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<<< < Предыдущая 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 2324 / 2824 25 26 27 28 > Следующая >>>

d 2 y

′

d n − 1 y

′

d 3 y

d x 2

d n y

d x

n − 1

d x 3

x t′

d x n

x t′

20.4. Higher order derivatives of implicit function

Assume that a function

y = f (x)

is given implicitly with equality F(x, y) = 0.

Executing differentiation with respect to x and solving the obtained equation relatively to y′ it is possible to determine the first derivative.

To find the second derivative it is necessary to differentiate the first derivative with respect to x and to put its values into obtained relation. While continuing differentiation the any order derivatives can be determined one after another. All of them will be expressed through the independent variable x and the function y itself.

20.4. Higher order differentials

Definition 3.24. The differential of the first-order differential of function

f (x) is

called the second-order differential of

twice

differentiable

function

y = f (x)

i.e.

In general case

d 2 y = d(dy).

d n y = d(d n−1 y).

The differentials of the function

y = f (x)

are calculated by the formulas:

y =

′′

y = f

′′′

, …,

y = f

(n)

f (x)(dx)

(x)(dx)

If x is some function of a variable t, then

d 2 y = f ′′(x)(dx)2 + f ′(x)x′′(t)(dt)2

and so on. Thus the higher order differentials don’t have invariance property.

Micromodule 20

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Determine the third-order derivative of the function y = x4 − − 2x2 + 3x − 5.

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Solution. Let’s find the third-order derivative step by step y′ = 4x3 − 4x + 3 , y′′ = 12x2 − 4 , y′′′ = 24x .

Example 2. Determine the second-order derivative of the function

y =

x2 −a2 −

ln(x +

x2 −a2 ) ,

a = const .

Solution. We receive

− a

+ x ( x

− a

a 2

(ln (x +

− a

)

−

)) =

(

))=

− a

+ x

2 x

−

a 2

− a

2 x

− a

x +

− a

x 2 − a 2 +

−

2(x +

)

− a

x 2 − a 2 + x 2

− 2

a 2 ( x 2 − a 2

+ x)

2 x 2 − a 2

(x + x 2 − a 2

)

x 2 − a 2

2x2 − a2

−

x2 −a2

−a

;

2 x2 − a2

x2 −a2

y '' = (

x 2 − a 2 )

(x 2 − a 2 )' =

2x =

2 x2 − a2

2 x 2 − a 2

x2 − a2

Example 3. Determine the n-order derivative of the functions:

а)	y = ex ; b) y = ax ; c)	y = sin x ; d) y =	1	.

Solution.			x

а) y′ = ex , y′′ = ex , y(n) = ex ;
b) y′ = ax ln a , y′′ = ax ln2 a , y(n) = ax lnn a ;
c)	y′ = cos x = sin(x + π ) ,	y′′ = − sin x = sin(x + 2			π ) ,
	2				2
y	′′′ = − cos x = sin(x + 3 π ) , y(n) = sin(x + n π ) ;
	2	2
d) y′ = − x−2 , y′′ = 2x−3 ,		y′′′ = −2 3x−4 , y(4) = 2 3 4x−5 , …,
y(n) = (−1)n 2 3 4 … nx−(n+1) = (−1)n n!x−(n+1) .

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Example 4. Determine the n-order derivative of the function

y =	3x +1	.

x2 +2x −3

Solution. Let’s decompose this function into partial fractions:

3x +1	=	3x +1	=	A		+	B	=	A(x + 3) + B(x − 1)	.
x2 +2x −3		(x −1)(x +3)		x −1			x +3		(x −1)(x + 3)

So,

3x +1= A(x +3) + B(x −1)

Suppose x = 1, then 4 = 4A, whence A = 1. If x = – 3, then – 8 = – 4B, B = 2. Thus

y =		1		+		2		.
			x −1				x +3

Finally according to case d) of the previous example, we obtain
y(n) = (−1)n n!		1			+ 2(−1)n n!				1		.
	(x − 1)n+1								(x +3)n+1

