专题3.1导数的概念及其意义、导数的运算九大题型（原卷版）_11.4-2_2026-11-04

      
      
专题3.1 导数的概念及其意义、导数的运算【九大题型】

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 2  %&'()
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 3  ,-./.01%203( ............................................................................................................3
 4  ,4-.567/.89:;67/.89 ............................................................................4
 5  </.=>
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 6  /.@
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 7  A@/.BC:DE? ....................................................................................................................5
 8  /.? ............................................................................................................................................6
 9  </.=FG? ........................................................................................................................6
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 y=f(g(x))
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 y=f(x)4 x=x0 Ü
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 1  
 !"#$
ă 1%2024·Ąą·ĆćĈĉ( lim
Δ𝑥→0
(2+Δ𝑥)3−23
Δ𝑥
= %   (
AĊ72
BĊ12
CĊ8
DĊ4
ºb 1-1%2024·ċČčĎ·ĆćĈĉ(ݔ)
𝑓(𝑥)4𝑥0Ü
‰𝑓′(𝑥0)]e lim
𝛥𝑥→0
𝑓(𝑥0)−𝑓(𝑥0−𝑚𝛥𝑥)
𝛥𝑥
Y¶
%    (
AĊ𝑚𝑓′(𝑥0)
BĊ−𝑚𝑓′(𝑥0)
CĊ−
1
𝑚𝑓′(𝑥0)
DĊ
1
𝑚𝑓′(𝑥0)
      
      
ºb 1-2%23-24 Îà·ċČďĐ·đ‹(â𝑓(𝑥)Ē4)
ēíî lim
Δ𝑥→0
𝑓(1)−𝑓(1−2Δ𝑥)
2Δ𝑥
= −1]e-.
𝑦= 𝑓(𝑥)57(1,𝑓(1))Ü/.01‰%    (
AĊ−1
BĊ−2
CĊ1
DĊ2
ºb 1-3%2024·ĄąĔĕĖ·ĆćĈĉ(ė !4𝑅5)
𝑓(𝑥)íî𝑓(0) = −1]#)
𝑓′(𝑥)íî𝑓′
(𝑥) > 𝑘> 1]e𝑓
1
𝑘−1 <
1
𝑘−1ĘęÄ6 p%    (
AĊ𝑓
1
𝑘−1 ≥
1
𝑘−1
BĊ𝑓
1
𝑘−1 ≤
1
𝑘−1
CĊ𝑓
1
𝑘−1 >
1
𝑘−1
DĊ𝑓
1
𝑘−1 <
1
𝑘−1
 2  %&'()
*+
ă 2%2024·ĚČě‹·ĆćĈĉ(ݔ)
𝑓(𝑥) = 2𝑥(𝑥−2)(𝑥−22)(𝑥−23)(𝑥−24)(𝑥−25)(𝑥−26)]e𝑓′(0)
= %    (
AĊ220
BĊ221
CĊ222
DĊ223
ºb 2-1%2024·ĚĜ·ÎĆ(ݔ𝑓(𝑥)‰ !4𝑅5ĝ)
]â𝑓′(𝑥)‰𝑓(𝑥))
]ė𝑓(𝑥) = 𝑓(2−𝑥)
+4𝑥−4]e𝑓′(2023) = %    (
AĊ1
BĊ−2023
CĊ2
DĊ2023
ºb 2-2%2024·ĞğċĠġĢ·ĆćĈĉ(â𝑓(𝑥) = sin𝑥]𝑓1(𝑥) = 𝑓′(𝑥)]𝑓2(𝑥) = 𝑓′1(𝑥)]⋯,𝑓𝑛+1(𝑥) =
𝑓′𝑛(𝑥)]
e
2024
𝑖=1 𝑓𝑖
π
6 Y¶%    (
AĊ0
BĊ 3
2
CĊ 3−1
2
DĊ
1
2
ºb 2-3%2024·ģĤĥ·ÎĆ(ݔ)
𝑓(𝑥),𝑔(𝑥) !Ħħ‰𝑅,𝑔′(𝑥)‰𝑔(𝑥))
]ē𝑓(𝑥) + 𝑔′(𝑥)
= 2]𝑓(𝑥)−𝑔′(2−𝑥) = 2]ė𝑔(𝑥)‰Ĩ)
]e𝑓(2023) + 𝑔′(2024) + 𝑔′(2025) = %     (
AĊ2
BĊ1
CĊ0
DĊ-1
 3  ,-./.01%203(
ă 3%2024·ĩĪīĬ·6Ć(ݔE.𝑙<-.𝑦= 𝑥3−𝑥4ĭ7ÜÑ/]e𝑙203‰%    (
AĊ
π
6
BĊ
π
4
CĊ
3π
4
DĊ
5π
6
ºb 3-1%2024·ĮįİĚ·ĆćĈĉ(ݔ-.𝑓(𝑥) = 2𝑥cos𝑥4𝑥= 0Ü/.‰𝑙]e𝑙01‰%    (
AĊln2
BĊ−ln2
CĊ1
DĊ−1
      
