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(Lu, Tiao)<br />
1482S62759189<br />
3,4.<br />
http://dsec.pku.edu.cn/˜tlu/index.html<br />
Tiao Lu (lutiaopku@sina.com)
Email<br />
<br />
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<br />
<br />
emaillutiaopku@sina.com,<br />
,””.<br />
Email,<br />
.<br />
,..<br />
Tiao Lu (lutiaopku@sina.com)
.<br />
,<br />
1,2,101<br />
1,2,101<br />
<br />
<br />
<br />
<br />
Tiao Lu (lutiaopku@sina.com)
..<br />
Tiao Lu (lutiaopku@sina.com)
()39, 40 \\<br />
()20(1)(3)(5)(7), 21(1), 22(1).<br />
.,.<br />
Tiao Lu (lutiaopku@sina.com)
y = sin−1 xsiny = x,y ∈ −π <br />
π<br />
2 , 2 y,<br />
y = cos−1 xcos y = x,y ∈ [0,π]y,<br />
y = tan−1 xtany = x,y ∈ −π <br />
π<br />
2 , 2 y,<br />
y = arcsinx,y = arccos x,y = arctanx<br />
Tiao Lu (lutiaopku@sina.com)
x → 0,<br />
arcsin x ∽ o(x) arctanx ∽ o(x)<br />
:<br />
arcsin x<br />
lim<br />
x→0 x<br />
y<br />
= lim<br />
y→0 siny<br />
= 1<br />
.<br />
:<br />
y = arcsinxx = siny.<br />
Tiao Lu (lutiaopku@sina.com)
Ferma.<br />
: Newton<br />
Leibniz<br />
<br />
– <br />
<br />
.<br />
Tiao Lu (lutiaopku@sina.com)
Tiao Lu (lutiaopku@sina.com)<br />
2006-10-9<br />
Tiao Lu (lutiaopku@sina.com)
1 <br />
<br />
x0<br />
<br />
<br />
2 <br />
3 <br />
Tiao Lu (lutiaopku@sina.com)
s <br />
f (t)<br />
t0 t <br />
( ) ( 0)<br />
t f t f <br />
v <br />
t t0<br />
t0<br />
<br />
v lim<br />
tt<br />
0<br />
f ( t)<br />
f ( t0)<br />
t t<br />
0<br />
<br />
x0<br />
<br />
<br />
o<br />
f ( t0<br />
) f (t)<br />
t0<br />
Tiao Lu (lutiaopku@sina.com) <br />
<br />
1 2<br />
s gt<br />
2<br />
t<br />
s
x0<br />
<br />
<br />
C : y f ( x)<br />
M <br />
M N M T<br />
( )<br />
MT <br />
k tan<br />
lim tan<br />
<br />
k <br />
<br />
<br />
M N <br />
lim<br />
xx0<br />
( ) ( 0)<br />
x f x f <br />
x x<br />
0<br />
tan<br />
<br />
y<br />
o<br />
y f (x)<br />
N<br />
T<br />
C M<br />
x<br />
<br />
Tiao Lu (lutiaopku@sina.com) <br />
0<br />
x<br />
( ) ( 0)<br />
x f x f <br />
x x0<br />
x
x0<br />
<br />
<br />
.<br />
2.1 (61)y = f (x)U(x0).<br />
f (x) − f (x0)<br />
lim<br />
x→x0 x − x0<br />
,f (x)x0(),<br />
f ′ (x0)y ′ |x=x0 dy<br />
dx |x=x0<br />
y = f (x)x0.<br />
,.<br />
Tiao Lu (lutiaopku@sina.com)
: f ′ (x0)<br />
<br />
<br />
<br />
<br />
dy<br />
dx |x=x0<br />
<br />
x0<br />
<br />
<br />
y = f (x)x = x0,y(x) = f (x0)<br />
<br />
df (x0)<br />
dx<br />
.<br />
Tiao Lu (lutiaopku@sina.com)
.<br />
<br />
<br />
<br />
<br />
limx→x0<br />
x0)<br />
<br />
x0<br />
<br />
<br />
f (x)−f (x0)<br />
x−x0 x = x0 + ∆x, (∆x<br />
f (x0 + ∆x) − f (x0)<br />
lim<br />
∆x→0 ∆x<br />
∆y = f (x0 + ∆x) − f (x0),<br />
2.1 (61)<br />
∆y<br />