Example 5. Determine the fifth-order derivative y(5) of the function
	y = (x2 − 3x + 4)e2x .
Solution. Let’s designate u = e2x ,						v = x2 − 3x + 4 .				Applying Leibniz's
formula we can obtaine
y(5) = ((x2 − 3x + 4)e2x )(5)					= (e2x )(5) (x2 − 3x + 4) +
+ C1 (e2x )(4)		(2x − 3) + C2 (e2x )(3)							2 .
5						5
Here we consider that (x2 − 3x + 4)(n)					= 0 if n > 2. Then we get

y(5) = 32e2x (x2 − 3x + 4) + 5 16e2x (2x − 3) + 10 8e2x 2 =

= e2x [32(x2 − 3x + 4) + 80(2x − 3) + 160] = e2x (32x2 + 64x + 48) .

Example 6. Find		d 2 y		, if x = 2 cos t , y			= 3sin t .
Example 6. Find		dx2		, if x = 2 cos t , y			= 3sin t .
		dx2
Solution. We have
	dy		=	(3sin t)t′	=	3 cos t	= −	3	ctg t ;
	dx		=	(2 cos t)′	=	−2 sin t	= −	2	ctg t ;
	dx			(2 cos t)′		−2 sin t		2
				t

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(−

ctg t)′

y =

2 sin2 t

t =

dx2

(2 cos t)t′

−2 sin t

Example 7. Find d3y, if

y = cos 3x .

Solution. Let’s determine derivatives

= −	3	.
	4 sin3 t

y′ = −3sin 3x , y′′ = −9 cos 3x , y′′′ = 27 sin 3x .

Therefore

d3 y = 27 sin 3x(dx)3 .

Example 8. Find d2y(0), if y = 4− x2 .

Solution. Let’s determine the second-order derivative in the point x = 0. We obtain

y′ = 4− x2 ln 4(−2x) ,

y′′ = −2 ln 4[(4− x2 )′x + 4− x2 ] = −2 ln 4 4− x2 [−2x2 ln 4 + 1] .

Thus

d 2 y(0) = y′′(0)dx2 = −2 ln 4dx2 .

Micromodule 20

CLASS AND HOME ASSIGNMENT

Find

d 2 y

dx2

y = (x3 − 2)4 .

y = x4 + 1 .

y = ex sin 2x .

4. y = 4x (x + 1) .

y = x x .

y = ln tg x .

y =

8. y =

2x − 1

y = (ln x)x .

x2 + 1

10.

y = sin3 x .

11.

y = 4 cos3 x − 3cos x .

Determine the n-order derivative of function:

12.

y = 3− x .

13.

y = ln x .

14. y = sin kx .

15.

y = sin x sin 2x .

16.

y =

− x

x .

17.

y =

18.

y =

x − 5

cos2

x2 + 3x + 2

− 4x +

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Determine the n-order derivative of function applying Leibniz's formula:

19.

y = (x3 − 2x + 5) sin x ,

n = 10 .

20.

y = (x2 + 4x − 3)2x ,

n = 8 .

21.

y = x ln(x2 − 3x + 2) ,

n = 6 .

Find

d 2 y

of implicit function:

dx2

22.

x3 + y3 = 3xy .

23. ex + x = ey + y .

24. cos(x + y) = x .

25.

= 2 px .

26.

y = sin(x + y) .

27. ln(x + y) = y − x .

28.

= y .

29.

= y .

Find

d 2 y

of parametric function:

dx2

30. y = 3t − t3 , x = 2t − t2 .

31. y = et sin t ,

x = et cos t .

32. y = ln t , x = t6 .

33. y = t2 / 2 , x = arctg t .

34. y = cos 2t , x = sin2 t .

35. y = ln(1+ t2 ) , x = arcctg t .

Answers

1. 12x(x3 − 2)2 (11x3 − 4).

2. 2x2 (x4 + 3)(x4 + 1)−3/ 2 .

5. xx (ln x + 1)2 + xx−1.

6. −4cos 2x / sin2 2x .

