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Национальный авиационный университет

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y′ = f ′(u) g′(x).

In other words, the derivative of the composite function y = f (u) where

u = g(x)

equals the product of the derivative of external function taken with

respect to internal argument

u , and that of internal function taken with respect

to independent variable x.

If u(x) is the

function

differentiable

point x,

the

following

rules are

fulfilled:

Table 17.2

1. (un )′ = nun−1u′

2. ( u )′ =

u′

2 u

3. (au )′ = au ln a u′

4. (eu )′ = eu u′

(loga u)′ =

u′

(ln u)′ =

1 u′

u ln a

(sin u)′ = cosu u′

(cosu)′ = − sin u u′

9. (tg u)′ =

′

10.

(ctg u)′ = −

u′

cos2 u

sin2 u

11.

(arcsin u)′ =

u′

12.

(arccos u)′ = −

u′

1− u2

(arctg u)′

u′

(arcctg u)′ = −

u′

13.

14.

1+ u2

15. (sh u)′ = ch u u′

16.

(ch u)′ = sh u u′

17.

(th u)′ =

′

18.

(cth u)′ = −

u′

ch2u

sh2u

Micromodule 17

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Using the definition of the derivative, find the derivative of the function y = x2 .

Solution. According to the definition of the derivative, we have

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(x2 )' = lim		f (x +	x) − f (x)		= lim	(x +	x)2 − x2	=
	x→0		x			x→0	x

= lim x→0	x2 + 2x x +		x2 − x2	= lim x→0 (2x +			x) = 2x
		x

Example 2. Find the derivative of the function y = cos x by the definition. Solution. We write the increment of function

y = cos(x +

x) − cos x.

Remember the first honorable limit

lim x→0

sin x

= 1

as well as formula

cos x − cos y = −2sin

x − y

sin

x + y

Then

(cos x)

= lim

x→0

cos(x + x) − cos x

= lim x→0

−2sin

2 sin(x

+ 2 )

− sin

x) = −1 sin x = − sin x.

= lim

lim

sin(x +

x→0

→0

Example 3. Find the derivative of the function

y = 3 x

at point

x = 0 .

Solution. We have

lim

x + x −3

lim

0 + x −3

x→0

= lim

= ∞.

x→0 ( x)23

So the limit increases infinitely when

x → 0. Therefore the derivative of the

function y = 3 x

at point

x = 0 does not exist. It might be said, as well, that

there exists no derivative of continuous function

y =| x |

within the complete

number scale at point

x = 0. Consequently, if a function is discontinuous at

pointі x, it does not follow necessarily that it has the derivative at this point.

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However, if the function is differentiable (if it has the derivative) at point x , it is continuous at this point then.

Find the derivatives of the functions:

Example 4. y = 4x5 −3x4 +1 .

Solution.

y′ = (4x5 − 3x4 + 1)′ = (4x5 )′ − (3x4 )′ + (1)′ = 4(x5 )′ − 3(x4 )′ = 4 5x4 − 3 4x3 = 20x4 − 12x3.

Example 5. y =

−

Solution.

−2

−1

2 )′ =

−2

)′ − (x

−1/ 2

)′ =

y ' =

−

= (3x

− x

3(x

= 3 (−2)x−2−1 − (−1/ 2)x−1/ 2−1 = = −6x−3 + (1/ 2)x−3/ 2

= −

2 x3

Example 6. y =

sin x

Solution.

′

(x2 )′ sin x − x2 (sin x)′

2x sin x − x2 cos x

y ' =

(sin x)

sin

sin x

Example 7. y = arcsin x tg x.

Solution.

′

(arcsin x tgx)

′

= (arcsin x) tgx +

+ arcsin x (tg x)

′

1− x2 tg x + arcsin x cos2 x .

Example 8. Find

y' (0),

y = ex x .

Solution. y′ = (ex )x + ex ;

y '(o) = e0 0 + e0 = 1.

Applying the rule of differentiation of composite function, find the derivatives of the functions:

Example 9. y = (x2 +1)3 .

Solution. Let us make a designation u = x2 +1, then y = (u(x))3 . By the rule of differentiation of the composite function, it follows

y′ = (u3 )′ = 3u2u′ = 3(x2 + 1)2 (x2 + 1)′ = 3(x2 + 1)2 2x = = 6x(x2 + 1)2 .

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Example 10. y = sin3 x .

