Bài tập toán cao cấp part 6 doc - Pdf 18

8.2. Vi phˆan 79
C
´
AC V
´
IDU
.
V´ı du
.
1. T´ınh vi phˆan df nˆe
´
u
1) f(x) = ln(arctg(sin x)); 2) f(x)=x

64 −x
2
+64arcsin
x
8
.
Gia

i. 1)
´
Ap du
.
ng c´ac t´ınh chˆa
´
tcu

a vi phˆan ta c´o


64 −x
2
dx +64d

arcsin
x
8

= x
d(64 − x
2
)
2

64 − x
2
+

64 −x
2
dx +64·
d

x
8


1 −
x

1) f(x)=xe
−x
,nˆe
´
u x l`a biˆe
´
ndˆo
.
clˆa
.
p;
2) f(x)=sinx
2
nˆe
´
u
a) x l`a biˆe
´
ndˆo
.
clˆa
.
p,
b) x l`a h`am cu

amˆo
.
tbiˆe
´
ndˆo

−x
)dx
= −(xde
−x
+ e
−x
dx)dx −e
−x
dx
2
= xe
−x
dx
2
− e
−x
dx
2
−e
−x
dx
2
=(x − 2)e
−x
dx
2
.
80 Chu
.
o

= −e
−x
− e
−x
+ xe
−x
=(x − 2)e
−x
v`a theo cˆong th´u
.
c (8.6) ta c´o
d
2
f =(x − 2)e
−x
dx
2
.
2) a) Phu
.
o
.
ng ph´ap I. Theo di
.
nh ngh˜ıa vi phˆan cˆa
´
p hai ta c´o
d
2
f = d[d sin x

o h`am cˆa
´
p hai f

xx
ta c´o
f

x
=2x cos x
2
,f

xx
= 2 cos x
2
− 4x
2
sin x
2
v`a theo (8.6) ta thu du
.
o
.
.
c
d
2
f = (2 cos x
2

2
− 4x
2
sin x
2
)dx
2
. 
V´ı d u
.
3.
´
Ap du
.
ng vi phˆan d
ˆe

t´ınh gˆa
`
nd´ung c´ac gi´a tri
.
:
1)
5

2 −0, 15
2+0, 15
; 2) arcsin 0, 51; 3) sin 29

.


(x
0
)∆x

f(x
0
+∆x) ≈ f(x
0
)+f

(x
0
)∆x
8.2. Vi phˆan 81
T`u
.
d
´o , d ˆe

t´ınh gˆa
`
nd´ung c´ac gi´a tri
.
ta cˆa
`
n thu
.
.
chiˆe


i t´ınh.
2
+
Cho
.
ndiˆe

m M
0
(x
0
) sao cho gi´a tri
.
cu

a h`am v`a cu

ada
.
o h`am cˆa
´
p
1cu

a n´o ta
.
idiˆe

mˆa

˜a cho l`a gi´a tri
.
cu

a h`am
y =
5

2 − x
2+x
ta
.
idiˆe

m x =0, 15. Ta d˘a
.
t x
0
=0;∆x =0, 15. Ta c´o
y

=
−4
5

2 −x
2+x
5(4 − x
2
)

cˆa
`
nt´ınh l`a gi´a tri
.
cu

a h`am ta
.
idiˆe

m
0, 51; t´u
.
cl`ay(0, 51).
D˘a
.
t x
0
=0, 5; ∆x =0, 01. Khi d´o ta c´o
arcsin(x
0
+∆x ≈ arcsinx
0
+ (arcsinx)

x=x
0
∆x
⇒ arcsin(0, 5+0, 01) ≈ arcsin0, 5 + (arcsinx)


