toan 1- chuong 7

Dao hamVi phan

Cac dinh ly ve ham kha vi

Ch÷ìng 7: PH�P T�NH VI PH�N H�M SÈ MËTBI�N SÈ

� m Thanh Ph÷ìng, Ngæ M¤nh T÷ðng

Ng y 15 th¡ng 12 n«m 2010

� m Thanh Ph÷ìng, Ngæ M¤nh T÷ðng Ch÷ìng 7: PH�P T�NH VI PH�N H�M SÈ MËT BI�N SÈ

Dao hamVi phan

Cac dinh ly ve ham kha viNëi dung ch½nh

�¤o h m cõa h m sè

Vi ph¥n cõa h m sè

C¡c �ành lþ v· h m kh£ vi

Cæng thùc Taylor, Maclaurint

Dao hamVi phan

Dinh nghiaCac phep toan dao hamDao ham cap cao

�ành ngh¾a �¤o h m

�ành ngh¾a

H m sè f(x) x¡c �ành trong l¥n cªn cõa �iºm x0.

f ′ (x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

f ′ (x0) �÷ñc gåi l �¤o h m cõa h m f (x) t¤i �iºm x0.

V½ dö: T¼m �¤o h m cõa h m sè y = cosx t¤i �iºm x0.

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

cos(x0 + ∆x)− cos x0∆x

= − lim∆x→0

sin(x0 + ∆x

)· sin ∆x

= − sin(x0)

Dao hamVi phan

�ành ngh¾a

f ′ (x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

= − lim∆x→0

sin(x0 + ∆x

)· sin ∆x

= − sin(x0)

Dao hamVi phan

�ành ngh¾a

f ′ (x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

= − lim∆x→0

sin(x0 + ∆x

)· sin ∆x

= − sin(x0)

Dao hamVi phan

�ành ngh¾a

f ′ (x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

= − lim∆x→0

sin(x0 + ∆x

)· sin ∆x

= − sin(x0)

Dao hamVi phan

�ành ngh¾a

f ′ (x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

= − lim∆x→0

sin(x0 + ∆x

)· sin ∆x

= − sin(x0)

Dao hamVi phan

�ành ngh¾a �¤o h m mët ph½a

�ành ngh¾a �¤o h m ph£i

f ′+(x0) = lim∆x→0+

f (x0 + ∆x)− f (x0)

f ′+(x0) �÷ñc gåi l �¤o h m ph£i cõa h m f (x) t¤i �iºm x0.

f ′−(x0) = lim∆x→0−

f (x0 + ∆x)− f (x0)

f ′−(x0) �÷ñc gåi l �¤o h m ph£i cõa h m f (x) t¤i �iºm x0.

Dao hamVi phan

f ′+(x0) = lim∆x→0+

f (x0 + ∆x)− f (x0)

f ′−(x0) = lim∆x→0−

f (x0 + ∆x)− f (x0)

Dao hamVi phan

f ′+(x0) = lim∆x→0+

f (x0 + ∆x)− f (x0)

f ′−(x0) = lim∆x→0−

f (x0 + ∆x)− f (x0)

Dao hamVi phan

f ′+(x0) = lim∆x→0+

f (x0 + ∆x)− f (x0)

f ′−(x0) = lim∆x→0−

f (x0 + ∆x)− f (x0)

Dao hamVi phan

f ′+(x0) = lim∆x→0+

f (x0 + ∆x)− f (x0)

f ′−(x0) = lim∆x→0−

f (x0 + ∆x)− f (x0)

Dao hamVi phan

�ành lþ

H m sè y = f (x) câ �¤o h m t¤i �iºm x0 khi v ch¿ khi nâ câ �¤oh m tr¡i v �¤o h m ph£i t¤i �iºm x0 v hai �¤o h m n y b¬ngnhau.