      
ºb 3-2%2024·ģıIJij·6Ć(ėE.𝑦= 𝑘𝑥+ 𝑛<-.𝑦= ln𝑥+
1
𝑥Ñ/]e k ĴGĵĶp%    (
AĊ −∞, 1
4
BĊ[4, + ∞)
CĊ[−4, + ∞)
DĊ 1
4 , + ∞
ºb 3-3%2024·ķĐ·ĆćĈĉ(â7𝑃p)
𝑓(𝑥) = 𝑥3−
1
2𝑓′(1)𝑥+ 𝑓′(2)[\5ĸJ67]7𝑃Ü/.
203‰𝛼]e3𝛼ĴGĵĶp%    (
AĊ 0, 3π
4
BĊ 0, π
2 ∪3π
4 ,π
CĊ π
2 , 3π
4
DĊ 0, π
2 ∪
3π
4 ,π
 4  ,4-.567/.89:;67/.89
ă 4%2024·mn·ĆćĈĉ(ݔ)
𝑓(𝑥) = 𝑥3−𝑓′(1)
2
⋅ln𝑥+ 3]e-.𝑦= 𝑓(𝑥)4(𝑒,𝑓(𝑒))Ü/.01
‰%    (
AĊ𝑒2−
1
2𝑒
BĊ3𝑒2−
1
2𝑒
CĊ𝑒2−
1
𝑒
DĊ3𝑒2−
1
𝑒
ºb 4-1%2024·ĮĹĺĻ·ĆćĈĉ(-.𝑦= 𝑥e𝑥+2𝑥−24𝑥= 0Ü/.89p%    (
AĊ3𝑥+ 𝑦+ 2 = 0
BĊ2𝑥+ 𝑦+ 2 = 0
CĊ2𝑥−𝑦−2 = 0
DĊ3𝑥−𝑦−2 = 0
ºb 4-2%2024·mn·ĆćĈĉ(;ĭ7†ˆ-.𝑦= 𝑓(𝑥) = 𝑥2−|𝑥| +1A@/.]eÁA@/.8
9‰%    (
AĊ𝑦= 𝑥c𝑦= −𝑥
BĊ𝑦= −3𝑥c𝑦= 3𝑥
CĊ𝑦= 𝑥c𝑦= −3𝑥
DĊ𝑦= −𝑥c𝑦= 3𝑥
ºb 4-3%2024·įļĜĽ·6Ć(;Þjĭ7ˆ-.𝑦= e𝑥−2 +1/.]e/.89‰%    (
AĊ𝑦= 𝑥
BĊ𝑦= 2𝑥
CĊ𝑦=
1
e2𝑥
DĊ𝑦= e𝑥
 5  </.=>
?
ă 5%2024·ľČķĿ·xĆ(ݔ-.𝑦= 𝑎𝑥e𝑥+ ln𝑥47(1,𝑎e)Ü/.89‰𝑦= 3𝑥+ 𝑏]e%    (
AĊ𝑎= e]𝑏= −2
BĊ𝑎= e]𝑏= 2
CĊ𝑎= e−1]𝑏= −2
DĊ𝑎= e−1]𝑏= 2
ºb 5-1%2024·ĮĹŀĐ·ÎĆ(ݔ-.𝑦= 𝑥ln𝑥+ 𝑎e−𝑥47𝑥= 1Ü/.89‰2𝑥−𝑦+ 𝑏= 0]e𝑏=
%    (
AĊŁ1
BĊŁ2
CĊŁ3
DĊ0
ºb 5-2%2024·mn·ĆćĈĉ(ݔ)
𝑓(𝑥) = 𝑎𝑥2 +𝑏ln𝑥[\47(1,𝑓(1))Ü/.8ł‰
𝑦= 3𝑥−1.e𝑎−𝑏G‰%    (
      