=

0, 75 ≈ 0, 88 v`a do d´o
arcsin0, 51 ≈
π
6
+0, 011 ≈ 0, 513.
3) Sˆo
´
sin 29

l`a gi´a tri
.
cu

a h`am y = sin x khi x =
π
180
×29. Ta d˘a
.
t
x
0
=
π
180
×30 =
π
6
; y

π
6
= −
π
180
.Dod´o
sin 29

≈ y

π
6

+ y


π
6

· ∆x =
1
2
+

3
2


π
180

cos
2
x
)
3. f(x) = arccos(2
x
). (DS. −
2
x
ln2dx

1 −e
2x
)
4. f(x)=x
3
lnx.(DS. x
2
(1 + 3lnx) dx)
5. f(x) = cos
2
(

x). (DS. −2 cos

x ·sin

x ·
dx
2

dx)
8.2. Vi phˆan 83
10. f(x)=e

1
cos
x
.(DS.
−tgx ·e

1
cos x
cos x
dx)
11. f(x)=2
−x
2
.(DS. −2xe
−x
2
ln2dx)
12. f(x) = arctg

x
2
+ 1. (DS.
2xdx
2+x
2
)


(arcsinx)
2
dx).
T´ınh vi phˆan cˆa
´
ptu
.
o
.
ng ´u
.
ng cu

a c´ac h`am sau
15. f(x)=4
−x
2
; d
2
f ?(DS. 4
−x
2
2ln4(2x
2
ln4 −1)(dx)
2
)
16. f(x)=


7/2
(dx)
4
)
19. f(x)=xlnx, d
5
f ?(DS. −
6
x
4
(dx)
5
, x>0)
20. f(x)=x sin x; d
10
f ?(DS. (10 cos x − x sin x)(dx)
10
)
Su
.

du
.
ng cˆong th´u
.
cgˆa
`
nd
´ung
∆f ≈ df

o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
25. y = tg45

10

.(DS. 0,99)
26. y = ln(10, 21). (DS. 1,009)
27. y = sin 31

.(DS. 0,51)
28. y = arcsin0, 54. (D
S. 0,57)
29. y = arctg(1, 05). (DS. 0,81)
30. y =(1, 03)
5
.(DS. 1,15)
8.3 C´ac di
.
nh l´yco
.
ba

nvˆe
`

i) f(x) liˆen tu
.
ctrˆen doa
.
n [a, b].
ii) f(x) c´o da
.
o h`am h˜u
.
uha
.
n trong (a, b).
iii) f(a)=f(b).
Khi d
´otˆo
`
nta
.
idiˆe

m ξ : a<ξ<bsao cho f(ξ)=0.
D
-
i
.
nh l´y Lagr˘ang (Lagrange). Gia

su
.


(ξ) (8.12)
hay l`a
f(b)=f(a)+f

(ξ)(b −a). (8.13)
Cˆong th´u
.
c (8.12) go
.
i l`a cˆong th´u
.
csˆo
´
gia h˜u
.
uha
.
n.
8.3. C´ac di
.
nh l´y co
.
ba

nvˆe
`
h`am kha

vi 85
D

2
=0, ngh˜ıa l`a c´ac da
.
o h`am khˆong dˆo
`
ng th`o
.
i
b˘a
`
ng 0.
iv) ϕ(a) = ϕ(b).
Khi d
´ot`ımdu
.
o
.
.
cdiˆe

m ξ ∈ (a, b) sao cho:
f(b) − f(a)
ϕ(b) − ϕ(a)
=
f

(ξ)
ϕ

(ξ)

ho
.
.
p riˆeng cu

adi
.
nh l´y Lagrange v´o
.
idiˆe
`
ukiˆe
.
n f(a)=f(b).
C
´
AC V
´
IDU
.
V´ı du
.
1. Gia

su
.