�ành ngh¾a �¤o h m væ còng

N¸u lim∆x→0

f (x0 + ∆x)− f (x0)

∆x=∞ th¼ ta nâi f (x) câ �¤o h m væ

còng t¤i �iºm x0.

Dao hamVi phan

�ành lþ

H m sè y = f (x) câ �¤o h m t¤i �iºm x0 khi v ch¿ khi nâ câ �¤oh m tr¡i v �¤o h m ph£i t¤i �iºm x0 v hai �¤o h m n y b¬ngnhau.

�ành ngh¾a �¤o h m væ còng

N¸u lim∆x→0

f (x0 + ∆x)− f (x0)

∆x=∞ th¼ ta nâi f (x) câ �¤o h m væ

còng t¤i �iºm x0.

Dao hamVi phan

V½ dö

T¼m f ′+(0); f ′−(0) bi¸t f (x) =

1x , x 6= 0

0, x = 0

f ′+(0) = lim∆x→0+

f (0 + ∆x)− f (0)

∆x= lim

∆x→0+

e1/∆x − 0∆x

= +∞

f ′−(0) = lim∆x→0−

f (0 + ∆x)− f (0)

∆x= lim

∆x→0−

e1/∆x − 0∆x

�¤o h m tr¡i v �¤o h m ph£i khæng b¬ng nhau n¶n khæng tçn t¤i�¤o h m t¤i �iºm x = 0

Dao hamVi phan

V½ dö

T¼m f ′+(0); f ′−(0) bi¸t f (x) =

1x , x 6= 0

0, x = 0

f ′+(0) = lim∆x→0+

f (0 + ∆x)− f (0)

= lim∆x→0+

e1/∆x − 0∆x

= +∞

f ′−(0) = lim∆x→0−

f (0 + ∆x)− f (0)

∆x= lim

∆x→0−

e1/∆x − 0∆x

Dao hamVi phan

V½ dö

T¼m f ′+(0); f ′−(0) bi¸t f (x) =

1x , x 6= 0

0, x = 0

f ′+(0) = lim∆x→0+

f (0 + ∆x)− f (0)

∆x= lim

∆x→0+

e1/∆x − 0∆x

= +∞

f ′−(0) = lim∆x→0−

f (0 + ∆x)− f (0)

∆x= lim

∆x→0−

e1/∆x − 0∆x

Dao hamVi phan

V½ dö

T¼m f ′+(0); f ′−(0) bi¸t f (x) =

1x , x 6= 0

0, x = 0

f ′+(0) = lim∆x→0+

f (0 + ∆x)− f (0)

∆x= lim

∆x→0+

e1/∆x − 0∆x

= +∞

f ′−(0) = lim∆x→0−

f (0 + ∆x)− f (0)

= lim∆x→0−

e1/∆x − 0∆x

Dao hamVi phan

V½ dö

T¼m f ′+(0); f ′−(0) bi¸t f (x) =

1x , x 6= 0

0, x = 0

f ′+(0) = lim∆x→0+

f (0 + ∆x)− f (0)

∆x= lim

∆x→0+

e1/∆x − 0∆x

= +∞

f ′−(0) = lim∆x→0−

f (0 + ∆x)− f (0)

∆x= lim

∆x→0−

e1/∆x − 0∆x

Dao hamVi phan

V½ dö

T¼m f ′+(0); f ′−(0) bi¸t f (x) =

1x , x 6= 0

0, x = 0

f ′+(0) = lim∆x→0+

f (0 + ∆x)− f (0)

∆x= lim

∆x→0+

e1/∆x − 0∆x

= +∞

f ′−(0) = lim∆x→0−

f (0 + ∆x)− f (0)

∆x= lim

∆x→0−

e1/∆x − 0∆x

Dao hamVi phan

V½ dö

T¼m f ′(x) bi¸t f (x) = x2 − 3|x |+ 2

f (x) =

{x2 − 3x + 2, x ≥ 0x2 + 3x + 2, x < 0

⇒ f ′(x) =

{2x − 3, x > 02x + 3, x < 0

T¤i �iºm x = 0: f ′+(0) = −3; f ′−(0) = 3.�¤o h m tr¡i v �¤o h m ph£i khæng b¬ng nhau suy ra khæng tçnt¤i �¤o h m t¤i �iºm x = 0.

