Дифференциальное исчисление. - 22 стр.

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22 §3. ðÒÏÉÚ×ÏÄÎÙÅ É ÄÉÆÆÅÒÅÎÃÉÁÌÙ ×ÙÓÛÉÈ ÐÏÒÑÄËÏ×
ðÏÄÓÔÁ×ÌÑÑ ÎÁÊÄÅÎÎÙÅ ÐÒÏÉÚ×ÏÄÎÙÅ É ÞÉÓÌÁ × ÆÏÒÍÕÌÕ ìÅÊÂÎÉÃÁ ÐÒÉ n = 5,
ÐÏÌÕÞÁÅÍ
y
(5)
= (uv)
(5)
=
= C
0
5
u
(5)
v + C
1
5
u
(5−1)
v
0
+ C
2
5
u
(5−2)
v
00
+ C
3
5
u
(5−3)
v
000
+ C
4
5
u
0
v
(5−1)
+ C
5
5
uv
(5)
=
= 1 ·120 ·e
x
+ 5 ·120x ·e
x
+ 10 ·60x
2
·e
x
+ 10 ·20x
3
·e
x
+ 5 ·5x
4
·e
x
+ 1 ·x
5
·e
x
=
= (120 + 600x + 600x
2
+ 200x
3
+ 25x
4
+ x
5
)e
x
.
ðÒÉÍÅÒ 9. ÷ÙÞÉÓÌÉÔØ 55-À ÐÒÏÉÚ×ÏÄÎÕÀ ÆÕÎËÃÉÉ y = (x
2
− 17) cos 3x.
òÅÛÅÎÉÅ. ðÏÌÁÇÁÑ u = cos 3x É v = x
2
− 17, ÎÁÈÏÄÉÍ (ÓÍ. ÐÒÉÍÅÒ 6):
u
(n)
= 3
n
cos
3x + n
π
2
,
v
0
= 2x, v
00
= 2, v
000
= v
(4)
= v
(5)
= . . . = v
(54)
= v
(55)
= 0.
óÌÅÄÏ×ÁÔÅÌØÎÏ, ×ÓÅ ÓÌÁÇÁÅÍÙÅ × ÆÏÒÍÕÌÅ ìÅÊÂÎÉÃÁ, ÓÏÄÅÒÖÁÝÉÅ ÐÒÏÉÚ×ÏÄ-
ÎÙÅ ÆÕÎËÃÉÉ v ×ÙÛÅ ×ÔÏÒÏÇÏ ÐÏÒÑÄËÁ, ÂÕÄÕÔ ÒÁ×ÎÙ ÎÕÌÀ. ÷ÙÞÉÓÌÑÅÍ ËÏ-
ÜÆÆÉÃÉÅÎÔÙ ÐÒÉ ÆÕÎËÃÉÑÈ v, v
0
É v
00
:
C
0
55
= 1, C
1
55
= 55, C
2
55
=
55 · 54
2!
= 1485.
ðÏÄÓÔÁ×ÌÑÑ ÎÁÊÄÅÎÎÙÅ ×ÙÒÁÖÅÎÉÑ × ÆÏÒÍÕÌÕ ìÅÊÂÎÉÃÁ ÐÒÉ n = 55, ÐÏÌÕ-
ÞÁÅÍ:
y
(55)
= (uv)
(55)
= C
0
55
u
(55)
v + C
1
55
u
(54)
v
0
+ C
2
55
u
(53)
v
00
=
= 1 · 3
55
cos
3x + 55 ·
π
2
· (x
2
− 17) + 55 · 3
54
cos
3x + 54 ·
π
2
· 2x+
+ 1485 · 3
53
cos
3x + 53 ·
π
2
· 2 =
= 3
55
(x
2
− 17) sin(3x) − 110x · 3
54
cos(3x) −2970 ·3
53
sin(3x).
3.4. äÉÆÆÅÒÅÎÃÉÁÌÙ ×ÙÓÛÉÈ ÐÏÒÑÄËÏ×
ðÕÓÔØ ÆÕÎËÃÉÑ y = f(x) ÄÉÆÆÅÒÅÎÃÉÒÕÅÍÁ × ËÁÖÄÏÊ ÔÏÞËÅ x ÎÅËÏÔÏÒÏÇÏ
ÐÒÏÍÅÖÕÔËÁ. ôÏÇÄÁ ž ÄÉÆÆÅÒÅÎÃÉÁÌ dy ×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ
dy = f
0
(x)dx
É ÎÁÚÙ×ÁÅÔÓÑ ÄÉÆÆÅÒÅÎÃÉÁÌÏÍ ÐÅÒ×ÏÇÏ ÐÏÒÑÄËÁ ÆÕÎËÃÉÉ f (x).
äÉÆÆÅÒÅÎÃÉÁÌ d(dy) ÏÔ ÄÉÆÆÅÒÅÎÃÉÁÌÁ dy ÎÁÚÙ×ÁÅÔÓÑ ÄÉÆÆÅÒÅÎÃÉÁÌÏÍ
×ÔÏÒÏÇÏ ÐÏÒÑÄËÁ ÆÕÎËÃÉÉ f(x) É ÏÂÏÚÎÁÞÁÅÔÓÑ d
2
y, ÔÏ ÅÓÔØ
d
2
y = f
00
(x)(dx)
2
.
22                         §3. ðÒÏÉÚ×ÏÄÎÙÅ É ÄÉÆÆÅÒÅÎÃÉÁÌÙ ×ÙÓÛÉÈ ÐÏÒÑÄËÏ×