11. − 9cos3x .

12. (−1)n 3− x lnn 3 .

13. (−1)n−1(n − 1)!/ xn.

14.

15.

16. n! .

sin(kx + nπ / 2).

2[cos(x + nπ / 2) −

3 cos(3x

+ nπ/ 2)] .

17.

(−1)n n![1/(x + 1)n+1 − 1/(x +2)n+1] .

18. (−1)n n![2/(x −1)n+1 − 1/(x − 3)n+1] .

19. (30x2 + 700)cos x − (x3 − 272x + 5)sin x. 20.														2x ln6 2[ln2 2(x2 + 4x − 3) +16ln2(x + 2) + 42] .
21.		24(6x		− 11)	+		24(6x − 23)			.	22.	2xy(3xy − x3 − y3 − 1)					. 23.	(e y − e x )(e x+ y − 1)			.
21.			(x −	1)6	+			(x − 3)6		.	22.		( y 2 − x)3				. 23.	(e y + 1)	3		.
			(x −	1)6				(x − 3)6					( y 2 − x)3					(e y + 1)	3
			− cos(x + y)						p 2				− cos		x + y
24.			− cos(x + y)			.		25. −	p 2		. 26.		− cos		2	. 27.	− 4(x + y) /(x + y − 1)			3	.
24.			sin3 (x + y)			.		25. −	y3		. 26.		4 sin		x + y	. 27.	− 4(x + y) /(x + y − 1)				.
													4 sin		2
															2
30.		3/(4(1− t)) . 31.						2e−t /(cost − sin t)3 . 32.						− 1/(6t12 ) .			33. 3t 4 + 4t 2 + 1 .		34. 0.
35.	2(1 + t 2 ) .
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Micromodule 20

SELF-TEST ASSIGNMENT

20.1. Find d2y at point x .
	0
20.1.1. y = x x − 3,	x0 = 12.
20.1.3. y = x2 2x + 3,		x0	= 11.
20.1.5. y = (ln x)	2x − 1,	x0 = 5.
20.1.7. y = sin3 x cos5 x,		x	= π .
		0	3
			3
20.1.9. y = sin3 x cos7 x,		x	= π .
		0	4
			4
20.1.11. y = sin3 x tg5 x,		x	= π .
		0	4
			4

20.1.13.	y = (x −1) 3 5x + 2, x0 = 5.
20.1.15.	y = (x + 1)2 3 3x + 5,	x0 = 1.
20.1.17.	y = (x − 1)5 3 2x − 2,	x0 = 5.

20.1.19. y =

3x + 2

, x

= 2.

20.1.21. y =

− 4

20.1.23.

y =

sin3

cos x

20.1.25.

y =

sin4

cos2 x

20.1.27. y = ctg5 x,

= π .

20.1.29. y = x4

2x + 7,

= 9.

20.1.2. y = x2 x − 5,	x	= 6.
20.1.4. y = (2x − 1)2	0
	x + 2,		x = 7.
20.1.6. y = (ln x) 2x + 1,			0
		x0 = 12.
20.1.8. y = sin2 x cos4 x, x			= π .
		0	6

20.1.10. y = sin4 x cos6 x,		x0 =			π		.

			6
20.1.12. y = sin2 x tg4 x,		x	=	π		.

		0	3

20.1.14. y = (2x − 1) 4 3x + 4, x0 = 4.

20.1.16. y = (3x − 1)3 4 7x + 2, x0 = 2.

20.1.18. y = (4x − 1)3 5 x − 2, x0 = 3.

20.1.20. y =		x + 5		,			x		= 4.
20.1.20. y =				,			x		= 4.
	(x − 2)3						0
	(x − 2)3
20.1.22. y =	4	x2 − 9			,		x		= 5.
20.1.22. y =					,		x		= 5.
		x2					0
		x2
20.1.24. y =	cos4 x		,			x	=			π	.
20.1.24. y =			,			x	=				.
	sin x					0			3
	sin x								3
20.1.26. y = tg4 x,				x		=		π		.
20.1.26. y = tg4 x,				x		=				.
				0			6
							6

20.1.28.	y = (x + 1)3 3 x + 2, x0 = 6.
20.1.30.	y = (2x − 3)2 x + 3, x0 = 1.