      
AĊ1
BĊ2
CĊ3
DĊ4
ºb 5-3%2024·mn·ĆćĈĉ(ݔ-.𝑓(𝑥) = 𝑒𝑥47𝑃(0,𝑓(0))Ü/.…p-.𝑔(𝑥) = ln(𝑎𝑥)6
@/.]e𝑎G‰%    (
AĊ
𝑒
3
BĊ
𝑒
2
CĊ𝑒2
DĊ𝑒3
3
 6  /.@
?
ă 6%2024·mn·ĆćĈĉ(ė-.𝑦= (1−𝑥)e𝑥=A@;7𝐴(𝑎,0)/.]e𝑎ĴGĵĶp%    (
AĊ(−∞,−1) ∪(3, + ∞)
BĊ(−3,1)
CĊ(−∞,−3)
DĊ(−∞,−3) ∪(1, + ∞)
ºb 6-1%2024·mn·ĆćĈĉ(ݔ)
𝑓(𝑥) = −𝑥3 +3𝑥]e;7(−3,−9)†ˆ-.𝑦= 𝑓(𝑥)/.
@
‰%    (
AĊ0
BĊ1
CĊ2
DĊ3
ºb 6-2%2021·mn·L(ė;7(𝑎,𝑏)†ˆ-.𝑦= e𝑥A@/.]e%    (
AĊe𝑏< 𝑎
BĊe𝑎< 𝑏
CĊ0 < 𝑎< e𝑏
DĊ0 < 𝑏< e𝑎
ºb 6-3%2024·mn·ĆćĈĉ(ė;7𝑃(𝑚,0)<-.𝑓(𝑥) =
𝑥+1
e𝑥Ñ/E.Ń= 2 @]e𝑚ĴGĵĶ
p%    (
AĊ(−∞, + ∞)
BĊ(−∞,−3) ∪(1, + ∞)
CĊ(−1,3)
DĊ(−∞,−1) ∪(3, + ∞)
 7  A@/.BC:DE?
ă 7%2024·ÔńŅņ·ĆćĈĉ(<-.𝑓(𝑥) =
1
2𝑥2 +𝑥+
1
4𝑏−
1
2c𝑔(𝑥) = 2ln(𝑥−1) + 1ŇÑ/E.𝑙<
E.𝑥+ 2𝑦+ 𝑎−1 = 0DE]e𝑏=%    
(
AĊ-8
BĊ-3
CĊ4
DĊ6
ºb 7-1%2024·ňʼnŊň·xĆ(ė)
𝑓(𝑥) = ln𝑥+ 𝑥<𝑔(𝑥) =
2𝑥−𝑚
𝑥−1 [\=6@ŋ/.]ēŌŋ
/.<E.𝑦= 2𝑥+ 1BC]eō
𝑚= %     (
AĊ
17
8
BĊ
17
6
CĊ
17
4
DĊ
17
2
ºb 7-2%2024·ŎċŏĐ·ĆćĈĉ()
𝑓(𝑥) = 𝑎𝑥+ sin𝑥[\5Ē4A@ÑŐDE/.]eō
𝑎
ĴGĵĶp%    (
      
