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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

201

(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.

202

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 .

203

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

204

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

205

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 .

206

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) .

207

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 .

208

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 .

209

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

210

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