高数学研教育导数与微分

导数与微返关dyydyydxydy系limx0

dy导数的定设函数yfx)在点x0的,当自变量x在x0处取得增量x(点x0x仍在该邻域内)时,相应地函数y取得增量yf(x0x)f(x0如果y与x之比当x0时的极限存在,则称数yfx)在点x0处可导,并称这个极0yfx)在点x处的导数,记为0

x

,

xx0或df

limylimf(

x)f(x0x单侧导

x0

左导数

x

fxfx0x

xlimf(

x)f(x00右导数

fxlimfxfx x xlimf(

x)f(x0xfx0都.

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

（常数和基本初等函数的导 (x)(cosx)sinx(cotx)csc2x(cscx)cscxctgx(ex)ex x) (lnx) xln x) x)求导法

11x1x

(arccosx)(arccotx)

11x211x函数的和,差,积,商的求导 uvu cu

uvuv

uv v

uv反函数的求导如果函数x(y)的反函数为y f(x),则f(x)

复合函数的求导设yf(u),而u(x),则对复合函数y fdydydu

或yx)f(u对数求多个函数相乘和幂指函数u(x)v(x)的情形隐函数求导y若参数方程(t)确定ydy

(t高阶导数(二阶和二阶以上的导数统称为高阶导数二阶

f(x)limf(xx)f(x),也可记作y

d2 d2f( 一般,函数fx)的n1fx)的n阶导数,记f(n

(

y(n

dn dnf( 微分设函数yfx)在某区间内有定义x0及x0属于此区间,且yf(x0x)f(x0)Axo(x)(其中A是与x无关的常数),则称函数y f(在点x0可微,并且称Ax为函数y f(x)在点相应于自变量增量x的微分,记作 或df(x0x 0dyx

Axfx0 (微分的实质0定理:函数f(x)在点x0可微的充要条f(x)在点x0处可导,且A f(x0微分的计dyf( 微分无论x是自变量还是中间变量,微分形式dydyf(例1.设fxxx1x2)x100),求f解:f(0limfxf

xlim(x1)(x2)(x或fxx1x100x,其中x由x1x2,x100的积及和构成,且00.从而可得:f0100!例 设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 设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 确定a，使曲线yax与曲线ylnx相切 设切点为(x0,y0曲线yax的斜率k1a曲线ylnx的斜率 1x2x a，又切点必须x01x0 ay0 y 0 lny 0例 已知y f((x))，求y解设ux)，则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)]解设umarcsinx则yeu(cosusinudyeu(cosusinu)eu(sinucosu2eucosu2emarcsinxcos(m xdu 1xdydydu 2m

em2

xcos(marcsinx例 设y解u

11x1x2,

14

1x11x1xy1arctanu1lnu u1arctanu1ln(u1)ln(u2y

1

) 2(1u2 u u1x1x21x

1x

2x21x1x

) (xy (x例

x2t

,

t0y5t24t 解:当t时t不可导,故t0时不能 求导

dxdy不存在 lim

5(t)24t

lim5t4t02sgn(t

t

2tlimt[54sgn(t)]故

t

设函数yf(x由方程 yyx(x0,yd2所确定,

解两边取对数

lnyx

lnx,即ylnyxlny两边对x

y'lnyy1y'lnxx 即(1lnyylnx

ylnx1lny

1(lny1)(lnx1)

(1lnny(lny1)2x(lnxn 设f(x)xx(x2),求f(解:先去掉绝对值,x2(x2),xf(x)

x2(x2),0x当x0时

x2(x

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

x2(x0x0

x

x0 f(0)limf(x)f

x2(x x0f(0)

x

x0 当x2或x0时

f(x)3x24当0x2时当x2时

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处不可导3x24xx2,或xf(x)0,x 4x2

( 设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练习 设yx(sinx)cosx,求 设yf(sin2xf(cos2x),求 设y (x),(x)0且(x)1，(x)(xx),x)可导，求3 设y x2

x0),求 设yx(sinx)cosx,求解利用对数求导法ln|y|lnxcosxlnsin1y(lnxcosxlnsinx)1y

sinxlnsinx

cos2 sinyy1

cos2x sinxlnsinx1

sinx

cos2x(sinx)cosxx

sinxlnsinx

sin设yf(sin2xf(cos2x),求解

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