      
AĊ{0,1}
BĊ{0}
CĊ[0,1)
DĊ[1, + ∞)
ºb 7-3%2024·ÔńŇ·6Ć(ݔ !4𝑅5)
𝑓(𝑥)[ő¶E.𝑥= 𝑎(𝑎＞0)µÀ，ē¥𝑥≥𝑎
²，𝑓(𝑥) = 𝑒𝑥
𝑒2𝑎，;7𝑃(𝑎,0)ˆ-.𝑦= 𝑓(𝑥)A@/.,ėÁA@/.ŐÑDE,eŌ)
𝑓(𝑥)FęG‰
%    ( 
AĊ𝑒−1
2
BĊ𝑒−1
CĊ𝑒−3
2
DĊ𝑒−2
 8  /.?
ă 8%2024·ĩĪ·ĆćĈĉ(ݔE.𝑦= 𝑘𝑥+ 𝑏Œp-.𝑦= ln𝑥/.]…p-.𝑦= −ln(−𝑥)/.]
e%œœ(
AĊ𝑘=
1
e]𝑏= 0
BĊ𝑘= 1]𝑏= 0
CĊ𝑘=
1
e]𝑏= −1
DĊ𝑘= 1]𝑏= −1
ºb 8-1%2024·ŔņĘŕ·6Ć(01‰1E.𝑙<-.𝑦= ln(𝑥+ 𝑎)cŖ𝑥2 + 𝑦2 =
1
2ŇÑ/]eō
𝑎
G‰%   (
AĊ0ŗ2
BĊ−2ŗ0
CĊ－1ŗ0
DĊ0ŗ1
ºb 8-2%2024·ċijĹZ·ĆćĈĉ(ė-.𝑓(𝑥) = 𝑎𝑥(𝑎> 1)<-.𝑔(𝑥) = log𝑎𝑥(𝑎> 1)=ēŃ=6·
ŋ7]ē4ŋ7Ü/.Ñó]eō
𝑎G‰%   (
AĊe
BĊe2
CĊe
1
e
DĊ e
ºb 8-3%2024·mn·ĆćĈĉ(ݔ)
𝑓(𝑥) = ln𝑥<𝑔(𝑥)[\¶E.𝑦= 𝑥µÀ]E.𝑙<𝑔(𝑥),ℎ(𝑥)
= e𝑥+1−1[\ħÑ/]e𝑙203‰%    (
AĊ
𝜋
6
BĊ
𝜋
4
CĊ
𝜋
3
DĊ
3𝜋
4
 9  </.=FG?
ă 9%2024·ÔńŘĻ·ĆćĈĉ(ݔ)
𝑓(𝑥) =
1+2tan 𝜔𝑥
2 −tan2 𝜔𝑥
2
1+tan2 𝜔𝑥
2
(𝜔> 0)]ė∃𝑥0 ∈−π
4 , π
3 úá𝑓(𝑥)[
\47(𝑥0,𝑓(𝑥0))Ü/.<𝑥řBC]e𝜔FęGp%    (
AĊ2
BĊ
3
2
CĊ1
DĊ
3
4
ºb 9-1%23-24 x5·ňʼn·ŚśŜŝ(ݔ)
𝑓(𝑥) = 𝑎𝑒𝑥<𝑔(𝑥) = ln𝑥+ 1Ē4/.]eō
 a 
FęG%    (
AĊ
2
𝑒
BĊ
1
𝑒
CĊ
1
2𝑒
DĊ
1
3𝑒
      
      
ºb 9-2%2024·mn·ĆćĈĉ(4BŞE3ÞjÄ𝑥𝑂𝑦‹]â-.𝑦=
𝑥+𝑎
e𝑥(𝑎> 1)4𝑥= 0Ü/.‰𝑙]
e𝑙<A@ÞjřşĶ[ƒŞŸFęG‰%    (
AĊ1
BĊ
3
2
CĊ2
DĊ
5
2
ºb 9-3%2024·Ŏċ·ĆćĈĉ(ݔE.𝑦= 𝑎𝑥+ 𝑏<-.𝑦= ln(e𝑥)Ñ/]e𝑎+ 𝑏FęG‰%    (
AĊ
1
2
BĊ1
CĊ 2
DĊ 3
6:g
1Ċ%2024·ŠįšĻ·ÎĆ(ݔ)
𝑓(𝑥) = 𝑥2 +
1
𝑥]e lim
Δ𝑥→0
𝑓(1+Δ𝑥)−𝑓(1)
2Δ𝑥
= %   (
AĊ1
BĊ
1
2
CĊ2
DĊ4
2Ċ%23-24 Îà·ĚĜ·ŚśŜŝ(ė lim
Δ𝑥→0
𝑓(−2+Δ𝑥)−𝑓(−2−Δ𝑥)
Δ𝑥
= −2]e𝑓′(−2) = %    (
AĊ1
BĊ-1
CĊ2
DĊ-2
3Ċ%2024·ĩĪŢĐ·xĆ(ݔ)
𝑓(𝑥) = ln𝑥+ 𝑥,𝑔(𝑥)p)
𝑓(2𝑥+ 1))
]e𝑔(0) = %    (
AĊ1
BĊ2
CĊ3
DĊ4
4Ċ%2024·5ţŤC·ÎĆ(ÛťŦŧĬK,ÑŨũŪūŬŭŮ]ůŰűŲjŨũKųđŴŵ:âŨ
ũŬŭůŰ»𝑊<²Ã t ĉ𝑊= 𝑓(𝑡)]− 
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
ĘęŶŷ4[𝑎,𝑏]Áś²ÃrŨũŬŭŮ`
ŸūŹ]ݔŴŵđr]o:źAŨũŬŭůŰ»<²Ãĸà[ş½.eàèŻ™żp%   (
  