Dao hamVi phan

V½ dö

T¼m f ′(x) bi¸t f (x) = x2 − 3|x |+ 2

f (x) =

{x2 − 3x + 2, x ≥ 0x2 + 3x + 2, x < 0

⇒ f ′(x) =

{2x − 3, x > 02x + 3, x < 0

Dao hamVi phan

V½ dö

T¼m f ′(x) bi¸t f (x) = x2 − 3|x |+ 2

f (x) =

{x2 − 3x + 2, x ≥ 0x2 + 3x + 2, x < 0

⇒ f ′(x) =

{2x − 3, x > 02x + 3, x < 0

Dao hamVi phan

V½ dö

T¼m f ′(x) bi¸t f (x) = x2 − 3|x |+ 2

f (x) =

{x2 − 3x + 2, x ≥ 0x2 + 3x + 2, x < 0

⇒ f ′(x) =

{2x − 3, x > 02x + 3, x < 0

T¤i �iºm x = 0: f ′+(0) = −3; f ′−(0) = 3.

�¤o h m tr¡i v �¤o h m ph£i khæng b¬ng nhau suy ra khæng tçnt¤i �¤o h m t¤i �iºm x = 0.

Dao hamVi phan

V½ dö

T¼m f ′(x) bi¸t f (x) = x2 − 3|x |+ 2

f (x) =

{x2 − 3x + 2, x ≥ 0x2 + 3x + 2, x < 0

⇒ f ′(x) =

{2x − 3, x > 02x + 3, x < 0

Dao hamVi phan

B£ng �¤o h m cì b£n

Dao hamVi phan

B£ng �¤o h m cì b£n

Dao hamVi phan

C¡c ph²p to¡n �¤o h m

�¤o h m cõa têng hi»u, t½ch th÷ìng:

1 (αu)′ = αu′

2 (u ± v)′ = u′ ± v ′

3 (u.v)′ = u.v ′ + u′.v

u′v − v ′u

�¤o h m cõa h m hñp:f = f (u), u = u(x)⇒ f ′(x) = f ′(u) · u′(x)�¤o h m cõa h m ng÷ñc:

x ′(y) =1

y ′(x)

Dao hamVi phan

�¤o h m cõa têng hi»u, t½ch th÷ìng:1 (αu)′ = αu′

2 (u ± v)′ = u′ ± v ′

3 (u.v)′ = u.v ′ + u′.v

u′v − v ′u

x ′(y) =1

y ′(x)

Dao hamVi phan

�¤o h m cõa têng hi»u, t½ch th÷ìng:1 (αu)′ = αu′

2 (u ± v)′ = u′ ± v ′

3 (u.v)′ = u.v ′ + u′.v

u′v − v ′u

x ′(y) =1

y ′(x)

Dao hamVi phan

�¤o h m cõa h m cho bði ph÷ìng tr¼nh tham sè

H m y = y(x) �÷ñc cho bði ph÷ìng tr¼nh tham sè:{

x = x(t)y = y(t)

Gi£ sû h m x = x(t) câ h m ng÷ñc t = t(x). Khi �â h my = y(t)=y(t(x)) l h m cõa y theo bi¸n x .

y ′(x) =dy

y ′(t)dt

x ′(t)dt=

y ′(t)

x ′(t)⇒ y ′(x) =

y ′(t)

x ′(t)�¤o h m cõa

h m ©nH m y = y(x) vîi x ∈ (a, b) cho ©n bði ph÷ìng tr¼nh F (x , y) = 0�º t¼m �¤o h m h m ©n, ta �¤o h m hai v¸, coi x l bi¸n, y l h m theo x

