导数的概念

(Lu, Tiao)
1482S62759189
3,4.
http://dsec.pku.edu.cn/˜tlu/index.html
Tiao Lu (lutiaopku@sina.com)

Email
emaillutiaopku@sina.com,
,””.
Email,
.
,..
Tiao Lu (lutiaopku@sina.com)

.
,
1,2,101
1,2,101
Tiao Lu (lutiaopku@sina.com)

..
Tiao Lu (lutiaopku@sina.com)

()39, 40 \\
()20(1)(3)(5)(7), 21(1), 22(1).
.,.
Tiao Lu (lutiaopku@sina.com)

y = sin−1 xsiny = x,y ∈ −π
π
2 , 2 y,
y = cos−1 xcos y = x,y ∈ [0,π]y,
y = tan−1 xtany = x,y ∈ −π
π
2 , 2 y,
y = arcsinx,y = arccos x,y = arctanx
Tiao Lu (lutiaopku@sina.com)

x → 0,
arcsin x ∽ o(x) arctanx ∽ o(x)
:
arcsin x
lim
x→0 x
y
= lim
y→0 siny
= 1
.
:
y = arcsinxx = siny.
Tiao Lu (lutiaopku@sina.com)

Ferma.
: Newton
Leibniz
–
.
Tiao Lu (lutiaopku@sina.com)

Tiao Lu (lutiaopku@sina.com)
2006-10-9
Tiao Lu (lutiaopku@sina.com)

1
x0
2
3
Tiao Lu (lutiaopku@sina.com)

s
f (t)
t0 t
( ) ( 0)
t f t f
v
t t0
t0
v lim
tt
0
f ( t)
f ( t0)
t t
0
x0
o
f ( t0
) f (t)
t0
Tiao Lu (lutiaopku@sina.com)
1 2
s gt
2
t
s

x0
C : y f ( x)
M
M N M T
( )
MT
k tan
lim tan
k
M N
lim
xx0
( ) ( 0)
x f x f
x x
0
tan
y
o
y f (x)
N
T
C M
x
Tiao Lu (lutiaopku@sina.com)
0
x
( ) ( 0)
x f x f
x x0
x

x0
.
2.1 (61)y = f (x)U(x0).
f (x) − f (x0)
lim
x→x0 x − x0
,f (x)x0(),
f ′ (x0)y ′ |x=x0 dy
dx |x=x0
y = f (x)x0.
,.
Tiao Lu (lutiaopku@sina.com)

: f ′ (x0)
dy
dx |x=x0
x0
y = f (x)x = x0,y(x) = f (x0)
df (x0)
dx
.
Tiao Lu (lutiaopku@sina.com)

.
limx→x0
x0)
x0
f (x)−f (x0)
x−x0 x = x0 + ∆x, (∆x
f (x0 + ∆x) − f (x0)
lim
∆x→0 ∆x
∆y = f (x0 + ∆x) − f (x0),
2.1 (61)
∆y
lim
∆x→0 ∆x
Tiao Lu (lutiaopku@sina.com)

x0
:y = f (x) = 3xx = 1.
:,limx→1
,
f (x)−f (1)
x−1 .
f (x) − f (1) 3x − 3 × 1 3(x − 1)
lim = lim = lim = 3
x→1 x − 1 x→1 x − 1 x→1 x − 1
y ′ (1) = 3.
Tiao Lu (lutiaopku@sina.com)

x0
:y = f (x) = 3x − 3x = 1.
:,limx→1
f (x)−f (1)
x−1 .
f (x) − f (1) (3x − 3) − (3 × 1 − 3) 3(x − 1)
lim = lim
= lim = 3
x→1 x − 1 x→1 x − 1 x→2 x − 1
,
y ′ (1) = 3.
Tiao Lu (lutiaopku@sina.com)

x0
y = f (x) = 3xy = f (x) = 3x − 3x = 1.
y=3x
3
1
y
1
y=3x−3
x = 1,,
.,
,.
Tiao Lu (lutiaopku@sina.com)
x

,
x0
/( 62),
f (x) − f (x0)
lim
x→x0−0 x − x0
,,,
f ′ − (x0).,f ′ + (x0).
/.
Tiao Lu (lutiaopku@sina.com)

x0
.f (x)x0⇔f (x)
x0,.
Tiao Lu (lutiaopku@sina.com)

x0
(4, 62):f (x) = |x|x = 0.
y=|x|
Tiao Lu (lutiaopku@sina.com)
y
x

x0
: f (x) = |x|, x = 0.
f ′ −x−0
− (0) = limx→0−0 x−0 = −1;
f ′ x−0
− (0) = limx→0+0 x−0 = 1.
,f (x) = |x|x = 0.
Tiao Lu (lutiaopku@sina.com)

x0
,x = 0.y
, x,x = 0,
−1,,1.
?.
Tiao Lu (lutiaopku@sina.com)