20.2. Find	d 2 y	of parametric function:
	dx2

20.2.1. x = 5(2t − sin 2t), y = 10 sin2 t.

20.2.2. x =	1	, y = tg t − t.
	cos t

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20.2.3. x =

y = ctg t + t.

sin t

20.2.4. x = lg sin t,

y = lg cos t.

20.2.5. x = sin lg t,

y = tg lg t.

20.2.6. x = cos t + t sin t, y = sin t − t cos t.

20.2.7.

x = cos 2t + 2t sin 2t, y = sin 2t − 2t cos 2t.

20.2.8. x = ln(1+ t2 ),

y = t − arctg t.

20.2.9. x = arcsin et ,

y =

1− e2t .

20.2.10.

x = sin et , y = cos et .

20.2.11.

x = ln ctg t,

y =

sin 2t

20.2.12.

x = ln t,

y =

t − 1

t + 1

20.2.13.

x = ln(1+ t),

y = arctg t .

20.2.14.

x = arctg et ,

y =

e2t + 1

20.2.15.x = ln(1+ 4t2 ), y = 2t − arctg 2t.

20.2.16.x = sin3 et , y = cos3 et .

20.2.17.	x = 3t cos t,			y = 3t sin t.
20.2.18.	x = arc ctg t,			y = log3 (t2 + 1).
20.2.19.	x = cos 2t − ln ctg t, y = sin 2t.
20.2.20.	x = tg 2t , y = ln cos3 2t .
20.2.21.	x = arccos 2t, y =					1− 4t2 .
20.2.22.	x = arcsin t,			y =		1− t2 .
20.2.23.	x =	1	tg2 t,	y =		1	.
		2			cos t

20.2.24.	x = ctg2 et ,			y =		1		.
					sin et

20.2.25.	x =	t2 + 1,		y = ln(t + t2 + 1).
20.2.26.	x = ln(1+ t4 ), y = arctg(t2 ).

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20.2.27.		1
	x = ln ctg(1+ t), y =			.
			sin(1+ t)
20.2.28.	x = tg et ,	y = ln cos2 et .
20.2.29.	x = arctg t,	y = log2 (t2 + 1).
20.2.30.	x = ln(1+ t6 ), y = arctg(t3 ).

20.3. Determine the n-order derivative of function applying Leibniz's formula:

20.3.1.y = e2x−1 (x3 − 3x + 4) , n=10.

20.3.2.y = (x3 − 2x + 1) sin 2x , n=12 .

20.3.3.y = 2x (3x3 − 7x + 5) , n=15.

20.3.4.y = (x3 + 2x2 + 3) ln x , n=8.

20.3.5.y = (x2 + 4x − 5) ln x , n=10.

20.3.6.y = (x2 + 5x − 12) 2x , n=9.

20.3.7.y = (x3 + 5x2 − 2) sin x , n=11.

20.3.8.y = (x3 − 2x2 − 3) cos x , n=9.

20.3.9.y = (x2 − 4x2 − x + 3) cos 2x , n=8.

20.3.10.y = (x3 + 2x + 3) ln x , n=7.

20.3.11.y = (x2 − 5x) ln(x + 1) , n=8.

20.3.12.y = (2x2 − 7x) ln(x − 2) , n=10.

20.3.13.y = (4x2 − 9x) ln(x − 2) , n=6.

20.3.14.y = (2x2 − 3x −11) 3x , n=9.

20.3.15.y = (4x2 − x + 8) 4− x , n=10.

20.3.16.y = (2x3 − 4x2 − 1) 5− x , n=7.

20.3.17. y = (x2 + 3x − 7) 6− x , n=8.

20.3.18.y = e− x−1 (x3 − x2 + 2) , n=10.

20.3.19.y = 2− x−1 (x3 − 6x + 3) , n=7.

