Higher_Mathematics_Part_1

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We have

	PN = y , QN = MN tgα =				′	′
	PN = y , QN = MN tgα =				x f (x) =	f (x)dx = dy.
So, the differential of function				f (x)	under given values of х and		x equals
the increment QN of secant MQ ordinate, which is drawn to the curve							y = f (x) at
point M , when argument receives increment x.
		19.2. General properties of the differential
Let	u(x) ,	v(x) be differentiable functions. Then the following equalities
are true:

1.	dC = 0 ( C = const ).			2. d(u +v) = du +dv .
3.	d(uv) = udv +vdu .			4. d(Cu) = Cdu .
5.	d (u ) = vdu − udv , v ≠ 0.			6. df (u) = f '(u)du, u = u(x).
	v	v	2
		v

19.3.The application of differentials in calculus of approximation and in error theory

At small values of x the following formula is true:

y ≈ dy

f (x + x) ≈ f (x) + f ' (x) x

Let y = f (x) , and the value х is defined approximately, in other words with

some absolute error x , then a value of function у will have its own absolute error y too. The relative error of computation of the value of function y can be

approximately calculated with the help of differential:

y	≈	dy	.
y		y

221

19.4. Tangent and normal

We remember that tangent equation, constructed to the graph of function y = f (x) at point P0 (x0 , y0 ) is the following:

y − y0 = f ′(x0 )(x − x0 ) ,

(3.10)

where y0 = f (x0 ) . Equation of normal is

y − y0 = −	1	(x − x0 ) .	(3.11)
	f ' (x0 )

Micromodule 19

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Find the differential of function y = x2 +2x at point x = 2 . Solution. The first method. As the differential is the main, linear with respect

to x , part of increment of function at point

x , let’s find the increment of given

function at point

x = 2 . Therefore

y = y(2 + x) − y(2) = (2 +

x)2 + 2(2 +

x) − 8 = 6 x + ( x)2 .

The linear part of increment is the expression 6

x . Finally, dy(2) = 6

x .

The second method. As dy = f ' (x)dx , then we have

y' = 2x + 2 ;

y' (2) = 6 ;

dy(2) = 6dx.

Example 2. Find the differential of function

y =

1− x2 + ln sin x.

Solution. We have

dy = d( 1−x2

+ln sin x) = ( 1− x2

+ ln sin x)'dx =

= (

− x

+ ctg x)dx .

1− x2

Example 3. Calculate approximate value ( with the help of differential) of

15 .

Solution. Let’s consider a function f (x) =

x . Let

x =16 , x +

x = 15 ,

x = −1 . Then

x + x ≈

x +

x , or

15 ≈ 4 +

(−1) = 31 =

3.875 .

2 4

222

Example 4. Calculate approximate value of

tg 54° .

Solution. Let’s consider a function

f (x) = tg x . Then we receive:

tg(x + x) ≈ tg x + (tg x)′

x ;

tg(x +

x) ≈ tg x +

x .

cos2

Let’s convert degrees to radians: 54° = 54π

= 3π .

π , x + x =

180

Let x =

3π ,

x =

3π − π

, then

tg 3π

≈ tg

=1+

≈ 1,314 .

cos2 π

Finally,

tg 54° ≈ 1,314.

Example 5. Find tangent equation and equation of normal, drawn to the curve y = x ln x +1 at point with ordinate 1.

Solution. By conditions y0 =1. Let’s find an abscissa of touching point. We have

1= x ln x +1, x ln x = 0 ; x ≠ 0 . Therefore ln x = 0 , then	x0 =1 .
Let’s find derivative
y′ = (x ln x + 1)′ = ln x + (ln x)′ x = ln x + 1.
Then y′(1) = ln1+ 1 = 1. Substituting values of x0 , y0 and	y′(x0 ) in

formula of tangent (3.10) and normal (3.11), we will obtain tangent equation

y − 1 = 1 (x − 1) , then y = x , and equation of normal	y − 1 = − 1	(x −1)	, or
y = 2 − x.	1
y = 2 − x.

Example 6. Find an angle between the tangent, constructed to the curve

y = x3 − 2x2 − 5x at point x =1 , and positive direction of	Ox .
Solution. Let’s find derivative y′ = 3x2 − 4x − 5 . Then	y′(2) = 12 − 8 − 5 = −1 .

According to the geometrical interpretation of derivative, solve the equation tg α =−1 . We obtain α = 34π (let’s point out that 0 ≤ α < π ).