AĊ4[𝑡1,𝑡2]Áś²Ãr]oŨũŬŭŮ`Ÿ÷źŨũŹÍ
BĊ4𝑡2²Ž]oŨũŬŭŮ`Ÿ÷źŨũŹÍ
CĊ4𝑡3²Ž]o:źAŨũŬŭůŰŇãŲjÍ
DĊoŨũ4[0,𝑡1]][𝑡1,𝑡2]][𝑡2,𝑡3]Áxś²Ã‹]4[𝑡1,𝑡2]ŬŭŮ`ŸFū
5Ċ%2024·ĮĹ·ĆćĈĉ(-.𝑦= 𝑥3
3 −247(−1,𝑎)Ü/.89‰%    (
      
      
AĊ3𝑥+ 3𝑦+ 4 = 0
BĊ3𝑥+ 3𝑦−4 = 0
CĊ3𝑥−3𝑦+ 4 = 0 DĊ3𝑥−3𝑦−4 = 0
6Ċ%2024·ĚČ·ĆćĈĉ(ݔ)
𝑓(𝑥) = (𝑎−3)𝑥3 + (𝑎−2)𝑥2 + (𝑎−1)𝑥+ 𝑎ėµĸJ𝑥0 ∈R]-.𝑦= 𝑓(𝑥)
47(𝑥0,𝑓(𝑥0))c(−𝑥0,𝑓(−𝑥0))Ü/.ŐÑBCŗĄ']eō
𝑎= %    (
AĊ0
BĊ1
CĊ2
DĊ3
7Ċ%2024·mn·ĆćĈĉ(ė;7(𝑚,𝑛)†ˆ)
𝑦= 2𝑥+
1
𝑥(𝑥> 0)[\A@/.]eq=%    (
AĊ0 < 2𝑚+
1
𝑚< 𝑛
BĊ0 < 𝑛< 2𝑚
CĊ2𝑚< 𝑛< 2𝑚+
1
𝑚
DĊ𝑛< 2𝑚
8Ċ%2024·ŔņŔĻ·ÎĆ(ėµ)
𝑓(𝑥) = 2𝑥−sin𝑥[\5ĸJ67Ü/.𝑙1])
𝑔(𝑥) = 𝑚𝑒𝑥+ (𝑚−2)
𝑥[\5žĒ467Ü/.𝑙2]úá𝑙1 ⊥𝑙2]e𝑚ĴGĵĶp%    (
AĊ −𝑒
2 ,0
BĊ 0, 𝑒
2
CĊ(−1,0)
DĊ(0,1)
Î:ſ
9Ċ%2024·ŠĹ·ÎĆ(àè)
[\<E.𝑦= 𝑥+ 1Ñ/=%    (
AĊ𝑦= e𝑥
BĊ𝑦= ln𝑥
CĊ𝑦= sin𝑥+ 1
DĊ𝑦= 𝑥3 +1
10Ċ%2024·ĚĜƀň·ĆćĈĉ(ݔ)
𝑓(𝑥)]𝑔(𝑥) !Ħ‰𝑅]𝑔′(𝑥)‰𝑔(𝑥))
]ē𝑓(𝑥) + 𝑔′(𝑥)
= 4]𝑓(𝑥)−𝑔′(4−𝑥) = 4]ė𝑔(𝑥)‰Ĩ)
]e%    (
AĊ𝑓(4) = 3
BĊ𝑔′(2) = 0
CĊ𝑓(1) +𝑓(3) = 8
DĊ𝑓(1) = 𝑓(−3)
11Ċ%2024·ċijĹZ·ĆćĈĉ(;BŞr67 P ˆ-.𝑦= |ln𝑥|A@ŐÑDE/.𝑙1:𝑙2]/7‰𝑃1:
𝑃2(𝑃1:𝑃2ãĄ')]âE.𝑙1:𝑙2PÏ< y řƁ¶7 A:B]e%    (
AĊ𝑃1:𝑃2A7ƂÞjƃŸ‰ G
BĊE.𝑃1𝑃201‰ G
CĊ.ś AB Ƅ‰ G
DĊ △𝐴𝐵𝑃ŞŸĴGĵĶ‰(0,1)
x:‚
12Ċ%2024·ČƅƆƇ·ĆćĈĉ(ݔ)
𝑓(𝑥) =
ln𝑥+𝑎
𝑥]ė𝑓′(1) = 2]e𝑎=      
.
13Ċ%2024·ƈĹƉƊ·ĆćĈĉ(-.𝑓(𝑥) = 𝑥3−ln𝑥47(1,𝑓(1))Ü/.<ÞjřĶ[ƒŞŸ‰
      