Dao hamVi phan

C¡c v½ dö

V½ dö 1: T¼m �¤o h m cõa h m ng÷ñc h m f (x) = x + x3

f (x) l h m 1− 1 tr¶n R , �¤o h m f ′(x) = 1+ 3x2 6= 0,∀x . Do �â

1y ′(x)

1 + 3x2

V½ dö 2: T¼m �¤o h m cõa h m cho bði ph÷ìng tr¼nh tham sè:x = a · cos3t, y = b · sin3t, t ∈ (0, π/2).

x ′(t) = −3acos2t sin t 6= 0,∀t ∈ (0, π/2)

y ′(t) = 3bsin2t cos t

y ′(x) =y ′(t)

x ′(t)=

3bsin2t cos t−3acos2t sin t

= −b

atan t

Dao hamVi phan

C¡c v½ dö

f (x) l h m 1− 1 tr¶n R , �¤o h m f ′(x) = 1+ 3x2 6= 0, ∀x . Do �â

1y ′(x)

1 + 3x2

x ′(t) = −3acos2t sin t 6= 0,∀t ∈ (0, π/2)

y ′(x) =y ′(t)

x ′(t)=

= −b

atan t

Dao hamVi phan

C¡c v½ dö

1y ′(x)

1 + 3x2

x ′(t) = −3acos2t sin t 6= 0,∀t ∈ (0, π/2)

y ′(x) =y ′(t)

x ′(t)=

= −b

atan t

Dao hamVi phan

C¡c v½ dö

1y ′(x)

1 + 3x2

x ′(t) = −3acos2t sin t 6= 0, ∀t ∈ (0, π/2)

y ′(x) =y ′(t)

x ′(t)=

= −b

atan t

Dao hamVi phan

C¡c v½ dö

V½ dö 3: T¼m y ′(x) bi¸t y = y(x) l h m ©n x¡c �ành bði ph÷ìngtr¼nh e2x+y = x3 + cos y

e2x+y (2 + y ′(x)) = 3x2 − y ′(x) · sin y ⇒ y ′(x) =3x2 − 2e2x+y

e2x+y + sin y

Dao hamVi phan

C¡c v½ dö

e2x+y (2 + y ′(x)) = 3x2 − y ′(x) · sin y

⇒ y ′(x) =3x2 − 2e2x+y

e2x+y + sin y

Dao hamVi phan

C¡c v½ dö

e2x+y (2 + y ′(x)) = 3x2 − y ′(x) · sin y ⇒ y ′(x) =3x2 − 2e2x+y

e2x+y + sin y

Dao hamVi phan

�ành ngh¾a

�¤o h m cõa h m y = f (x) l mët h m sèCâ thº l§y mët l¦n núa cõa �¤o h m c§p mët ta �÷ñc kh¡i ni»m�¤o h m c§p 2

f ′′(x) =(f ′(x)

)′Ti¸p töc qu¡ tr¼nh tr¶n ta câ �¤o h m c§p n

f (n)(x) =(f (n−1)(x)

Dao hamVi phan

Cæng thùc leibnitz

Gi£ sû y = f .gDòng quy n¤p ta chùng minh �÷ñc cæng thùc:

(f · g)(n) =n∑

C kn f

(k) · g (n−k)

⇔ (f · g)(n) = C 0n f

(0) · g (n) + C 1n f

(1) · g (n−1) + · · ·+ Cnn f

(n) · g (0)

Trong �â quy ÷îc f (0) = f ; g (0) = g .Ph÷ìng ph¡p t½nh �¤o h m c§p cao

1 Sû döng �¤o h m c§p cao cõa mët sè h m sè �¢ bi¸t2 Ph¥n t½ch th nh têng c¡c h m �ìn gi£n3 Sû döng cæng thùc leibnitz

Dao hamVi phan

V½ dö

T½nh y (n)(x) bi¸t y =1

x2 − 4

Gi£i:

(x − 2)(x + 2)=

x − 2− 1

Sû döng cæng thùc(

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc y (n) =(−1)nn!