ðÏÄÓÔÁ×ÌÑÑ ÎÁÊÄÅÎÎÙÅ ÐÒÏÉÚ×ÏÄÎÙÅ É ÞÉÓÌÁ × ÆÏÒÍÕÌÕ ìÅÊÂÎÉÃÁ ÐÒÉ n = 5,
ÐÏÌÕÞÁÅÍ

 y (5) = (uv)(5) =
     = C50u(5) v + C51u(5−1)v 0 + C52u(5−2)v 00 + C53u(5−3)v 000 + C54u0v (5−1) + C55uv (5) =
= 1 · 120 · ex + 5 · 120x · ex + 10 · 60x2 · ex + 10 · 20x3 · ex + 5 · 5x4 · ex + 1 · x5 · ex =
                                     = (120 + 600x + 600x2 + 200x3 + 25x4 + x5)ex .
     ðÒÉÍÅÒ 9. ÷ÙÞÉÓÌÉÔØ 55-À ÐÒÏÉÚ×ÏÄÎÕÀ ÆÕÎËÃÉÉ y = (x2 − 17) cos 3x.
     òÅÛÅÎÉÅ. ðÏÌÁÇÁÑ u = cos 3x É v = x2 − 17, ÎÁÈÏÄÉÍ (ÓÍ. ÐÒÉÍÅÒ 6):
                             (n)     n
                                                    π
                            u = 3 cos 3x + n               ,
                                                     2
         v 0 = 2x, v 00 = 2, v 000 = v (4) = v (5) = . . . = v (54) = v (55) = 0.
óÌÅÄÏ×ÁÔÅÌØÎÏ, ×ÓÅ ÓÌÁÇÁÅÍÙÅ × ÆÏÒÍÕÌÅ ìÅÊÂÎÉÃÁ, ÓÏÄÅÒÖÁÝÉÅ ÐÒÏÉÚ×ÏÄ-
ÎÙÅ ÆÕÎËÃÉÉ v ×ÙÛÅ ×ÔÏÒÏÇÏ ÐÏÒÑÄËÁ, ÂÕÄÕÔ ÒÁ×ÎÙ ÎÕÌÀ. ÷ÙÞÉÓÌÑÅÍ ËÏ-
ÜÆÆÉÃÉÅÎÔÙ ÐÒÉ ÆÕÎËÃÉÑÈ v, v 0 É v 00:
                 0         1           2   55 · 54
               C55  = 1, C55 = 55, C55   =         = 1485.
                                             2!
ðÏÄÓÔÁ×ÌÑÑ ÎÁÊÄÅÎÎÙÅ ×ÙÒÁÖÅÎÉÑ × ÆÏÒÍÕÌÕ ìÅÊÂÎÉÃÁ ÐÒÉ n = 55, ÐÏÌÕ-
ÞÁÅÍ:
 y (55) = (uv)(55) = C55
                      0 (55)
                         u v + C55  1 (54) 0   2 (53) 00
                                       u v + C55 u v =
             55
                            π  
                                       2             54
                                                                    π
      = 1 · 3 cos 3x + 55 ·       · (x − 17) + 55 · 3 cos 3x + 54 ·      · 2x+
                              2                                      2
                                    53
                                                   π
                         + 1485 · 3 cos 3x + 53 ·       ·2=
                                                    2
                 = 355(x2 − 17) sin(3x) − 110x · 354 cos(3x) − 2970 · 353 sin(3x).

3.4. äÉÆÆÅÒÅÎÃÉÁÌÙ ×ÙÓÛÉÈ ÐÏÒÑÄËÏ×

  ðÕÓÔØ ÆÕÎËÃÉÑ y = f (x) ÄÉÆÆÅÒÅÎÃÉÒÕÅÍÁ × ËÁÖÄÏÊ ÔÏÞËÅ x ÎÅËÏÔÏÒÏÇÏ
ÐÒÏÍÅÖÕÔËÁ. ôÏÇÄÁ ž ÄÉÆÆÅÒÅÎÃÉÁÌ dy ×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ
                                         dy = f 0(x)dx
É ÎÁÚÙ×ÁÅÔÓÑ ÄÉÆÆÅÒÅÎÃÉÁÌÏÍ ÐÅÒ×ÏÇÏ ÐÏÒÑÄËÁ ÆÕÎËÃÉÉ f (x).
   äÉÆÆÅÒÅÎÃÉÁÌ d(dy) ÏÔ ÄÉÆÆÅÒÅÎÃÉÁÌÁ dy ÎÁÚÙ×ÁÅÔÓÑ ÄÉÆÆÅÒÅÎÃÉÁÌÏÍ
×ÔÏÒÏÇÏ ÐÏÒÑÄËÁ ÆÕÎËÃÉÉ f (x) É ÏÂÏÚÎÁÞÁÅÔÓÑ d2 y, ÔÏ ÅÓÔØ
                                      d2 y = f 00 (x)(dx)2.

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