Дифференциальное исчисление. - 4 стр.

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4 §1. ðÒÏÉÚ×ÏÄÎÁÑ ÆÕÎËÃÉÉ
òÅÛÅÎÉÅ. ëÁÖÄÁÑ ÉÚ ÄÁÎÎÙÈ ÆÕÎËÃÉÊ Ñ×ÌÑÅÔÓÑ ÓÔÅÐÅÎÎÏÊ ÆÕÎËÃÉÅÊ,
ÐÏÜÔÏÍÕ ×ÓÅ ÐÒÏÉÚ×ÏÄÎÙÅ ÎÁÈÏÄÑÔÓÑ ÐÏ ÆÏÒÍÕÌÅ (x
α
)
0
= αx
α1
. éÍÅÅÍ:
x
17
0
= 17x
171
= 17x
16
;
x
πe
0
= (π e)x
πe1
;
1
x
0
=
x
1
0
= 1 ·x
11
= x
2
=
1
x
2
;
x
0
=
x
1
2
0
=
1
2
x
1
2
1
=
1
2
x
1
2
=
1
2
·
1
x
1
2
=
1
2
x
;
3
x
2
0
=
x
2
3
0
=
2
3
x
2
3
1
=
2
3
x
1
3
=
2
3
·
1
x
1
3
=
2
3
3
x
;
1
5
x
7
0
=
1
x
7
5
0
=
x
7
5
0
=
7
5
x
7
5
1
=
7
5
x
12
5
=
7
5
·
1
x
12
5
=
7
5
5
x
12
.
1.3. ðÒÏÉÚ×ÏÄÎÁÑ ÓÕÍÍÙ, ÒÁÚÎÏÓÔÉ, ÐÒÏÉÚ×ÅÄÅÎÉÑ É ÞÁÓÔÎÏÇÏ
ðÒÏÉÚ×ÏÄÎÙÅ ÓÕÍÍÙ, ÒÁÚÎÏÓÔÉ, ÐÒÏÉÚ×ÅÄÅÎÉÑ É ÞÁÓÔÎÏÇÏ Ä×ÕÈ ÆÕÎËÃÉÊ
u = u(x) É v = v(x) ÎÁÈÏÄÑÔÓÑ ÐÏ ÓÌÅÄÕÀÝÉÍ ÆÏÒÍÕÌÁÍ:
(u + v)
0
= u
0
+ v
0
, (u v)
0
= u
0
v
0
,
(uv)
0
= u
0
v + uv
0
, (cu)
0
= cu
0
, c ¡ ÞÉÓÌÏ,
u
v
0
=
u
0
v uv
0
v
2
, v 6= 0.
ðÒÉÍÅÒ 5. îÁÊÔÉ ÐÒÏÉÚ×ÏÄÎÕÀ ÆÕÎËÃÉÉ
1
x
3
5 ln x.
òÅÛÅÎÉÅ.
1
x
3
5 ln x
0
=
1
x
3
0
(5 ln x)
0
=
x
3
0
5 (ln x)
0
=
= 3x
4
5
1
x
=
3
x
4
5
x
=
3 + 5x
3
x
4
.
ðÒÉÍÅÒ 6. îÁÊÔÉ ÐÒÏÉÚ×ÏÄÎÕÀ ÆÕÎËÃÉÉ
2 ch x
3
+
cth x
4
.
òÅÛÅÎÉÅ.
2 ch x
3
+
cth x
4
0
=
2 ch x
3
0
+
cth x
4
0
=
2
3
(ch x)
0
+
1
4
(cth x)
0
=
=
2
3
sh x +
1
4
·
1
sh
2
x
=
2 sh x
3
1
4 sh
2
x
.
ðÒÉÍÅÒ 7. f (x) = 5x
3
3x
2
2x + 7, ÎÁÊÔÉ f
0
(0), f
0
(2), f
0
(1).
4                                                             §1. ðÒÏÉÚ×ÏÄÎÁÑ ÆÕÎËÃÉÉ

