10.26. y =

lg

(x2 +2x)

10.27. y =

(x −8)8

.

.

(x +1)8

lg (2x −3)

52x

10.28. y =

.

10.29. y = ln

1+cos3x

1−cos3x

2 + 2 +52x

находим:

y′

1

ln y = x ln x;

= x′ln x + x(ln x)′ = ln x + x x

= ln x +1;

y

11.10.

y =

x3

(x2 +1)2

.

11.11. y =

x(

x2 +1

)

.11.12. y = xex .

x(x −1)

1− x2

11.16. (ctg2x)x2 .

11.17. (sin 5x)1+x2 .

11.18. y = (ln 5x)x .

11.19. y = xarcsin x .

11.20. y = (ln x)2x .

11.21. y = xarctgx .

11.22. y = xctgx .

11.23. y = (cos5x)x2

.11.24. y = (tgx)2x .

11.25. y =

x(x2 −1)

x

2x

11.27. y = (ctgx)

sin x

. 11.26. y =

.

.

x2 +

3

2

11.28. y = x2tgx .

11.29. y = xe5 x .

1

11.30. y = x

2x

.

ски x =10t,

y = t2 .

Решение.

Находим xt′ =10, yt′ = 2t . По формуле y′x =

yt′

xt′

находим производную y′x =

2t

=

1 t .

10

5

12.15. x = t (1−sin t), y = t cost .

12.16. x = e2t ,

y = cos 2t .

12.17. x = e−t sin t,

y = et cost .

12.18. x = a −t

,

y =

t

.

a +t

a +t

12.19. x = t3 +3t +2, y = 3t5 +5t3 +3. 12.20. x = t2 , y =

t3

.

2 −t

12.21. x = 3cos3 3t, y = 4sin2 t .

12.22. x = t2 +1,

y = et2 .

12.23. x = t2 +t +1,

y = t3 +t .

12.24. x = sin

t

,

y = cost .

2

1

12.25. x = t −sin t,

y =1−cost .

12.26. x = ctgt,

y =

.

cos2

2t

t

12.27. x = t cost, y = 3t sin t .

12.28. x = cos

,

y =t −sin t .

2

12.29. x = cos at, y = sin at .

12.30. x = tgt +ctgt, y =3ln ctgt .

3x

2

−4xy

2

−2x

2

2yy

′

+5

+ y

′

= 0 , т. е.

y

′

=

−3x2 +4xy2 −5

.

1−4x2 y

Задание 13. Найти производные неявных функций.

13.1. y2 −2x = 0

13.2. x2 + y2 −5 = 0

13.3. ln y −2x = 0 .

13.4. ex + x + y = 0 .

13.5. xy2 −4 = 0 .

13.6. arctgy − y + x = 0.

13.7. ex +ey

−3xy

−1 = 0.

13.8. x2 + y2 −4x −10y +4 = 0 .

13.9. 5x2 y +3xy −2y2 +2 = 0 .

2

2

2

13.10. x3

+ y3

= a3 .

13.11. x

= y

.

13.12. y = tg (x + y).

y

x

13.13. ex −ey = y − x .

13.14. 6x2 +3xy +2y2 +3 = 0 .

13.15. x3

+ y3 −4xy = 0 .

13.16. y +ln y = x +3 .

13.17. x2

+5xy + y2 −7 = 0 .

13.18.

+

=

.

x

y

3

13.22. e−y −ey +3xy = 0 .

13.23. xy − yx

= 0 .

13.24. exy − x3 − y3 = 4 .

13.25. xy −arctg

x

= 3 .

y

13.26. 3

+ 3

−9 = 0 .

13.27. 2x +2y

= 2x+y .

x

y

13.28. x2 +2xy2 +3y4 =

6 .

13.29.

x2

−

y2

=1.

13.30. 3x3 +ln 3y − x2ey

= 0 .

9

8

y′

(y′x )

′

2

3t2 3

y′x =

t

,

y′′xx =

t

xt′ = 2t,

yt′ = 3t

y′x = 2t = 2 t.

xt′

xt′

3

3

3

Имеем (y′x )t′ =

, следовательно

y′′xx =

2

=

.

2

2t

4t

y′′ =

−

ex + y ′

= −

(ex + y′) x −(ex + y)

.

x

x

2

Имеем x0 =1,

y0 = −e .

Находим y′

M : y′

M = −

e1 −e

= 0 .

1

e1 +0

1−

e1

−e

Вычисляем y′′

M

= −(

)

2 (

)= −e .

1

14.1. y = x5 +52x ,

x

= 0.

14.2. y = x2 ln 1 ,

x

=1.

0

x

0

14.3. y = ln ctg4x,

x

=

π

.

14.4. y = arccos 2 ,

x

= 4.

16

0

x

0

x = t −sin t,

x = e

t2

,

14.5.

t0 =π .

