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Математический анализ Раздел: Дифференциальное исчисление Тема: Производная функции. Лектор Пахомова Е.Г. 20 1 0 г. Глава II . Дифференциальное исчисление функции одной переменной.

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Лектор Пахомова Е.Г.

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5260913

:

:

..

2010 .


5260913

II.

, .

5.

1. .

y=f(x) x0 .

x0 x , x0+xD(f).

f(x0)=f(x0+x) f(x0).


5260913

. y=f(x) x0 x, x0 ( ), ..

:

y=f(x) x0 ()

( ).

:

y=f(x) x0 ,

y=f(x) x0 .


5260913

1 ( - ).

y=f(x) x0 .

2 ( - ).

y=f(x) x0 , f(x) .

. x0 .

, y=|x| -, x0=0.


5260913

x0f(x0) , D1 D(f).

y=f(x)

y=f(x) f(x).

. ,

(sinx)=cosx, (cosx)=sinx, x

(ex)=ex , (ax)=axlna , x


5260913

2.

1) .

y=f(x) x , f(x) y x.

.

) S=S(t) , t.

S(t0) t0.

) q=q(t) , t.

q(t0) t0, .. t0.

) m=m(x) [a;x].

m(x) x0, .. x0.


5260913

2) .

, M0 .

, , .

M0 M0M1, M1 M0, .

, M0 , .


5260913

y=f(x).

M0(x0;f(x0)) - M0N.

, : f(x0) y=f(x) M0(x0;f(x0)).

( ).

y=f(x) M0(x0;f(x0))


5260913

.

1), M0 , M0, M0.

.. k1k2=1, y=f(x) M0(x0;f(x0))

, f(x0)0.

f(x0)=0, y=f(x) M0(x0;f(x0))

y=f(x0),

x=x0.


5260913

2) y=f(x) M0(x0;f(x0)) M0N, M0M1 Ox.

, y=f(x) M0(x0;f(x0)) , y=f(x) x0.

M0 y=f(x) Ox 90 x0, x0 f(x) II ,


5260913

3.

1) , ..

C=0, .

2) () () , ..

3) :

. . ,


5260913

, .

: .

5) :

6) (t) t, f(u) u=(t), y=f((t)) t,

( ).

7) 3 ( ).

y=f(x) x0, f(x0)0. x=(y), y0=f(x0)


5260913

.

1), (sinx)=cosx, (cosx)=sinx, (ex)=ex,

2) , ,


5260913

( , . ).

-.


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