求大一上学期高数导数公式全部 谢谢

2025-03-15 07:48:29
推荐回答(1个)
回答1:

高等数学公式
1导数公式:
(tgx)sec2x(ctgx)csc2x(secx)secxtgx(cscx)cscxctgx(ax)axlna
1
(logax)
xlna
2基本积分表:
(arcsinx)
1
x2
1
(arccosx)
x21
(arctgx)
1x2
1
(arcctgx)
1x2
tgxdxlncosxCctgxdxlnsinxC
secxdxlnsecxtgxCcscxdxlncscxctgxC
dx1x
arctgCa2x2aadx1xa
lnx2a22axaCdx1ax
a2x22alnaxCdxx
arcsinCa2x2
a

2
n
dx2
cos2xsecxdxtgxCdx2
sin2xcscxdxctgxC
secxtgxdxsecxCcscxctgxdxcscxC
ax
adxlnaC
x
shxdxchxCchxdxshxC
dxx2a2
ln(xx2a2)C

2
Insinxdxcosnxdx
n1
In2n

x2a22
xadxxaln(xx2a2)C
22x2a2222
xadxxalnxx2a2C
22x2a2x222
axdxaxarcsinC
22a
2
2
3三角函数的有理式积分:
2u1u2x2du
sinx, cosx, utg, dx
21u21u21u2
1/14
4一些初等函数: 5两个重要极限:
exex
双曲正弦:shx
2exex
双曲余弦:chx
2
shxexex
双曲正切:thx
chxexexarshxln(xx21)archxln(xx21)
11x
arthxln
21x
6三角函数公式: ·诱导公式:

lim
sinx
1
x0x
1
lim(1)xe2.718281828459045...xx
7·和差角公式: 8 ·和差化积公式:
sin()sincoscossincos()coscossinsintg()
tgtg1tgtgctgctg1
ctg()
ctgctg
sinsin2sin

22
sinsin2cossin
22
coscos2coscos
22
coscos2sinsin
22
cos

2/14

9·倍角公式:
sin22sincos
cos22cos2112sin2cos2sin2ctg21
ctg2
2ctg2tg
tg2
1tg2
10·半角公式:
sin33sin4sin3cos34cos33cos3tgtg3tg3
13tg2
sintg

2

coscos            cos222
1cos1cossincos1cossin
  ctg
1cossin1cos21cossin1cos
abc
2R 12·余弦定理:c2a2b22abcosC sinAsinBsinC

2
11·正弦定理:
13·反三角函数性质:arcsinx

2
arccosx   arctgx

2
arcctgx
14高阶导数公式——莱布尼兹(Leibniz)公式:
(uv)
(n)
k(nk)(k)
Cnuvk0
n
u(n)vnu(n1)v
n(n1)(n2)n(n1)(nk1)(nk)(k)
uvuvuv(n)
2!k!
15中值定理与导数应用:
拉格朗日中值定理:f(b)f(a)f()(ba)f(b)f(a)f()

F(b)F(a)F()
16曲率:
当F(x)x时,柯西中值定理就是拉格朗日中值定理。
3/14
弧微分公式:dsy2dx,其中ytg平均曲率:K

:从M点到M点,切线斜率的倾角变化量;s:MM弧长。s
yd
M点的曲率:Klim.
23s0sds(1y)
直线:K0;1
半径为a的圆:K.
a
17定积分的近似计算:
b
矩形法:f(x)
ab
ba
(y0y1yn1)

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