   òÅÛÅÎÉÅ. ëÁÖÄÁÑ ÉÚ ÄÁÎÎÙÈ ÆÕÎËÃÉÊ Ñ×ÌÑÅÔÓÑ ÓÔÅÐÅÎÎÏÊ ÆÕÎËÃÉÅÊ,
ÐÏÜÔÏÍÕ ×ÓÅ ÐÒÏÉÚ×ÏÄÎÙÅ ÎÁÈÏÄÑÔÓÑ ÐÏ ÆÏÒÍÕÌÅ (xα )0 = αxα−1. éÍÅÅÍ:
        0
   x17 = 17x17−1 = 17x16;
         0
   xπ−e = (π − e)xπ−e−1;
   0
    1           0                              1
           = x−1 = −1 · x−1−1 = −x−2 = − 2 ;
    x                                           x
   √ 0       1 0
                    1 1        1 1      1 1        1
      x = x 2 = x 2 −1 = x− 2 = · 1 = √ ;
                     2          2        2 x2     2 x
   √ 0  2  0 2 2               2 1     2 1        2
    3
       x2 = x 3 = x 3 −1 = x− 3 = · 1 = √                ;
                       3           3       3 x3     33x
          0  0 
       1         1         − 57
                                0     7 − 7 −1     7 − 12    7 1            7
    √        =     7  =  x         = −   x  5   = −   x 5 = −  ·      = −  √    .
     5
       x7        x5                    5            5         5 x 125      5
                                                                          5 x12

1.3. ðÒÏÉÚ×ÏÄÎÁÑ ÓÕÍÍÙ, ÒÁÚÎÏÓÔÉ, ÐÒÏÉÚ×ÅÄÅÎÉÑ É ÞÁÓÔÎÏÇÏ

   ðÒÏÉÚ×ÏÄÎÙÅ ÓÕÍÍÙ, ÒÁÚÎÏÓÔÉ, ÐÒÏÉÚ×ÅÄÅÎÉÑ É ÞÁÓÔÎÏÇÏ Ä×ÕÈ ÆÕÎËÃÉÊ
u = u(x) É v = v(x) ÎÁÈÏÄÑÔÓÑ ÐÏ ÓÌÅÄÕÀÝÉÍ ÆÏÒÍÕÌÁÍ:
                        (u + v)0 = u0 + v 0,        (u − v)0 = u0 − v 0 ,
                      (uv)0 = u0v + uv 0 ,        (cu)0 = cu0 , c ¡ ÞÉÓÌÏ,
                                   u 0 u0v − uv 0
                                          =              , v 6= 0.
                                     v            v2
     ðÒÉÍÅÒ 5. îÁÊÔÉ ÐÒÏÉÚ×ÏÄÎÕÀ ÆÕÎËÃÉÉ x13 − 5 ln x.
     òÅÛÅÎÉÅ.
                   0  0
      1                     1                 0        −3 0
                                                            − 5 (ln x)0 =
                                                         
        3
           − 5 ln x    =      3
                                  −   (5 ln x)  =    x
      x                     x
                                                                 1       3    5      3 + 5x3
                                                 = −3x−4 − 5 = − 4 − = −                      .
                                                                 x      x     x         x4
     ðÒÉÍÅÒ 6. îÁÊÔÉ ÐÒÏÉÚ×ÏÄÎÕÀ ÆÕÎËÃÉÉ 2 ch3 x + cth4 x .
     òÅÛÅÎÉÅ.
                      0              0            0
      2 ch x cth x              2 ch x           cth x        2           1
              +           =                +               = (ch x)0 + (cth x)0 =
          3         4              3               4          3           4
                                                                       
                                               2          1         1        2 sh x      1
                                            = sh x + · − 2                 =        −         .
                                               3          4       sh x          3     4 sh2 x
     ðÒÉÍÅÒ 7. f (x) = 5x3 − 3x2 − 2x + 7, ÎÁÊÔÉ f 0(0), f 0(2), f 0 (−1).