P ( x)=(x + 3)(x + 2)(x − 1).
Ch´u
.


m x
1
= −3, x
2
= −2,
x
3
= 1. Trong c´ac khoa

ng (−3, −2) v`a (−2, 1) h`am P (x) kha

vi v`a
tho

a m˜an c´ac diˆe
`
ukiˆe
.
ncu

adi
.
nh l´y Rˆon v`a:
P ( −3) = P(−2)=0,
P ( −2) = P(1) = 0.
Do d
´o theo di
.
nh l´y Rˆon, t`ım du

.
nh l´y Rˆon cho doa
.
n[ξ
1

2
] v`a h`am P

(x), ta
la
.
it`ımdu
.
o
.
.
cdiˆe

m ξ ∈ (ξ
1

2
) ⊂ (−3, 1) sao cho P

(ξ)=0.
86 Chu
.
o
.


(x).
f

(x)=
1

1 −x
2
→ f

(x) < ∞,x∈ (−1, 1)
(Lu
.
u´yr˘a
`
ng khi x = ±1da
.
o h`am khˆong tˆo
`
nta
.
inhu
.
ng diˆe

ud´o
khˆong a

nh hu

iˆe

m ξ. Ta c´o:
arcsin1 − arcsin( −1)
1 − (−1)
=
1

1 −ξ
2

π
2



π
2

2
=
1

1 −ξ
2


1 −ξ
2
=

hai d
iˆe

m.
V´ı du
.
3. H˜ay kha

o s´at xem c´ac h`am f(x)=x
2
− 2x +3v`a ϕ(x)=
x
3
− 7x
2
+20x − 5 c´o tho

a m˜an diˆe
`
ukiˆe
.
ndi
.
nh l´y Cauchy trˆen doa
.
n
[1, 4] khˆong ? Nˆe
´
uch´ung tho


.
iii) c˜ung tho

a m˜an v`ı:
g

(x)=3x
2
− 14x +20> 0,x∈ R.
iv) Hiˆe

n nhiˆen ϕ(1) = ϕ(4).
8.3. C´ac di
.
nh l´y co
.
ba

nvˆe
`
h`am kha

vi 87
Do d´o f(x)v`aϕ(x) tho

a m˜an di
.
nh l´y Cauchy v`a ta c´o
f(4) −f(1)
ϕ(4) −ϕ(1)


d
ˆay chı

c´o ξ
1
=2l`adiˆe

m trong
cu

a(1, 4). Do d´o: ξ =2.
V´ı d u
.
4. Di
.
nh l´y Cauchy c´o ´ap du
.
ng du
.
o
.
.
c cho c´ac h`am f(x)=cosx,
ϕ(x)=x
3
trˆen doa
.
n[−π/2,π/2] hay khˆong ?
Gia

vˆa
.
y


(0)]
2
+[f

(0)]
2
= 0. Do d´odiˆe
`
ukiˆe
.
n iii) khˆong du
.
o
.
.
c tho

a m˜an. Ta
x´et vˆe
´
tr´ai cu

a (8.14):
f(b) − f(a)
ϕ(b) −ϕ(a)

ˆo
´
iv´o
.
ivˆe
´
pha

i n`ay ta c´o:
lim
ξ→0


sin ξ

2

= lim
ξ→0
sin ξ
ξ
· lim
ξ→0


1


= ∞.
Diˆe

`
ukiˆe
.
ncu

adi
.
nh
l´y Rˆon khˆong ? Ta
.
i sao ? (Tra

l`o
.
i: Khˆong)
88 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
2. H`am y =3x
2
− 5 c´o tho

am˜andi
.