(x − 2)n+1− 1

(x + 2)n+1

Dao hamVi phan

V½ dö

x2 − 4

Gi£i:

(x − 2)(x + 2)=

x − 2− 1

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc y (n) =(−1)nn!

(x − 2)n+1− 1

(x + 2)n+1

Dao hamVi phan

V½ dö

x2 − 4

Gi£i:

(x − 2)(x + 2)=

x − 2− 1

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc y (n) =(−1)nn!

(x − 2)n+1− 1

(x + 2)n+1

Dao hamVi phan

V½ dö

x2 − 4

Gi£i:

(x − 2)(x + 2)=

x − 2− 1

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc y (n) =(−1)nn!

(x − 2)n+1− 1

(x + 2)n+1

Dao hamVi phan

V½ dö

x2 − 4

Gi£i:

(x − 2)(x + 2)=

x − 2− 1

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc y (n) =(−1)nn!

(x − 2)n+1− 1

(x + 2)n+1

Dao hamVi phan

V½ dö

T½nh y (100)(0) bi¸t y =1

x2 + 4

Gi£i:

(x − 2i)(x + 2i)=

x − 2i− 1

x + 2i

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc:

y (n) =(−1)nn!

(x − 2i)n+1− 1

(x + 2i)n+1

⇒ y (100) = (−1)100100!4i ·

(−2i)101 −1

(2i)101

)= 100!

4·2100

Dao hamVi phan

V½ dö

T½nh y (100)(0) bi¸t y =1

x2 + 4

Gi£i:

(x − 2i)(x + 2i)=

x − 2i− 1

x + 2i

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc:

y (n) =(−1)nn!

(x − 2i)n+1− 1

(x + 2i)n+1

⇒ y (100) = (−1)100100!4i ·

(−2i)101 −1

(2i)101

)= 100!

4·2100

Dao hamVi phan

V½ dö

T½nh y (100)(0) bi¸t y =1

x2 + 4

Gi£i:

(x − 2i)(x + 2i)=

x − 2i− 1

x + 2i

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc:

y (n) =(−1)nn!

(x − 2i)n+1− 1

(x + 2i)n+1

⇒ y (100) = (−1)100100!4i ·

(−2i)101 −1

(2i)101

)= 100!

4·2100

Dao hamVi phan

V½ dö

T½nh y (100)(0) bi¸t y =1

x2 + 4

Gi£i:

(x − 2i)(x + 2i)=

x − 2i− 1

x + 2i

1x + a

= (−1)nn!1

(x + a)n+1

Ta �÷ñc:

y (n) =(−1)nn!

(x − 2i)n+1− 1

(x + 2i)n+1

⇒ y (100) = (−1)100100!4i ·

(−2i)101 −1

(2i)101

)= 100!

4·2100

Dao hamVi phan

V½ dö

T½nh y (100)(x) bi¸t

y = sin2x

y = (3x2 + 1) ln x

y = (2x + 3) · cos 2x

Dao hamVi phan

Dinh nghiaCac phep toan vi phanUng dung

�ành ngh¾a

�ành ngh¾a kh£ vi

H m sè f (x) �÷ñc gåi l kh£ vi t¤i �iºm x0 n¸u

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

Khi �â, A ·∆x �÷ñc gåi l vi ph¥n cõa h m f (x) t¤i �iºm x0, kþhi»u df (x0) = A ·∆x

�ành lþ

H m sè y = f (x) kh£ vi t¤i x0 khi v ch¿ khi tçn t¤i f ′(x0).

