ЛЕКЦІЇ ² ВПРАВИ

,
4 0 . lim lim 1 x
⎡∆x
⎤
∆ y ⎛ ∆ x ⎞∆x
= t
= log
a 1 + = ⎢
x
⎥ =
∆x→0∆x ∆x→0x ⎜
x
⎟
⎝ ⎠ ⎢ ⎥
⎢⎣∆x→0⇒t
→0⎥⎦
Çîêðåìà,
( e
x
)
¢ = x . (7.4.3)
e
1
1 1
t
limlog a (1 t)
x t→0
x a
= + = .
ln
Ó öüîìó ëàíöþæêó ð³âíîñòåé ìè çàñòîñóâàëè ôîðìóëó
(6.3.12).
Îòæå,
Çîêðåìà,
1
( loga x)
¢ = .
x ln a
1
( ln x)
¢ = .
x
7.4.4. Ïîõ³äíà ñòåïåíåâî¿ ôóíêö³¿ y = x α
Íåõàé α º äîâ³ëüíå ä³éñíå ÷èñëî. Òîä³ îáëàñòü ³ñíóâàííÿ
ôóíêö³¿ çàëåæèòü â³ä α.
Ïîçíà÷èìî ÷åðåç Õ îáëàñòü ³ñíóâàííÿ ôóíêö³¿ ïðè ô³êñîâàíîìó
α. ³çüìåìî äîâ³ëüíå x ∈ X, àëå õ ≠ 0 (x = 0 áóëî
ðîçãëÿíóòî ðàí³øå). Ó â³äïîâ³äíîñò³ äî ñõåìè îá÷èñëåííÿ
ïîõ³äíî¿ (äèâ. ï. 7.2) áóäåìî ìàòè:
1 0 . y + ∆y =(x + ∆x) α , ∆x ≠ 0.
α α α
æ
α
x
ö
ç æ D ö
ç
çç
ç x
÷ ø çè ÷ ø .
2 0 . D y = ( x+Dx) - x = x ç1+ -1
7.4.3. Ïîõ³äíà ïîêàçíèêîâî¿ ôóíêö³¿ y = a x , x ∈ R,
a>0, a ≠ 1
1 0 . y + ∆y = a x + ∆x , ∆x ≠ 0.
2 0 . ∆y = a x + ∆x − a x = a x (a ∆x − 1).
3 0 .
D y a
=
Dx
x
x
( a
D -1)
Dx
.
3 0 .
4 0 .
æ
α
x
ö
α
æ ö
x
D
1 1
+ -
ç
y x
÷
D çèç è ø ÷ ø.
=
Dx
Dx
α
⎛
α ⎛ ∆x
⎞ ⎞
x
⎜1+ 1
x
⎟ −
⎡ ∆x
⎤
∆y
⎜⎝ ⎠ ⎟
t = ,
lim =
⎝ ⎠
lim
= ⎢
x
⎥ =
∆x
∆x
⎢
⎥
⎢⎣∆x→0⇒t
→0⎥⎦
∆x→0 ∆x→0
∆x
t
∆y 1 ,
1
4 0 x a − ⎡∆ x = t ⎤ x a − x
. lim = a lim = = a lim = a ln a
∆x→0 ∆x ∆x→0 ∆
⎢
x ∆x
→ 0 ⇒
⎥
t →
.
⎣
0⎦
t→0
t
Ó öüîìó ëàíöþæêó ð³âíîñòåé ìè çàñòîñóâàëè ôîðìóëó
(6.3.13).
Îòæå,
a
¢ = a a . (7.4.2)
x x
( ) ln
α
α−1 (1 + t) −1
α−1
= limx
=α⋅ x . (7.4.4)
t→0
t
 îñòàííüîìó ëàíöþæêó ð³âíîñòåé (7.4.4) ìè ñêîðèñòóâàëèñÿ
ôîðìóëîþ (6.3.14).
Îòæå, âñòàíîâëåíî ôîðìóëó
¢ = α × ¹ . (7.4.5)
α
α-
( x ) x 1 , x 0
206 207