      
      
.
14Ċ%2024·ÔńŇ·ĆćĈĉ(ݔ)
𝑦=
𝑥[\<)
𝑦= 𝑎𝑥%𝑎> 0ē𝑎≠1([\4ŋ7Ü=
Ñó/.]eŋ7Þj‰      
.
Ô:SŠ
15Ċ%23-24 Îà·įļƋą·đƌ(,àè)
)
.
(1)𝑓(𝑥) = 𝑥2 ⋅e𝑥Í
(2)𝑓(𝑥) =
log2𝑥
2𝑥Í
(3)𝑓(𝑥) =
sin𝑥
1+cos𝑥Í
(4)𝑓(𝑥) = ln(1−2𝑥).
16Ċ%23-24 ηmn·ƍƎŜŝ(%1(ݔ𝑓(𝑥+ ℎ)−𝑓(𝑥) = 2ℎ𝑥+ 5ℎ+ ℎ2]Ə.Ɛw/.8d,𝑓′
(𝑥)Í
%2(ݔ𝑔(𝑥+ ℎ)−𝑔(𝑥) = 3ℎ𝑥2 +3ℎ2𝑥+ ℎ3]Ə.Ɛw/.8d,𝑔′(𝑥)Ċ
17Ċ%2024·ĩĪƑĐ·ĆćĈĉ(ݔ)
𝑓(𝑥) = 𝑎𝑥3−2𝑥2−2𝑥+ 𝑎(𝑎≥0)Ċ
(1)¥𝑎= 1²]ėE.𝑦= −3𝑥+ 𝑏<-.𝑦= 𝑓(𝑥)Ñ/],𝑏Í
(2)ėE.𝑦= −2𝑥−2<-.𝑦= 𝑓(𝑥)¤=A·ŋ7],𝑎Ċ
18Ċ%2024·ŠĹƒĐ·xĆ(ݔ)
𝑓(𝑥) = 𝑥2−𝑎𝑥+ 1,𝑔(𝑥) = ln𝑥+ 𝑎(𝑎∈𝑅).
      
      
(1)ė𝑎= 1,𝑓(𝑥) > 𝑔(𝑥)4ƓÃ(0,𝑡)5Ɣƕ],ō
𝑡ĴGĵĶÍ
(2)ė)
𝑓(𝑥)c𝑔(𝑥)=/.],ō
𝑎ĴGĵĶ.
19Ċ%23-24 Îà·ċČ·đ‹(ݔ)
𝑓(𝑥) = (𝑥−𝑎)2]𝑔(𝑥) = −(𝑥−𝑏)2.
(1)¥𝑎= 1²],-.𝑦= 𝑓(𝑥)4𝑥= 0Ü/.89.
(2)ė𝑎+ 𝑏= 1]pƖĒ4E.𝑙<-.𝑦= 𝑓(𝑥)c𝑦= 𝑔(𝑥)ŇÑ/ƗėĒ4],ÊE.𝑙89%ėE.𝑙8
9Ƙ>
]e𝑎¼½(ÍėãĒ4]ƙƚƛ­.