.
ng minh r˘a
`
ng c´ac h`am f(x) = cos x, ϕ(x) = sin x tho

a m˜an
d
i
.
nh l´y Cauchy trˆen doa
.
n[0,π/2]. T`ım ξ ?(DS. ξ = π/4)
5. Ch´u
.
ng minh r˘a
`
ng h`am f(x)=e
x
v`a ϕ(x)=x
2
/(1 + x
2
) khˆong
tho

a m˜an d
i
.
nh l´y Cauchy trˆen doa
.

i B(2, 8).
(D
S. M(1, 1))
Chı

dˆa
˜
n. Du
.
.
a v`ao ´y ngh˜ıa h`ınh ho
.
ccu

a cˆong th´u
.
csˆo
´
gia h˜u
.
uha
.
n.
8.3.2 Khu
.

c´ac da
.
ng vˆo d
i

co
.
ba

ndˆe

khu
.

c´ac da
.
ng vˆo
di
.
nh
Da
.
ng vˆo di
.
nh 0/0
Gia

su
.

hai h`am f(x)v`aϕ(x) tho

a m˜an c´ac diˆe
`
ukiˆe

nta
.
i gi´o
.
iha
.
n(h˜u
.
uha
.
n ho˘a
.
cvˆoc`ung)
lim
x→a
f

(x)
ϕ

(x)
= k.
8.3. C´ac di
.
nh l´y co
.
ba

nvˆe
`

`
ukiˆe
.
n ii) v`a iii) cu

adi
.
nh l´y
trˆen dˆay c`on diˆe
`
ukiˆe
.
ni)du
.
o
.
.
cthaybo
.

idiˆe
`
ukiˆe
.
n:
i)

lim
x→a
f(x)=∞, lim

ng vˆo di
.
nh 0/0 (ho˘a
.
c
∞/∞)ta
.
id
iˆe

m x = a v`a f

, ϕ

tho

a m˜an c´ac diˆe
`
ukiˆe
.
n i), ii) v`a iii)
(tu
.
o
.
ng ´u
.
ng i)

, ii) v`a iii)) th`ı ta c´o thˆe


ta
biˆe
´
nd
ˆo

it´ıchf(x) · ϕ(x) th`anh:
i)
f(x)
1/ϕ(x)
(da
.
ng 0/0)
ii)
ϕ(x)
1/f(x)
(da
.
ng ∞/∞).
b) D
ˆe

khu
.

da
.
ng vˆo di
.


90 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
ho˘a
.
c
f(x) −ϕ(x)=ϕ(x)

f(x)
ϕ(x)
− 1

.
c) Da
.
ng vˆo di
.
nh 0
0
, ∞
0
,1


.
ng tru
.
`o
.
ng ho
.
.
p
n`ay ta c´o thˆe

biˆe
´
nd
ˆo

i F (x)dˆe

du
.
avˆe
`
da
.
ng vˆo d
i
.
nh 0 ·∞ d˜a n´oi trong
1) nh`o
.

cd`u quy t˘a
´
c Lˆopitan l`a mˆo
.
t cˆong cu
.
ma
.
nh d
e

t´ınh gi´o
.
iha
.
nnhu
.
ng n´o khˆong thˆe

thay to`an bˆo
.
c´ac phu
.
o
.
ng
ph´ap t´ınh gi´o
.
iha
.

2
− 1+lnx
e
x
−e
Gia

i. Ta c´o vˆo d
i
.
nh da
.
ng “0/0”.
´
Ap du
.
ng quy t˘a
´
c L’Hospital ta
thu d
u
.
o
.
.
c
lim
x→1
x
2

n
e
x
8.3. C´ac di
.
nh l´y co
.
ba

nvˆe
`
h`am kha

vi 91
Gia

i. Tac´ovˆodi
.
nh da
.
ng “∞/∞”.
´
Ap du
.
ng quy t˘a
´
c L’Hospital n
lˆa
`
n ta thu d

x→1
n!
e
x
=0. 
V´ı du
.
3. T´ınh lim
x→0+0
xlnx.
Gia

i. Ta c´o vˆo di
.
nh da
.
ng “0 ·∞”. Nhu
.
ng
xlnx =
lnx
1
x
v`a ta thu du
.
o
.
.
cvˆod
i

x
x
.
Gia

i. O
.

dˆay ta c´o vˆo di
.
nh da
.
ng “0
0
”. Nhu
.
ng
x
x
= e
xlnx
v`a ta thu du
.
o
.
.
cvˆodi
.
nh da
.