Dao hamVi phan

�ành ngh¾a

�ành ngh¾a kh£ vi

H m sè f (x) �÷ñc gåi l kh£ vi t¤i �iºm x0 n¸u

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

Khi �â, A ·∆x �÷ñc gåi l vi ph¥n cõa h m f (x) t¤i �iºm x0, kþhi»u df (x0) = A ·∆x

�ành lþ

H m sè y = f (x) kh£ vi t¤i x0 khi v ch¿ khi tçn t¤i f ′(x0).

Dao hamVi phan

Chùng minh

N¸u f (x) kh£ vi t¤i x0, khi �â:

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

⇒ f (x0 + ∆x)− f (x0)

∆x= A +

o(∆x)

⇒ ∃f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

o(∆x)

Ng÷ñc l¤i, n¸u tçn t¤i

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

⇒ f (x0+∆x)−f (x0)∆x − f ′(x0)→ 0. Suy ra f (x) kh£ vi t¤i x0

Dao hamVi phan

Chùng minh

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

⇒ f (x0 + ∆x)− f (x0)

∆x= A +

o(∆x)

⇒ ∃f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

o(∆x)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

Dao hamVi phan

Chùng minh

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

⇒ f (x0 + ∆x)− f (x0)

∆x= A +

o(∆x)

⇒ ∃f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

o(∆x)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

Dao hamVi phan

Chùng minh

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

⇒ f (x0 + ∆x)− f (x0)

∆x= A +

o(∆x)

⇒ ∃f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

o(∆x)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

Dao hamVi phan

Chùng minh

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

⇒ f (x0 + ∆x)− f (x0)

∆x= A +

o(∆x)

⇒ ∃f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

o(∆x)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

⇒ f (x0+∆x)−f (x0)∆x − f ′(x0)→ 0.

Suy ra f (x) kh£ vi t¤i x0

Dao hamVi phan

Chùng minh

f (x0 + ∆x)− f (x0) = A ·∆x + o(∆x)

⇒ f (x0 + ∆x)− f (x0)

∆x= A +

o(∆x)

⇒ ∃f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

∆x= lim

∆x→0

o(∆x)

f ′(x0) = lim∆x→0

f (x0 + ∆x)− f (x0)

Dao hamVi phan

T½nh ch§t cõa vi ph¥n

Vi ph¥n cõa h m f (x) t¤i �iºm x0: df (x0) = f ′(x0)dx

T½nh ch§t1 dα = 0, α ∈ R

2 d (αf ) = α · df , α ∈ R

3 d (f + g) = df + dg

4 d (f · g) = gdf + fdg

)= gdf−fdg

C¡c t½nh ch§t n y suy ra trüc ti¸p tø t½nh ch§t cõa �¤o h m.

Dao hamVi phan

2 d (αf ) = α · df , α ∈ R

3 d (f + g) = df + dg

4 d (f · g) = gdf + fdg

)= gdf−fdg

Dao hamVi phan

2 d (αf ) = α · df , α ∈ R

3 d (f + g) = df + dg

4 d (f · g) = gdf + fdg

)= gdf−fdg

Dao hamVi phan

Vi ph¥n cõa h m hñp

{y = y(u)u = u(x)

⇒ y = y(u(x))

dy = y ′(x)dx = y ′(u) · u′(x)dx = y ′(u)du

dy = y ′(x)dx ; dy = y ′(u)duHai cæng thùc n y câ d¤ng gièng nhau, khæng phö thuëc bi¸n �ëclªp x hay bi¸n h m u.Vi ph¥n c§p 1 câ t½nh b§t bi¸n.Vi ph¥n cõa h m cho bði ptts{

x = x(t)y = y(t)

⇒ dy = y ′(x)dx =y ′(t)

x ′(t)dx

Dao hamVi phan

{y = y(u)u = u(x)

⇒ y = y(u(x))

x = x(t)y = y(t)

⇒ dy = y ′(x)dx =y ′(t)

x ′(t)dx

Dao hamVi phan

{y = y(u)u = u(x)