14.6.

t0

=1.

y =1−cost,

y

=

4

−t2 ,

=

t

+ 1

sin 2t,

x = 2

(

t −sin t

)

,

π

14.7.

x

2

t0

= 0.

14.8.

t0

=

.

2

3

y = 2(1−cost),

y = cos t,

14.9. tg y = x +3y,

M (0, 0).

x = t +ln cost,

t0

=

π

.

14.10.

4

y = t −ln sin t,

14.11. x y2 −4 = 0,

M (1, 2).

x = 2t

−t

5

,

t0

14.12.

=1.

y =t3 −8t −1,

2

14.13. x2 + y2 −25 = 0, M (3,4).

x =t

+1,

t0

= 2.

14.14.

y = et

2

,

t

14.15. f (x)=

1−2x

,

x0 = 4.

14.16. x = e

sin t,

t0

=

π .

cost,

1+ 3 2x

y = et

4

14.17. f (x)=

14.18. f (x)= ln (1+4−2x ), x0 = 0.

x +2

, x0

=1.

x

14.19. f (x)= e−x cos3x, x0

= 0.

14.20.2y =1+ xy3 , x0

= 4, y0

=1.

14.21. yey = ex+1,

x0

= 0, y0 =1. 14.22.y = ln (x2 −4),

x0

= 3.

14.23. y = (x + y)3 −27(x − y),

x0

= 2, y0 =1.

14.24.

y2 = x +ln

y

,

x

= y

=1.

0

0

x

14.25.

ey = 4x −7 y,

x

= 1

,

y

= 0.

0

4

0

14.26.

arctg y = 5x +4y,

x0

=

y0

= 0.

14.27.

3y = 7 + xy3 ,

x

= −4,

y

=1.

0

0

14.28.

ln y −

y

= 2,

x = − 1 ,

y

=1.

x

0

2

0

14.29. x2 y2 + x = 5y,

x

= y

=

0.

0

0

14.30. y2 − x = cos y,

x

= −1,

y

= 0.

0

0

плоской кривой y = f (x) в точке M0 (x0 , y0 )

соответственно:

y − y0

= f ′(x0 ) (x − x0 ),

(12.1)

y − y0

= −

1

(x − x0 ).

(12.2)

f ′(x0 )

Находим производную

f (x):

f ′(x)= 3x2 −10x +10.

Вычисляем значение

f ′(x) в точке x0 =1:

f ′(1)= 3.

Подставляя эти значения в формулы (12.1) и ( 12.2) (учитывая,

что y0 = −1) получаем:

y −(−1)= 3 (x −1) или

3x − y −4 = 0 –

уравнение касательной;

y −(−1)

= −

1 (x −1)

или

x +3y +2 = 0 –

3

ной параметрическими уравнениями x = t, y = t2

в точке t0 = 2 .

Решение. Имеем x0

= 2,

y0 = 4 .

y′x =

yt′

xt′ =1,

yt′ = 2t y′x =

2t

= 2t.

;

1

xt′

Выражаем y′:

y′ =

8 −18x −4y

.

4x +12y +16

Вычисляем y

′

M

′

= −

8

.

при x0 = 2, y0 =1: yM

9

15.1.

y = 3x − x2 +7,

M (5;−3). 15.2. y = x3 +4x +6, M (−1;1).

15.3.

y = x e−x , M

(0;0). 15.4. y = x +ln(x2 +4), M (0;ln 4).

15.5. y = cos3 x, M (π;−1).

15.6. y = ln2 x,

M (e;1).

15.7. (x +2)2 +(y −6)2 = 34,

M (1; 1).

15.8. y = x2 −7x +3,

M (1;−3).

15.9. y2 + x y +sin y = 0, M (0;0).

x = t +1, y =

1

t0 =1.

x = t −sin t

t0

15.10.

,

15.11.

,

=π .

t −2

y =1−cost

2

π

x = t

+1

15.12. y = 4sin 6x,

;4

15.13.

,

t0

= 2.

M

2

.

2

y = et

t

15.14. x2 + y2 −25 = 0, M (3;4).

x = arcsin

15.15.

2

,

t0

= 0.

4 −t

2

y =

15.16. 2y =1+ xy3 ,

M (1;1).

15.17. xy2 −2 = 0,

M (2;1).

t

15.18. y2 = 25x −9,

M (1;4).

15.19.

x =

,

t0

= 3.

y

= 5

t

−2t

15.20. y = x +arctg y, M (1;1).

15.21. x = e

,

t0

=1.

y = e4t

2

15.22. y = 4ey +4x,

M (−1;0).

15.23. x = 5cos

t

,

t0 = π .

y = 3sin2 t

4

x

= arccost

15.24. y2 + x2

= sin y, M (0;0).

15.25.

,

t0

= 0.

=

1−t2

y

2

15.26.

x

+

y

= 2,

M (1;1).

15.27. x = 6t

−

4,

t0

=1.

y = 3t5