20.3.20.y = 3− x (2x2 + x + 3) , n=8.

20.3.21.y = (4x3 − x2 − 1) cos 2x , n=10.

20.3.22.y = (3x2 − 4x + 1) cos(x + 1) , n=12.

20.3.23.y = (5x2 − 3x + 2) sin(x − 2) , n=11.

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20.3.24.	y = (6x3 − 1) sin(x + 3) , n=15.
20.3.25.	y = (2 x3 − 4 x − 1) ln( x − 3) , n=10.

20.3.26.y = (3x3 + 2x + 3) ln(x + 3) , n=8.

20.3.27.y = (x2 − 2x − 1) ln(2x − 1) , n=11.

20.3.28.y = (x3 + 2x − 3) ln(2x + 1) , n=10.

20.3.29.y = e−2x (x3 − 6) , n=8.

20.3.30.y = 2− x (x3 − 4) , n=10.

Micromodule 21

BASIC THEORETICAL INFORMATION

BASIC THEOREMS OF DIFFERENTIAL CALCULUS

Fermat’s, Rolle’s, Lagrange’s, Cauchy’s theorems. Taylor’s, Maclaurin’s formulas. L’Hospital’s rule.

Literature: [4, part 5], [6, chapter 5, §5], [7, chapter 6, §19], [9], [10, chapter 4, §24], [11, chapter 4, §4], [12, chapter 4, §§1-7] .

21.1. Fermat’s, Rolle’s, Lagrange’s, Cauchy’s theorems

Those functions, which have the derivative on the current interval, are distinguished by the properties , which help to investigate the function behavior on the interval of differentiability.

Theorem 3.11 (Fermat’s theorem). Let the function f (x) be continuous on

the interval (a; b) and reaches it’s minimum and maximum values at the point on this interval. Therefore, if there’s a derivative f ′(c) at the point с, then f ′(c) = 0 .

This theorem has trivial geometric interpretation. If at the point x = c function f (x) reaches minimum or maximum values (Fig. 3.17 and 3.18), then the tangent to the graph of this function at the point (c; f (c)) is parallel to the x-axis.

Theorem 3.12 (Rolle’s theorem). If the function f (x) is continuous on the segment [a; b] , has the derivative at the each point on the interval (a; b) and at the ends of the interval reaches the same values f (a) = f (b) , then there’s at least one point c (a; b) such as f ′(c) = 0 .

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The geometric interpretation of this theorem is understood from the Fig. 3.19–3.21.

О	a с	b x	О a	с	b x
	Fig. 3.17			Fig. 3.18
y		y		y

О a	с	b x О a	с	b x	a	с1 с2 b x
	Fig. 3.19		Fig. 3.20			Fig. 3.21
		(Lagrange’s theorem). If the function				f (x) is continuous on
Theorem 3.13
		the segment [a; b] and differentiable at the interval (a; b) ,

then there’s at least one point on this interval c (a; b) , where the following statement takes place

	f (b) − f (a)	= f ′(c).
	b − a	= f ′(c).
	b − a

This formula also known as Lagrange’s formula. It can be transform as: f (x + x) − f (x) = f ′(x + θΔx) x , 0 < θ < 1.

The geometric interpretation of Lagrange’s theorem. If the function f (x) is

satisfied conditions of Lagrange’s theorem, then there’s at least one point on this
y			graph, where the tangent to the graph is parallel
y	В		to the bisecant, which unites the ends of the curve
	В		to the bisecant, which unites the ends of the curve
			A(a; f (a)) and B(b; f (b)) (Fig. 3.22).
			Some useful consequence follow Lagrange’s
			theorem:
А			1) if the derivative f ′(x) = 0 at the each point
А	c2 b		of the interval, then	f (x) = const;
О a c1		x	of the interval, then	f (x) = const;
О a c1		x	2) if f ′(x) = c	at each point of the interval,
			2) if f ′(x) = c	at each point of the interval,
Fig. 3.22			then f (x) = cx + d , i.e. the function is linear;
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