(5, 64):
x = 0.
f (x) =
x0
x, x < 0
ln(1 + x), x ≥ 0
Tiao Lu (lutiaopku@sina.com)

x, x < 0
: f (x) =
ln(1 + x), x ≥ 0
, f (0) = 0.
f ′ x−0
− (0) = limx→0−0 x−0 = 1;
f ′ ln(1+x)−0
− (0) = limx→0+0 x−0
,f (x) =
.
= 1.
x0
, x = 0
x, x < 0
ln(1 + x), x ≥ 0
Tiao Lu (lutiaopku@sina.com)
x = 0

625
f (x) =
x0
x, x < 0
ln(1 + x), x ≥ 0
x = 0,
.
Tiao Lu (lutiaopku@sina.com)

x0
f (x)(a,b),f (x)
(a,b).
f (x)[a,b],
,f (x)[a,b].
f (x),x
f ′ (x)..
f ′ (x)f (x),.
Tiao Lu (lutiaopku@sina.com)

x0
6 (63)f (x) = C C
:
(C) ′ = 0.
f (x + ∆x) − f (x) C − C
lim
= lim
∆x→0 ∆x ∆x→0 ∆x
(C) ′ = 0
Tiao Lu (lutiaopku@sina.com)
= 0

x0
7 (63)(sinx) ′ = cos x,
:
cos(x + ∆x) − cos x
lim
∆x→0 ∆x
(cos x) ′ = − sinx.
= lim
∆x→0
−2sin x + ∆x
∆x
2 sin 2
∆x
(table.pdf, p.11).
= sinx,lim∆x→0 = 1,
lim∆x→0 sin x + ∆x
2
lim
∆x→0
(cos x) ′ = − sinx.
−2sin x + ∆x
∆x
2 sin 2
∆x
sin ∆x
2
∆x
2
= − sinx
Tiao Lu (lutiaopku@sina.com)

8 (64)
:
ln(x + ∆x) − lnx
lim
∆x→0 ∆x
x0
lnx = 1
,x > 0
x
ln
= lim
∆x→0
x+∆x
x
∆x
1
= lim
∆x→0 x
ln 1 + ∆x
.:(see table.pdf,
p2),x → 0
ln(1 + x) ∽ x
Tiao Lu (lutiaopku@sina.com)
∆x
x
x
= 1
x

x0
9 (65)(x n ) ′ = nx n−1 ,n.
(x a ) ′ = ax a−1 ,a ∈ R.,
:
(x + ∆x)
lim
∆x→0
2 − x2 x 2 ′ = 2x
x
∆x
= lim
∆x→0
2 + 2x∆x + (∆x) 2
− x2 ∆x
2x∆x + (∆x)
= lim
∆x→0
2
= lim (2x + ∆x) = 2x
∆x ∆x→0
.:(
).
Tiao Lu (lutiaopku@sina.com)

: f (x)x⇒ f (x)x
: f (x)x⇒lim∆x→0 ∆x
lim∆x→0 ∆x = 0.,
f (x+∆x)−f (x)
lim∆x→0 ∆x ∆x
f (x+∆x)−f (x)
,
f (x + ∆x) − f (x) f (x + ∆x) − f (x)
lim
∆x = lim
lim ∆x = 0
∆x→0 ∆x
∆x→0 ∆x ∆x→0
,lim∆x→0 (f (x + ∆x) − f (x)) = 0,f (x)x.
,.f (x) = |x|x = 0.
?
Tiao Lu (lutiaopku@sina.com)

: f (x)x⇒ f (x)x
: f (x)x⇒lim∆x→0 ∆x
lim∆x→0 ∆x = 0.,
f (x+∆x)−f (x)
lim∆x→0 ∆x ∆x
f (x+∆x)−f (x)
,
f (x + ∆x) − f (x) f (x + ∆x) − f (x)
lim
∆x = lim
lim ∆x = 0
∆x→0 ∆x
∆x→0 ∆x ∆x→0
,lim∆x→0 (f (x + ∆x) − f (x)) = 0,f (x)x.
,.f (x) = |x|x = 0.
?. google,
.
Tiao Lu (lutiaopku@sina.com)

y f (x)
( x0 , y0)
tan ( 0)
o
x f
f ( x0)
0 , x , ) ;
( 0 y0
( 0 y0
f ( x0)
0 , x , ) ;
f ( x0)
0 , x , x ;
f x ) , x .
( 0
, ) x
f x ) ,
( 0
( 0 0 y
0
y
C
y
o
y f (x)
: y y0
f (
x0)(
x x0)
1
: y y0
( x x0)
( ( 0) 0)
f (
x )
x f
o x0
Tiao Lu (lutiaopku@sina.com)
0
y
M
x0
x0
x
T
( 0 , 0)
y x
x
x

11 (67)y = 4x 3 (1,4).
:y = 4x 3 y ′ = 12x 2 ,x = 1
y ′ |x=1 = 12.,y = 4x 3
(1,4)12 ,
:
y − 4 = 12(x − 1)
y − 4 = − 1
(x − 1)
12
Tiao Lu (lutiaopku@sina.com)