223

Example 7. Write the tangent equation and equation of normal to the curve x3 +2 y3 = 5xy at point M (2; 1) .

Solution. Substituting coordinates of point M in equation of the curve you will obtain right equality 10=10. Then point М will belong to this curve and

x0 = 2, y0 =1.

Let’s find derivative of function

y at point x0 . We have:

(x3 + 2y3 )′ = (5xy)′ , 3x2 + 6y2 y ' = 5( y + xy ') , (6y2 − 5x) y ' = 5y − 3x2 ,

y' =

5y − 3x3

y' (2) =

5 1− 3 8

Then:

6y 2 −

6 1− 5 2

y − 1 = 19 (x − 2) , or 19x − 4y − 34 = 0 is equation of tangent,

y − 1 = −

(x − 2) ,

or 4x + 19 y − 27 = 0 is equation of normal.

Example 8. Compose tangent equation for an ellipse, given by equations

x = 2 sin t ,

y = 3cost,

at point M0 (1; y0 ) , where

y0 > 0.

Solution. Equation

tangent is

y − y0

= f ' (x0 )(x − x0 ) . By conditions

x =1 . Let’s find an ordinate of touching point

y . t = π

and

t =

5π

are

solutions

the

equation

1= 2 sin t

t [0; 2π) . Taking

into

account

the

condition

> 0,

that is, cost > 0, we obtain

t =

π . Then

= cos π =

Next we shall find the derivative f ′(x0 ) of the parametrically given function:

y' (t) = −3sin t ,

y '(t) = 2cost.

Therefore

y x' =

−3sin t

= −

tgt ;

f ' (1) = − 3 tg π = −

3 .

2 cos t

Tangent equation is

y −

= −

(x − 1) ,

y = −

x +

224

Micromodule 19

CLASS AND HOME ASSIGNMENT

1. Find the increment and the differential of the function y = x2 −4x +3 , if:

а)	x =1 , x = 0,1 ; b)	x = 3 ,		x = 0,05 .
Find the differential of a function:
2. y = (4−x2 )2x . 3.		y =	sin x + tg2 x .				4. y = ln arcsin x .
5.	y = x2x+1 . 6. y =	arccos x		+arctg	1	.	7. y = 4 x + x log2 tg x .

		x2			x

Find the differential of an implicit function at the point M0 (x0 ; y0 ) :

8.x3 + y3 +3xy −15 = 0 , M0 (1; 2) .

9.ln x2 + y2 = arctg xy , M0 (0; 1) .

Calculate an approximate value with the help of the differential:

10. 3 131 . 11. (0, 95)6 . 12. sin 9° . 13. cos155° .

14. arctg1, 05 . 15. arccos 0, 02 . 16. ctg 85° .

Compose an equation of a tangent line and a normal to a curve, which is given by a parametric equation:

17.	x = t −sin t , y =1−cos t , at the point t =		π .
		π .	2
18.	x = cos3 t , y = sin3 t , at the point t =	π .
		4		π .
19.	x = 2 cos t −cos 2t , y = 2 sin t −sin 2t , at the point t =			π .
20.	x = tet , y = te−t , at the point x = e .			2
20.	x = tet , y = te−t , at the point x = e .
	0

Compose an equation of a tangent line and a normal to a curve at point М:

21.4x4 +6xy − y4 = 0 , M (1; 2) .

22.x2 (2x −y) = 2x −y3 , M (1;1) .

23.x2 / 3 + y2 / 3 = 8 , M (8; 8) .

	Answers
1. a) y = −0,19	, dy = −0,2 ; b) dy = 0,1,	y = 0,1025. 2.		2x[(4 − x2 )ln 2 − 2x]dx .
5. x2x (2x ln x + 2x +	1)dx . 8. −3dx / 5 . 9. dx.	10.	5,08. 11. 0,7.	12. 0,157. 13. – 0,908.

14. 0,81. 15. 1,55. 16. 0,087. 17. y = x + 2 − π / 2 ; y = − x + π / 2 . 18. y + x = 2 / 2 ;

225

y = x . 19. y + x = 3 ;	y = x + 1 . 20. y = 1/ e ; x = e . 21. 14x − 13y + 12 = 0 ;
13x + 14y − 41 = 0 . 22.	y = x ; y = − x + 2 . 23. y + x =16; y = x .

