Дифференцирование функции одной и нескольких переменных с приложениями. Мустафина Д.А - 11 стр.

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10
x
xvxuxxvxxu
x
y
y
xx
Δ
Δ
+
Δ
+
=
Δ
Δ
=
ΔΔ
)()()()(
limlim
00
=
=
x
xvxuvxvuxu
x
Δ
Δ+Δ+
Δ
)()())()()((
lim
0
=
=
Δ
Δ
Δ
+Δ+
Δ
+
=
Δ
x
xvxuvuuxvvxuxuxv
x
)()()()()()(
lim
0
=
Δ
Δ
Δ+
Δ
Δ
+
Δ
Δ
Δ
x
u
v
x
v
xu
x
u
xv
x
)()(lim
0
uvvuuvuvu
x
u
v
x
v
xu
x
u
xv
xxxx
+
=
+
+
=
Δ
Δ
Δ+
Δ
Δ
+
Δ
Δ
=
ΔΔΔΔ
0limlimlim)(lim)(
0000
3)
2
v
uvvu
v
u
=
, если v 0
Доказательство. Обозначим у=
v
u
. Тогда
=
Δ
Δ+
Δ+
=
Δ
x
xv
xu
xxv
xxu
y
x
)(
)(
)(
)(
lim
0
=
Δ
Δ+
Δ
+
=
Δ
x
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xu
vxv
uxu
x
)(
)(
)(
)(
lim
0
=
Δ+Δ
Δ
Δ
+
Δ
)())((
)()()()()()(
lim
0
xvvxvx
xvuxvxuxuvxvxu
x
=
Δ+
Δ
Δ
Δ
Δ
=
Δ+Δ
ΔΔ
=
ΔΔ
vvv
x
v
u
x
u
v
vxvxvx
vxuuxv
xx
2
0
2
0
lim
))()((
)()(
lim
2
0
2
00
lim
limlim
v
uvvu
vvv
x
v
u
x
u
v
x
xx
=
Δ+
Δ
Δ
Δ
Δ
=
Δ
ΔΔ
1.3 Производные основных элементарных функций
1) (x
m
)
= mx
m-1
;
6)
()
xx sincos =
13)
()
x
ch
thx
2
1
=
2)
()
xx
ee =
7)
()
chxshx =
14)
()
x
h
cth
2
1
=
3)
()
aaa
xx
ln=
8)
()
shxсhx =
15)
()
2
1
1
x
arctgx
+
=
4)
()
ax
x
a
ln
1
log =
10)
()
xx sincos =
16)
()
2
1
1
x
arcctgx
+
=
5)
()
x
x
1
ln =
11)
()
x
tgx
2
cos
1
=
17)
()
2
1
1
arcsin
x
x
=
5)
()
xx cossin =
12)
()
x
ctgx
2
sin
1
=
18)
()
2
1
1
arccos
x
x
=
                     Δy         u ( x + Δx) ⋅ v( x + Δx) − u ( x) ⋅ v( x)
         y ′ = lim      = lim                                             =
               Δx →0 Δx  Δx → 0                    Δx
       (u ( x) + Δu )(v( x) + Δv) − u ( x)v( x)
= lim                                           =
 Δx →0                    Δx
      v( x)u ( x) + u ( x)Δv + v( x)Δu + ΔuΔv − u ( x)v( x)        ⎛      Δu          Δv      Δu ⎞
= lim                                                       = lim ⎜ v( x)    + u ( x)    + Δv    ⎟=
 Δx→0                           Δx                            Δx →0⎝      Δx          Δx      Δx ⎠

               Δu                Δv               Δu
= v( x) lim        + u ( x) lim     + lim Δv lim     = u ′v + uv ′ + 0u ′ = u ′v + v ′u
        Δx → 0 Δ x         Δx →0 Δx  Δx → 0 Δx →0 Δx

               ′
           ⎛u⎞   u ′v − v ′u
        3) ⎜ ⎟ =             , если v ≠ 0
           ⎝v⎠       v2

                                                         u ( x + Δx) u ( x)
                                                                     −
                                    u                    v( x + Δx) v( x)
        Доказательство. Обозначим у= . Тогда y ′ = lim                      =
                                    v             Δx → 0          Δx
                u ( x) + Δu u ( x)
                            −
                v( x) + Δv v( x)           u ( x)v( x) + Δuv( x) − u ( x)v( x) − Δvu ( x)
        = lim                      = lim                                                  =
         Δx → 0          Δx         Δx → 0              Δx(v( x) + Δv)v( x)
                                                     Δu      Δv            Δu            Δv
                                                 v        −u       v lim      − u lim
           v( x)Δu − u ( x)Δv                        Δx      Δx = = Δx → 0 Δx     Δx → 0 Δx   u ′v − v′u
= lim                                  = lim                                                =
  Δx → 0 Δx (v 2 ( x )   + v( x)Δv)     Δx → 0       v 2 + vΔv         v 2 + v lim Δv             v2
                                                                                  Δx → 0

                         1.3 Производные основных элементарных функций

                                                                                      1
          1) (xm)′ = mxm-1;             6) (cos x )′ = − sin x         13) (thx )′ = 2
                                                                                    ch x
                                                                                        1
                  ′
          2) (e x ) = e x               7) (shx )′ = chx               14) (cth )′ = − 2
                                                                                      sh x

                                                                       15) (arctgx )′ =
                                                                                             1
                                        8) (сhx )′ = shx
                   ′
          3) (a x ) = a x ln a                                                             1+ x2

          4) (log a x )′ =                                             16) (arcctgx )′ = −
                                 1                                                             1
                                        10) (cos x )′ = − sin x
                              x ln a                                                         1+ x2

                                                                       17) (arcsin x )′ =
                                                                                               1
          5) (ln x )′ =                 11) (tgx )′ =
                          1                                 1
                          x                               cos 2 x                            1− x2

                                                                       18) (arccos x )′ = −
                                                                                                   1
                                        12) (ctgx )′ = −
                                                                1
          5) (sin x )′ = cos x
                                                             sin 2 x                           1− x2


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