Solution. Let us designate u = sin x , then

y = (u(x))3 . Hence,

y′ = (u3 )′ = 3u2u′ = 3sin2 x (sin x)′ = 3sin2 x cos x.

Example 11. y =

x4 + x3 +1 .

Solution. In this case

y =

where u = x4 + x3 +1 . Then

y ' = ( u ) ' =

u ' =

(x4

+ x3 + 1) '

4x3 + 3x2

2 u

2 x4 + x3 + 1 2 x4 + x3 + 1

Example 12. y =

cos2

Solution. We write the given function as follows y = cos−2 x . Then

u = cos x ,

y = u−2 ,

y′ = (u−2 )′ = −2u−3u′ = −2 cos−3 x(cos x)′ = 2 cos−3 x sin x.

Example 13. y = 5sin x .

Solution. We have the composite function

y = 5u

where u = sin x . Then

y′ = (5u )′ = 5u ln 5 u′ = 5sin x ln 5(sin x)′ = 5sin x ln 5cos x .

Example 14. y = tg ex .

Solution. We have the composite function

y = tg u , where u = ex . Then

y′ = (tg u)′ =

u′ =

(ex )′

cos2 u

cos2 ex

Example 15. y = arcsin x

arcsin x .

Solution. Having applied the product rule, we get

y = (arcsin

x)′

arcsin x + arcsin

arcsin x)′ =

( x)′

′

1− ( x)2

arcsin x + arcsin

x 2

arcsin x (arcsin x)

arcsin x + arcsin

2 x 1− x

arcsin x

1− x2

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Example 16. y = ln(x2 + 1) .
Solution. We have the composite function			y = ln u , where u = x2 +1.
Then
y′ = (ln u)′ = 1 u′ =	1	(x2 + 1)′ =		2x	.
y′ = (ln u)′ = 1 u′ =	x2 + 1	(x2 + 1)′ =			.
u	x2 + 1			x2 + 1

Example 17. y = arctg(log2 (x +cos x)) .

Solution. We have the composite function y = arctg u , where

u = log2 (x +

+cos x) is also composite, that is,

u = log2 w , w = x + cos x . In this case, we

find the derivative as follows:

′

= (arctg u)

′

(log2 w)′ =

+ u2

wln 2

(x + cos x)′

1− sin x

(1+ (log2 (x + cos x))2 )(x + cos x) ln 2

(1+ log22 (x + cos x))(x + cos x) ln 2

Example 18. y = arcctg2 (x3 − 2) .

Solution.

y′ = 2arcctg(x3 − 2) (arcctg(x3 − 2))′ = − 6x2 arcctg(x3 − 2) . 1+ (x3 − 2)2

Micromodule 17

CLASS AND HOME ASSIGNMENT

Find the derivatives of the functions:

y = 2 + x − x2 .

y = 3 4 x .

3. y = 3 +

− x3 .

y =

− x ln x .

а) y = arccos x 3x ;

b) y = 10x

/ x .

sin x

y = 4 tg x 6 x5 .

а) y =

cos x

;

b) y =

sin x

3 x2

y =

arcsin x

9. y = 2x sin x x4 .

10. y = arctg x +arcctg x .

arccos x

11.

y = (3x +

x )(

x − x) .

12. а)

y =

− x5

y =

2 − 7

; b)

7 x2

x5 + 1

+ 2

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13.

y = (x2 − 1)(x2 + 1)(x4 + 1) .

14.

y = (1+ x )(1+ 3 x ) .

15. а) y =

; b) y

3 x

. 16.

y = (x

−3x +2)(x

−5x +3) .

+ 4 x

17.

y =

18.

а) y =

−2x +4

; b) y =

3 x + 1

(x3 −1)(1−2x4 )

x + x

3 x2 − 1

19.

y = log2 x + 1/ ln x .

20.

y = tg x arctg x − 3arccos x .

Find the derivatives of the composite functions:

21. y = (3x + 2)5 .

24.y = arcsin(ln x)

27.y = 2arctg x .

30. y = 4 ctg x .

33. y = arccos 1−x

1+ x

35. y = sh(cos 1+x x ) .

22.	y = cos4 x .				23.	y = tg log3 x .
25.	y = arccos	x .			26.	y = 3x+ x .
28.	y = x ln sin 2x .				29.	y = ex tg(3 x + 1) .
31.	y = sin6 (cos 3x) .				32.	y = cos5 (log32 x) .
34. а) y = arctg			1− x	;	b) y =		arctg x −1	.
		1+ x					arctg x +1
36.	y = earccos(ln x) .

