2cosx(sinsin2x[f'(sin2x)f'(cos2 設(shè)y (x),(x)0且(x