xlnx
= e
0
=1. 
V´ı du
.
5. T´ınh lim
x→0

1+x
2

1
e
x
−1−x
92 Chu
.
o
.
ng 8. Ph´ep t´ınh vi phˆan h`am mˆo
.
tbiˆe
´
n
Gia

i. O
.


m˜ucu

al˜uy th`u
.
a ta thu du
.
o
.
.
cvˆodi
.
nh da
.
ng “0/0”.
´
Ap du
.
ng
quy t˘a
´
c L’Hospital ta thu d
u
.
o
.
.
c
lim
x→0
ln(1 + x

− 1)2x
=
2
1
=2. 
V´ı d u
.
6. T´ınh lim
x→
π
2

tgx

2 cosx
.
Gia

i. Tac´ovˆodi
.
nh da
.
ng “∞
0
”. Nhu
.
ng

tgx


´
c L’Hospital ta c´o
lim
x→
π
2
2ln tgx
1
cos x
= 2 lim
x→
π
2
1
cos
2
x ·tgx
+ sin x
cos
2
x
= 2 lim
x→
π
2
1
cos x
tg
2
x

lim
x→
π
2
2 cos x·ln tgx
= e
0
=1. 
V´ı d u
.
7. Ch´u
.
ng minh r˘a
`
ng gi´o
.
iha
.
n
1) lim
x→0
x
2
sin(1/x)
sin x
=0
8.3. C´ac di
.
nh l´y co
.


i. 1) Quy t˘a
´
c L’Hospital khˆong ´ap du
.
ng d
u
.
o
.
.
cv`ıty

sˆo
´
c´ac da
.
o
h`am [2x sin(1/x) − cos(1/x)]/ cos x khˆong c´o gi´o
.
iha
.
n khi x → 0.
Ta t´ınh tru
.
.
ctiˆe
´
p gi´o
.


sˆo
´
c´ac da
.
o h`am
1 −cos x
1 + cos x
=tg
2
(x/2)
khˆong c´o gi´o
.
iha
.
n khi x →∞.
Ta t´ınh tru
.
.
ctiˆe
´
p gi´o
.
iha
.
n n`ay
lim
x→∞
x − sin x
x + sin x

t`ım gi´o
.
iha
.
nnhu
.
ng d
iˆe
`
ud´o khˆong c´o ngh˜ıa l`a n´o c´o
thˆe

thay cho to`an bˆo
.
c´ac phu
.
o
.
ng ph´ap t`ım gi´o
.
iha
.
n. Cˆa
`
nlu
.
u´yr˘a
`
ng
quy t˘a

ukiˆe
.
ncˆa
`
n.
B
`
AI T
ˆ
A
.
P
´
Ap du
.
ng quy t˘a
´
c L’Hospital d
ˆe

t´ınh gi´o
.
iha
.
n:
1. lim
x→2
x
4
− 16

.
tbiˆe
´
n
3. lim
x→0
e
2x
− 1
sin x
.(DS. 2)
4. lim
x→0
1 −cos ax
1 + cos bx
.(D
S.
a
2
b
2
)
5. lim
x→0
e
x
−e
−x
− 2x
x −sin x

x→∞
ln(1 + x
2
)
ln[(π/2) − arctgx]
.(DS. −2)
10. lim
x→∞

x
2
−1
x
.(DS. −1)
11. lim
x→∞
x
ln(1 + x)
.(D
S. +∞)
12. lim
x→+0
ln sin x
ln sin 5x
.(D
S. 1)
13. lim
x→a
arcsin
x −a

x→∞
(x −x
2
ln(1 + 1/x)). (Ds. 1/2)
19. lim
x→0

1
x
2
− cotg
2
x

.(DS. 2/3)
20. lim
x→0
x
1/ln(e
x
−1)
.(DS. e)


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