⇒ y = y(u(x))

dy = y ′(x)dx ; dy = y ′(u)du

Hai cæng thùc n y câ d¤ng gièng nhau, khæng phö thuëc bi¸n �ëclªp x hay bi¸n h m u.Vi ph¥n c§p 1 câ t½nh b§t bi¸n.Vi ph¥n cõa h m cho bði ptts{

x = x(t)y = y(t)

⇒ dy = y ′(x)dx =y ′(t)

x ′(t)dx

Dao hamVi phan

{y = y(u)u = u(x)

⇒ y = y(u(x))

dy = y ′(x)dx ; dy = y ′(u)duHai cæng thùc n y câ d¤ng gièng nhau, khæng phö thuëc bi¸n �ëclªp x hay bi¸n h m u.

Vi ph¥n c§p 1 câ t½nh b§t bi¸n.Vi ph¥n cõa h m cho bði ptts{

x = x(t)y = y(t)

⇒ dy = y ′(x)dx =y ′(t)

x ′(t)dx

Dao hamVi phan

{y = y(u)u = u(x)

⇒ y = y(u(x))

dy = y ′(x)dx ; dy = y ′(u)duHai cæng thùc n y câ d¤ng gièng nhau, khæng phö thuëc bi¸n �ëclªp x hay bi¸n h m u.Vi ph¥n c§p 1 câ t½nh b§t bi¸n.

Vi ph¥n cõa h m cho bði ptts{x = x(t)y = y(t)

⇒ dy = y ′(x)dx =y ′(t)

x ′(t)dx

Dao hamVi phan

{y = y(u)u = u(x)

⇒ y = y(u(x))

x = x(t)y = y(t)

⇒ dy = y ′(x)dx =y ′(t)

x ′(t)dx

Dao hamVi phan

Ùng döng vi ph¥n t½nh g¦n �óng

y = y(x) l h m kh£ vi trong l¥n cªn cõa x0.

f (x0 + ∆x)− f (x0) = f ′(x0)∆x + o(∆x)

⇒ f (x)− f (x0) ≈ f ′(x0) (x − x0)

∆f ≈ df

Cæng thùc t½nh g¦n �óng nhí vi ph¥n c§p 1:

f (x) ≈ f (x0) + f ′(x0) (x − x0)

Thay v¼ t½nh gi¡ trà ∆f phùc t¤p, ta t½nh df �ìn gi£n hìn.

Dao hamVi phan

f (x0 + ∆x)− f (x0) = f ′(x0)∆x + o(∆x)

⇒ f (x)− f (x0) ≈ f ′(x0) (x − x0)

∆f ≈ df

f (x) ≈ f (x0) + f ′(x0) (x − x0)

Dao hamVi phan

f (x0 + ∆x)− f (x0) = f ′(x0)∆x + o(∆x)

⇒ f (x)− f (x0) ≈ f ′(x0) (x − x0)

∆f ≈ df

f (x) ≈ f (x0) + f ′(x0) (x − x0)

Dao hamVi phan

V½ dö

V½ dö 1

Cho f (x) = x3 + x2 − 2x + 1a. T½nh ∆f v df n¸u x thay �êi tø 2 �¸n 2.01b. T½nh ∆f v df n¸u x thay �êi tø 2 �¸n 2.05

Gi£ia. f (2) = 23 + 22 − 2.2 + 1 = 9

f (2.01) = (2.01)3 + (2.01)2 − 2. (2.01) + 1 = 9.140701

∆f = f (x0 + ∆x)− f (x0) = f (2.01)− f (2) = 0.140701

df = f ′(x0) (x − x0) =(3.22 + 2.2− 2

)× 0.01 = 0.14

b. T÷ìng tü, ∆f = 0.717625, df = 0.7Khi x thay �êi nhä ∆f v df c ng g¦n nhau.