lim<br />
∆x→0 ∆x<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
:y = f (x) = 3xx = 1.<br />
:,limx→1<br />
,<br />
f (x)−f (1)<br />
x−1 .<br />
f (x) − f (1) 3x − 3 × 1 3(x − 1)<br />
lim = lim = lim = 3<br />
x→1 x − 1 x→1 x − 1 x→1 x − 1<br />
y ′ (1) = 3.<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
:y = f (x) = 3x − 3x = 1.<br />
:,limx→1<br />
f (x)−f (1)<br />
x−1 .<br />
f (x) − f (1) (3x − 3) − (3 × 1 − 3) 3(x − 1)<br />
lim = lim<br />
= lim = 3<br />
x→1 x − 1 x→1 x − 1 x→2 x − 1<br />
,<br />
y ′ (1) = 3.<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
y = f (x) = 3xy = f (x) = 3x − 3x = 1.<br />
y=3x<br />
3<br />
1<br />
y<br />
1<br />
y=3x−3<br />
x = 1,,<br />
.,<br />
,.<br />
Tiao Lu (lutiaopku@sina.com) <br />
x
,<br />
<br />
<br />
<br />
<br />
<br />
x0<br />
<br />
<br />
/( 62),<br />
f (x) − f (x0)<br />
lim<br />
x→x0−0 x − x0<br />
,,,<br />
f ′ − (x0).,f ′ + (x0).<br />
/.<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
.f (x)x0⇔f (x)<br />
x0,.<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
(4, 62):f (x) = |x|x = 0.<br />
y=|x|<br />
Tiao Lu (lutiaopku@sina.com) <br />
y<br />
x
x0<br />
<br />
<br />
: f (x) = |x|, x = 0.<br />
f ′ −x−0<br />
− (0) = limx→0−0 x−0 = −1;<br />
f ′ x−0<br />
− (0) = limx→0+0 x−0 = 1.<br />
,f (x) = |x|x = 0.<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
,x = 0.y<br />
, x,x = 0,<br />
−1,,1.<br />
?.<br />
Tiao Lu (lutiaopku@sina.com)
(5, 64):<br />
x = 0.<br />
<br />
<br />
<br />
<br />
f (x) =<br />
<br />
x0<br />
<br />
<br />
x, x < 0<br />
ln(1 + x), x ≥ 0<br />
Tiao Lu (lutiaopku@sina.com)
x, x < 0<br />
: f (x) =<br />
ln(1 + x), x ≥ 0<br />
, f (0) = 0.<br />
f ′ x−0<br />
− (0) = limx→0−0 x−0 = 1;<br />
f ′ ln(1+x)−0<br />
− (0) = limx→0+0 x−0<br />
,f (x) =<br />
.<br />
= 1.<br />
<br />
x0<br />
<br />
<br />
, x = 0<br />
x, x < 0<br />
ln(1 + x), x ≥ 0<br />
Tiao Lu (lutiaopku@sina.com) <br />
x = 0
625<br />
<br />
<br />
<br />
<br />
f (x) =<br />
<br />
x0<br />
<br />
<br />
x, x < 0<br />
ln(1 + x), x ≥ 0<br />
x = 0,<br />
.<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
f (x)(a,b),f (x)<br />
(a,b).<br />
f (x)[a,b],<br />
,f (x)[a,b].<br />
f (x),x<br />
f ′ (x)..<br />
f ′ (x)f (x),.<br />
Tiao Lu (lutiaopku@sina.com)
x0<br />
<br />
<br />
6 (63)f (x) = C C <br />
:<br />
<br />
(C) ′ = 0.<br />
f (x + ∆x) − f (x) C − C<br />
lim<br />
= lim<br />
∆x→0 ∆x ∆x→0 ∆x<br />
(C) ′ = 0<br />
Tiao Lu (lutiaopku@sina.com) <br />
= 0
x0<br />
<br />
<br />
7 (63)(sinx) ′ = cos x,<br />
:<br />
cos(x + ∆x) − cos x<br />
lim<br />
∆x→0 ∆x<br />
(cos x) ′ = − sinx.<br />
= lim<br />
∆x→0<br />
−2sin x + ∆x<br />
<br />
∆x<br />