Micromodule 19

SELF-TEST ASSIGNMENT

19.1. Calculate an approximate value with the help of the first differential:

19.1.1. cos 61° .	19.1.2. e0,2 .		19.1.3. sin 33° .
19.1.4. arctg1, 05 .	19.1.5.	120 .	19.1.6.	3 340 .
19.1.7. 3 66 .	19.1.8.	5 33 .	19.1.9.	6 70 .
19.1.10. cos 85° .	19.1.11. sin 8° .		9.1.12. sin 28° .
19.1.13. arctg 0, 95 .	19.1.14. arctg 0, 9 .		19.1.15. e0,3 .
19.1.16. ln1, 05 .	19.1.17. ln 0, 97 .		19.1.18. ln1, 08 .
19.1.19. tg 47° .	19.1.20. ctg 50° .		19.1.21. (1, 02)5 .
19.1.22. arccos 0, 45 .	19.1.23. arcsin 0, 52 .		19.1.24. (1, 97)6 .
19.1.25. (2, 04)4 .	19.1.26. ln tg 48° .		19.1.27. ln tg 43° .
19.1.28. cos 86° .	19.1.29. sin 26° .		19.1.30. tg 40° .

19.2. Solve the tasks for composition of an equation of the normal and the tangent to a curve.

19.2.1. Write the equation of a normal and a tangent to the curve y = 13 x3 −12 x2 + 56 at the points, where the slope of the tangent is equal to 2.

19.2.2. Write the equation of a normal and a tangent to the ellipse	x2	+	y2	= 1
	9		4

at the points, where the slope of the normal line is equal to 12 .

19.2.3. Write			the	equation of	a normal	and a tangent to the ellipse
x = 3cost, y = 2sin t				at the points,	where the	tangent	is parallel	to the line
y =−	2	x +4 .
y =−		x +4 .
3
19.2.4. Write			the	equation of	a normal	and a	tangent to	the curve
y = 3x3 + 5x2 + 5x + 4				at the points, where the tangents are parallel to the line

y = 4x − 2 .

226

19.2.5. Write the

equation

of a

normal and a tangent to the

curve

y = 2 x3

+ 5x2 + 9x +

the

points,

where the tangents are parallel

to the

line y = x −2 .

19.2.6. Write

the

equation of a

normal and a tangent to the

curve

y = 1 x3

7x2 + 15x + 1

the

points,

where the slope of the tangent

line is

equal to 2.

19.2.7.

Write

the

equation of a normal and a tangent to the curve

y = x3 + 7x2 + 11x +

where the tangent is parallel to the line y = 3x −4 .

19.2.8.

Write

equation of a

normal and a tangent to the

curve

the

y = 5 x3

11x2 + 13x +

at the points, where the slope of the normal line is

1 .

equal to

−

19.2.9.

the

equation of a

normal and a tangent to the

curve

Write

y = 4 x3

5x2 + 3x +

at the points, where tangents and the x-axis form an

angle of1350 .

19.2.10.

Write

the

equation of a

normal and a tangent to the

curve

= x3 + 6x2 +

1 x + 5 at the points, where the slope of the normal line is equal

to 1 .

Write

the

equation of a

normal and a tangent to the

curve

19.2.11.

y = 1 x3

+ 7x2 + 10x + 130

at the points, where the tangents are parallel to the

line y = −3x +2 .

19.2.12.

Write

the

equation of a

normal and a tangent to the

curve

y = x3 + 6x2 + 2 x +

at the points, where the tangents are parallel to the line

y = −3x + 1 .

19.2.13.

Write

the

equation of a

normal and a tangent to the

curve

y = 3x3 + 6x2 − x + 1

at the points, where the slope of the tangent lines is equal

to −4 .

Write

the

equation of a

normal and a tangent to the

curve

19.2.14.

y = 3x2 + 6x2 − 6x + 25

at the points, where the tangents and the x- axis form

the angle 1350 .

227

19.2.15.

Write

the

equation of a normal and

tangent

the

line

y =

x2 −3x +6

at the point with the abscissa x

= 3 .

19.2.16.

Write

the

equation of

normal

and

tangent

to the curve

y = (x +1)3

3−x

at the point with the abscissa

= 2 .

19.2.17. Write the equation of

normal

and a

tangent

the

astroid x = 2

2 cos3 t,

y = 2 2 sin3 t , if t

= π .

19.2.18 Write

the

equation

normal

and

tangent

to the

cycloid

x = 2(t −sin t),

y = 2(1−cos t) , if

t =

π .