Dao hamVi phan

V½ dö

V½ dö 1

Gi£ia. f (2) = 23 + 22 − 2.2 + 1 = 9

f (2.01) = (2.01)3 + (2.01)2 − 2. (2.01) + 1 = 9.140701

∆f = f (x0 + ∆x)− f (x0) = f (2.01)− f (2) = 0.140701

df = f ′(x0) (x − x0) =(3.22 + 2.2− 2

)× 0.01 = 0.14

Dao hamVi phan

V½ dö

V½ dö 1

Gi£ia. f (2) = 23 + 22 − 2.2 + 1 = 9

f (2.01) = (2.01)3 + (2.01)2 − 2. (2.01) + 1 = 9.140701

∆f = f (x0 + ∆x)− f (x0) = f (2.01)− f (2) = 0.140701

df = f ′(x0) (x − x0) =(3.22 + 2.2− 2

)× 0.01 = 0.14

Dao hamVi phan

V½ dö

V½ dö 2

T½nh g¦n �óng 4√17

Gi£i Ta x²t h m sè f (x) = 4√x . �p döng cæng thùc t½nh g¦n �óng

ta câ:4√

x0 + ∆x ∼= 4√x0 +

4 4√

x03∆x

Chån x0 = 16, ∆x = 1 ta câ:4√17 ∼= 4

√16 +

4 4√163

.1 = 2 +1

4.23= 2.031

Dao hamVi phan

V½ dö

V½ dö 2

ta câ:4√

x0 + ∆x ∼= 4√x0 +

4 4√

x03∆x

Chån x0 = 16, ∆x = 1 ta câ:4√17 ∼= 4

√16 +

4 4√163

.1 = 2 +1

4.23= 2.031

Dao hamVi phan

V½ dö

V½ dö 2

ta câ:4√

x0 + ∆x ∼= 4√x0 +

4 4√

x03∆x

Chån x0 = 16, ∆x = 1 ta câ:4√17 ∼= 4

√16 +

4 4√163

.1 = 2 +1

4.23= 2.031

Dao hamVi phan

Vi ph¥n c§p cao

df (x) = f ′(x)dx l mët h m theo bi¸n x

. Vi ph¥n (n¸u câ) cõadf (x) �÷ñc gåi l vi ph¥n c§p 2 cõa h m y = f (x)

d2f (x) = d (df ) = d(f ′(x)dx

)= dxd

(f ′(x)

)′dx = f ′′(x)dxdx = f ′′(x)dx2

T÷ìng tü, vi ph¥n c§p n l vi ph¥n (n¸u câ) cõa vi ph¥n c§p n− 1

dnf (x) = f (n)(x)dxn

Dao hamVi phan

Vi ph¥n c§p cao

df (x) = f ′(x)dx l mët h m theo bi¸n x . Vi ph¥n (n¸u câ) cõadf (x) �÷ñc gåi l vi ph¥n c§p 2 cõa h m y = f (x)

d2f (x) = d (df ) = d(f ′(x)dx

)= dxd

(f ′(x)

)′dx = f ′′(x)dxdx = f ′′(x)dx2

Dao hamVi phan

Vi ph¥n c§p cao

df (x) = f ′(x)dx l mët h m theo bi¸n x . Vi ph¥n (n¸u câ) cõadf (x) �÷ñc gåi l vi ph¥n c§p 2 cõa h m y = f (x)

d2f (x) = d (df ) = d(f ′(x)dx

)= dxd

(f ′(x)

)′dx = f ′′(x)dxdx = f ′′(x)dx2

Dao hamVi phan

Cac dinh ly ve ham kha viPhat bieu cac dinh ly

�ành lþ Rolle: Cho h m y = f (x) thäa m¢n li¶n töc tr¶n [a, b], kh£vi trong (a, b) v f (a) = f (b). Khi �â tçn t¤i mët �iºm c ∈ (a, b)sao cho f ′(c) = 0.

toan 1- chuong 7

Education