2 sin 2<br />
∆x<br />
(table.pdf, p.11). <br />
<br />
= sinx,lim∆x→0 = 1,<br />
lim∆x→0 sin x + ∆x<br />
2<br />
lim<br />
∆x→0<br />
(cos x) ′ = − sinx.<br />
−2sin x + ∆x<br />
<br />
∆x<br />
2 sin 2<br />
∆x<br />
sin ∆x<br />
2<br />
∆x<br />
2<br />
= − sinx<br />
Tiao Lu (lutiaopku@sina.com)
8 (64)<br />
:<br />
ln(x + ∆x) − lnx<br />
lim<br />
∆x→0 ∆x<br />
<br />
<br />
<br />
<br />
<br />
x0<br />
<br />
<br />
lnx = 1<br />
,x > 0<br />
x<br />
ln<br />
= lim<br />
∆x→0<br />
x+∆x<br />
x<br />
∆x<br />
1<br />
= lim<br />
∆x→0 x<br />
ln 1 + ∆x<br />
<br />
.:(see table.pdf,<br />
p2),x → 0<br />
ln(1 + x) ∽ x<br />
Tiao Lu (lutiaopku@sina.com) <br />
∆x<br />
x<br />
x<br />
= 1<br />
x
x0<br />
<br />
<br />
9 (65)(x n ) ′ = nx n−1 ,n.<br />
(x a ) ′ = ax a−1 ,a ∈ R.,<br />
:<br />
(x + ∆x)<br />
lim<br />
∆x→0<br />
2 − x2 x 2 ′ = 2x<br />
<br />
x<br />
∆x<br />
= lim<br />
∆x→0<br />
2 + 2x∆x + (∆x) 2<br />
− x2 ∆x<br />
2x∆x + (∆x)<br />
= lim<br />
∆x→0<br />
2<br />
= lim (2x + ∆x) = 2x<br />
∆x ∆x→0<br />
.:(<br />
).<br />
Tiao Lu (lutiaopku@sina.com)
: f (x)x⇒ f (x)x<br />
: f (x)x⇒lim∆x→0 ∆x<br />
lim∆x→0 ∆x = 0.,<br />
<br />
f (x+∆x)−f (x)<br />
lim∆x→0 ∆x ∆x <br />
f (x+∆x)−f (x)<br />
,<br />
<br />
f (x + ∆x) − f (x) f (x + ∆x) − f (x)<br />
lim<br />
∆x = lim<br />
lim ∆x = 0<br />
∆x→0 ∆x<br />
∆x→0 ∆x ∆x→0<br />
,lim∆x→0 (f (x + ∆x) − f (x)) = 0,f (x)x.<br />
,.f (x) = |x|x = 0.<br />
?<br />
Tiao Lu (lutiaopku@sina.com)
: f (x)x⇒ f (x)x<br />
: f (x)x⇒lim∆x→0 ∆x<br />
lim∆x→0 ∆x = 0.,<br />
<br />
f (x+∆x)−f (x)<br />
lim∆x→0 ∆x ∆x <br />
f (x+∆x)−f (x)<br />
,<br />
<br />
f (x + ∆x) − f (x) f (x + ∆x) − f (x)<br />
lim<br />
∆x = lim<br />
lim ∆x = 0<br />
∆x→0 ∆x<br />
∆x→0 ∆x ∆x→0<br />
,lim∆x→0 (f (x + ∆x) − f (x)) = 0,f (x)x.<br />
,.f (x) = |x|x = 0.<br />
?. google,<br />
.<br />
Tiao Lu (lutiaopku@sina.com)
y f (x)<br />
( x0 , y0)<br />
<br />
tan ( 0)<br />
o<br />
x f <br />
f ( x0)<br />
0 , x , ) ;<br />
( 0 y0<br />
( 0 y0<br />
f ( x0)<br />
0 , x , ) ;<br />
f ( x0)<br />
0 , x , x ;<br />
f x ) , x .<br />
( 0<br />
, ) x<br />
f x ) ,<br />
( 0<br />
( 0 0 y<br />
0<br />
y<br />
C<br />
<br />
y<br />
o<br />
y f (x)<br />
: y y0<br />
f (<br />
x0)(<br />
x x0)<br />
1<br />
: y y0<br />
( x x0)<br />
( ( 0) 0)<br />
f (<br />
x )<br />
x f<br />
o x0<br />
Tiao Lu (lutiaopku@sina.com) <br />
0<br />
y<br />
M<br />
x0<br />
x0<br />
x<br />
T<br />
( 0 , 0)<br />
y x<br />
x<br />
x
11 (67)y = 4x 3 (1,4).<br />
:y = 4x 3 y ′ = 12x 2 ,x = 1<br />
y ′ |x=1 = 12.,y = 4x 3 <br />
(1,4)12 ,<br />
<br />
:<br />
y − 4 = 12(x − 1)<br />
y − 4 = − 1<br />
(x − 1)<br />
12<br />
Tiao Lu (lutiaopku@sina.com)