# Дифференциальное исчисление функций нескольких переменных. Скляренко В.А - 50 стр.

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• ## Математический анализ

10. f(x, y)= sin (x + y
2
), x
0
= 1~2 π, y
0
= 0.
11. f(x, y)=
º
y + cos x, x
0
= 0, y
0
= 0.
12. f(x, y)= ln (e + ye
x
), x
0
= 1, y
0
= 0.
13. f(x, y)= ch (x y
2
),1 x
0
= 1, y
0
= 1.
14. f( x, y)= e
e
x+ y
1
, x
0
= 0, y
0
= 0.
15. f(x, y)=
sh x
º
y
, x
0
= 0, y
0
= 1.
16. f(x, y)=
x
x + y
, x
0
= 0, y
0
= 1.
17. f(x, y)= ln (1 + sh (x + y)), x
0
= 0, y
0
= 0.
18. f(x, y)=
y
cos x
, x
0
= 0, y
0
= 1.
19. f(x, y)= ln (y + x
2
), x
0
= 0, y
0
= 1.
20. f(x, y)=
3
»
y + sh x, x
0
= 0, y
0
= 1.
21. f(x, y)=
x
1 x y
, x
0
= 0, y
0
= 0.
22. f(x, y)= e
sin(x+ y)
, x
0
= 0, y
0
= 0.
23. f(x, y)= ln (1 + sin (x y)), x
0
= 2, y
0
= 2.
24. f(x, y)= sin (y + x
2
), x
0
= 1, y
0
= 1.
25. f(x, y)= ln (cos (x y)), x
0
= 0, y
0
= 0.
26. f(x, y)=
4
º
2 x + 2 y + xy, x
0
= 0, y
0
= 0.
27. f(x, y)=
1
cos (x + y)
, x
0
= 1, y
0
= 1.
28. f(x, y)=
e
y
( x 1)
2
, x
0
= 2, y
0
= 0.
29. f(x, y)=
3
»
cos (x y), x
0
= 1, y
0
= 1.
30. f(x, y)= ln (y+ ch x), x
0
= 0, y
0
= 0.
Задача 7. Используя формулу Тейлора, разложить функцию u = u(x, y)
по степеням x x
0
, y y
0
.
1. u = 2x
4
2x
2
y
2
xy
3
+ y
4
+ 4x
3
+ 2xy x 3; x
0
= 1, y
0
= 3.
2. u = 4x
4
4x
2
y
2
3xy
3
y
4
+ x
3
x
2
y+ 2y
3
xy 4x; x
0
= 2, y
0
= 1.
3. u = 3x
4
xy
3
4xy
2
+ y
3
+ 3x; x
0
= 2, y
0
= 2.
50
 10. f(x, y) = sin (x + y2 ) , x0 = 1~2 π, y0 = 0.
º
11. f(x, y) = y + cos x, x0 = 0, y0 = 0.
12. f(x, y) = ln (e + yex ), x0 = 1, y0 = 0.
13. f(x, y) = ch (x − y2 ) ,1 x0 = 1, y0 = −1.
x+y
14. f(x, y) = e e      −1 ,    x0 = 0, y0 = 0.
sh x
15. f(x, y) = º , x0 = 0, y0 = 1.
y
x
16. f(x, y) =      , x = 0, y0 = 1.
x+y 0
17. f(x, y) = ln (1 + sh (x + y)), x0 = 0, y0 = 0.
y
18. f(x, y) =        , x = 0, y0 = 1.
cos x 0
19. f(x, y) = ln (y + x2 ), x0 = 0, y0 = 1.
»
20. f(x, y) = 3 y + sh x, x0 = 0, y0 = 1.
x
21. f(x, y) =         , x = 0, y0 = 0.
1−x− y 0
22. f(x, y) = esin(x+y) , x0 = 0, y0 = 0.
23. f(x, y) = ln (1 + sin (x − y)), x0 = 2, y0 = 2.
24. f(x, y) = sin (y + x2 ) , x0 = 1, y0 = −1.
25. f(x, y) = ln (cos (x − y)), x0 = 0, y0 = 0.
º
26. f(x, y) = 4 2 − x + 2 y + xy, x0 = 0, y0 = 0.
1
27. f(x, y) =              , x = 1, y0 = −1.
cos (x + y) 0
ey
28. f(x, y) =                  , x0 = 2, y0 = 0.
(x − 1)
2

»
29. f(x, y) =    3
cos (x − y), x0 = 1, y0 = 1.
30. f(x, y) = ln (y + ch x), x0 = 0, y0 = 0.

Задача 7. Используя формулу Тейлора, разложить функцию u = u(x, y)
по степеням x − x0 , y − y0 .
1. u = 2x4 − 2x2 y2 − xy3 + y4 + 4x3 + 2xy − x − 3;                 x0 = 1, y0 = −3.
2. u = −4x4 − 4x2 y2 − 3xy3 − y4 + x3 − x2 y+ 2y3 − xy− 4x;                   x0 = 2, y0 = −1.
3. u = −3x4 − xy3 − 4xy2 + y3 + 3x;                    x0 = −2, y0 = −2.

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