37. а) y = (x2 −1) arccos

;

b) y =

x2 +1

38.

y = ln(ln2 x) .

arccos x

39.

y = cos ln

− 1

40.

y =

sin x

41.

y =

tg x − x

+ 1

3 cos x

4 x

42.

y =

ln(x + arctg x)

43. y =

44.

y =

sin3 x

arctg ln x

cos4 x

3 arccos 4x

45.

y = 3sin 5x cos4 x .

46.

y = 4 sin

cos

47.

y = sin(cos x) cos(sin x).

48.

y = ln(x +

x2 −1) .

x2 −4

−3x

49.

y = 3

+4log3 cos 2x.

50.

y =ln(e

+ e

−1)

+arccose

4 x3 −1

51.

y = (1+ x2 )

1−x2 .

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Answers

1− 2x.

. 3.

−

−3x2 .

4. −

− 3

cos x

− ln x −1.

6. 12x + 5sin 2x .

44 x3

sin2 x

3cos2 x6 x

7. а)

−

2xsin x + cos x .

2x x3(ln 2 xsin x + xcos x + 4sin x) .

2 1− x2 arccos2 x

10. а)

11.

x − 6x + 1 .

12. а)

− 10x 4

13.

8x 7 .

14.

4x3 − 24x 2 + 40x − 19 .

(x5 + 1)2

19.

−

21. 15(3x + 2)4 .

22. −4cos3 xsin x.

23.

x ln 2

x ln2 x

x ln 3 cos2 (log3 x)

24.

25. −

27.

2arctg x ln 2

28. ln sin 2x + 2xctg2x .

x 1−ln2 x

1 + x2

2 x −x2

31. −18sin5(cos3x)cos(cos3x)sin 3x.

33.

34. а)

−1

35. −chcos

x(1+ x)

2 1− x2

+ 1

×sin

/(1+ x)2 .

36.

−earccosln x

39.

sin ln ex − 1

2e2x

. 44.

sin2 x (3+ sin2 x) .

x + 1

1 − ln2 x

ex + 1 1− e2x

cos5 x

48.

x2 − 1

Micromodule 17

SELF-TEST ASSIGNMENTS

17.1. Find the derivatives of the first order y = f (x).

17.1.1. а) y = cos2 x + sin(tg x) ;

y = ln2 arcsin

x ;

y = 2sin x+cos2 x ;

y = 5 (2x +1) arcctg x .

17.1.2. а)

y = 3 ctg x + tg x2

;

y = log3

;

1−x2

y = 10

ln x

3tan x ;

y = 5 arctg(ln2 x) − 1 .

17.1.3. а)

y = sin

x − 2 sin3 x ;

b) y = ln arccos

;

y = earctg x cos 2x ;

y = 4 log3 sin(x3 + 1) .

17.1.4. а)

y = x2 /(1+ cos2 2x) ;

y = 3 ln cos

x − 2

;

y = ln sin(3x x2 ) ;

y = 3 x

cos3 (tg x) .

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17.1.5. a) y = cos5 (sin 3x) ;	b)	y = (1+ cos2 x)5 sin 4x ;
c) y = ln(x +arccos 1−x2 ) ;	d)	y = ln5 arctg	1	.

			x

17.1.6. а) y = e x2

x / sin x 2 ;

c)y = 102− tg4 x ;

17.1.7.а) y = tg2 x − 2 ctg x2 ;

c)y = 25x 4cos x ;

17.1.8.а) y = x cos2 x − ctg 4x ;

c)y = arctg5 (e2x x) ;

17.1.9.а) y = x3 /(1+sin4 x) ;

2 sin 5x

;

y = 3

17.1.10. а)

y = sin

;

1− x

−x2

sin(3x −2) ;

y = e

17.1.11. а)

y =

2 +sin

1+ x2

;

1−x2

y = 2

ln x x3 ;

17.1.12. а)

y =

5 ctg2 (5 +1/ x)

;

y = tg(2cos x ) ln(x3x ) ;

17.1.13. а)

y =

sec2 (1+ x2 )

−1

;

cos x

y = 5 1−tg2 x

ctg 3x ;

b)y = arctg ln x ; ln arctg x

d)y = 6 e− x + 1 sin(4x + 1) .

b)y = (2 +ln2 sin x)3 ;

d)y = log32 arcsin(x2 ) .

b)y = arcsin3 ln sin 2x ;

d)	y =	1+ ln 2	x	.
		x3 + 2

b) y = log2 arctg(1−x2 ) ;

d) y = 1+sh		1+ x3		.
			1−x3

b) y = lg	x +		x2 −2	;
		3x

d)	y = cos(sin3 (x tg x)) .
b)	y =	1−arctg x	;

		ln2 x

d)y = tg4 (ch x) − ch(tg x2 ) .

b)y = ln arccos(2x −5) ;

d) y = c h2 (x2 − 1) − c h x .

b)	y = log24 arcsin(3x3 ) ;
d)	y = 4 2−x +1 cos 4 x .

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17.1.14.а) y = tg(cos(5 ctg x)) ;

c)y = 3sin x sin3 x +3 ;

17.1.15. а)	y = cos(sin					x tg x ) ;
c)	y = 6arctg			x	−2tg x ;
17.1.16. а)	y = tg2			1	−5 ctg 3x ;
			x2

c)	y = x	3 −x2			/ 2	−cos 2		x	;
		e
17.1.17. a)	y = (tg			3x ) / sin			x ;

c)y = 2arccos x cos2 x ;

17.1.18.а) y = 4 tg(x / 4) +ctg 4x ;

c)y = tg(4ln x +3tgx ) ;

17.1.19.а) y = cos3 5x −8 sin3 4x ;

c)y = 2x2 / x2 ;

17.1.20. а)	y = 5 sin4 x −cos x ;
c)	y = 2ln arcsin x ;

17.1.21.а) y = ctg3 (6 2−x tg2 x ) ;

c)y = e3 sin 5x−5 cos2 x ;

17.1.22. а) y = cos	x	;
	sin x

c) y = cos e4 x−1/ x ;

b) y = 3 ln arctg				x ;
d) y =	log2 (x +1/ x)				.
d) y =		2x3			.
		2x3
b) y = ln5 arctg				x +1	;
b) y = ln5 arctg				x −1	;
				x −1
d) y =	4	x −1	.
d) y =	4		.
		x +1
b) y = log34 sin				1+ x3 ;

d)	y = arc tg2 (x ln x) .
b)	y = arcsin4 ln ln x ;
d)	y = ln(x +ln(x + 1−x2 ) .
b)	y = arccos2 ln sin x ;

d) y =	2−arcctg			x	.

		log34 x
b) y = ln ln cos ln tg x ;
d) y = 7 log33 sin			1+ x .
b) y =	ln(1+ln2			x)		;
		log2 x

d) y =	sh2	(1+ x2 )		−2 th 2x .
		ch x

b)	y =	arcsin ln x	;
		ln(x2 +1)
d)	y = tg(4ln x +7ctg x ) .

b)y = logx 3+log34 x ;

d)y = ln5 arctg xx −+11 .

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17.1.23. а)

y =

;

y = x ln(x + 1−x2 ) ;

sin2 x −cos(x2 )

y = 2(x3−2) / sin x ;

y = tg2

−5 ctg4 2x .

x2 +1

17.1.24. а)

y = cos

sin x

;

y = sin ln tg x −ln ctg

;

x +1

y = cos 2x /(e

) ;

y =

1+ln4 x

log5 x

17.1.25. а)

y =

sin x

;

y = x2 (cos ln x −sin ln x) ;

3 cos x +cos3

y = 5

;

y = x ln(

4 − x

) .

17.1.26. а)

y = cos

1+ x3

+ cos x ;

y = log32 (ln x −logcos x 2) ;

y = 103x− 4 (3x − 4)10 ;

y = arccos

x −1

17.1.27. а)

y = 3 tg(x / 3) +cos

sin x ;

y = log33 log22 ln(5x −2) ;

y = 3sin2 3x

3cos x

;

y =

arcsin ln x .

x 2

17.1.28. а)

y = cos

;

y = ln(sin

x +

sin

ln x ) ;

x cos x

y = x cos(2arctan

1− x2

) ;

d) 102x−5 (7x2 − 1)8 .

17.1.29. a)

y =

cos3 (x 2 − 1/ x 2 )

;

y = ln cos log7 ctg ln x ;

cos x

y = e5 arctan

/ ln x ;

y = tg

sin 2x + x3 .

17.1.30. а)

y = x sin3 5x +cos ec2 x ;

y = ln3 sin(x cos x) ;

y = arcsin 8sin x

/ 2cos x ;

y = 3

tan x .

x ln x

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