19.2.19. Write

the

equation

normal

and

tangent

the

parabola

y2 −y +2x −4 = 0 at the point with the abscissa x

= −4 .

19.2.20.

Write

the

equation of

normal

and

tangent

to the curve

x2 +2xy +2y4 = 5 at the point

M0 (1; 1) .

19.2.21. Write

the

equation

normal

and

tangent to the

cycloid

x = 3(t −sin t),

y = 3(1−cos t) , if

3π

19.2.22.

Write

the

equation of

normal

and

tangent

to the curve

y = 3x3 −6x2 −6x +

at the points, where the tangents and the x- axis form

the angle 1350 .

19.2.23.

Write

the

equation of

normal

and

tangent

to the curve

y = 3x3 −6x2 −x +1 at the points, where the tangents are parallel to the line

y = −4x +5 .

19.2.24.

Write

the

equation of

normal

and

tangent

the

curve

x4 +3xy2 +3y4 =1 at the point M0 (−1;1) .

19.2.25.

Write

the

equation of

normal

and

tangent

the

curve

y =

−5x

+9x +

at the points, where the tangents are parallel to the

bisectrix of the first coordinate quarter (angle).

19.2.26.

Write

the

equation of

normal

and

tangent

the

curve

y =

−3x

+6x −

at the points, where the tangents are parallel

to the

line y = −2x +3 .

19.2.27.

Write

the

equation

a normal

and a tangent to the

curve x = t2 ,

y = t3 , if t = 2 .

228

19.2.28. Write the equation of a normal and a tangent to the curve

y =	4	x	3	−5x	2	+3x +	1	at the points, where the tangents are parallel to the
	3						3

bisectrix of the second coordinate quarter (angle).

19.2.29. Write the equation of a normal and a tangent to the curve

x3 −3xy2 + y3 = −3 at the point			M0 (1; 2) .
19.2.30. Write	the	equation	of a normal and a tangent to the curve
y = x3 −5x2 + x −	1	at the points, where the normal is parallel to the line
	27

y = 12 x +1 .

Micromodule 20

BASIC THEORETICAL INFORMATION.

HIGHER ORDER DERIVATIVES AND DIFFERENTIALS

Higher order derivatives of functions given explicitly, implicitly, or parametrically. Leibniz's formula. Higher order differentials.

Literature: [3, chapter 4, items 4.1—4.2], [4, part 5], [6, chapter 5, § 4], [7, chapter 6, § 18], [9], [10, chapter 4, § 12, 20], [12, chapter 3, §§ 22—24].

20.1. Higher order derivatives

Suppose differentiable function y = f (x) is defined on an interval (a; b), then its derivative f ′(x) , called also the first-order derivative or the first derivative, is a function of x too. The function f ′(x) may have derivative at the interval (a; b) too.

Definition 3.22. The later derivative is called the second derivative (or the second-order derivative).

It is denoted with one of the following symbols:

y'', f ''		d 2 y			d 2	f		d	dy
	(x),			,			,			.
		dx	2		dx2
								dx dx

The second derivative has the following mechanical matter. If a mass point occurs in accordance with the law S = f (t) , then derivative S′ is a velocity at

the given time point and S′′ is an acceleration in the same time point t.

Definition 3.23. Derivative of the second-order derivative, if it is existed, is called the third-order derivative.

229

The first derivative of the (n – 1)-th derivative, if it is existed, is called the n- th derivative:

		( y ( n − 1) ) ′ , or y	( n )	=	d	d n − 1 y		′
y ( n )	=
					d x	d x	n − 1

Derivatives having the order higher than one are called the higher order derivatives.

20.2 Leibniz’s formula

Suppose y = uv , where u(x) and v(x) are the n times differentiated functions. Then

k		n!
where Cn	=	(n−k)!k!,

(uv)(n) = ∑ Cnk u(n−k )v(k ) , k =0

n ! = 1 2 3 k ( n − 1) n , 0 ! = 1 .

Particularly,

(u v ) '	= u ' v + u v ' (n = 1) ,
(u v ) ′′ = u ′′v + 2 u ′v ′ + u v ′′		(n = 2),
(u v ) ′′ = u ′′′v +	3u ′′v ′ + 3u ′v ′′ +	u v ′′′ (n = 3).

20.3. Calculation of the higher order derivatives of parametrically given functions

If some function is given parametrically by equations x = x(t) , y = y(t) , then derivatives are calculated by the following formulas:

						dy ′
dy		y′		d 2 y		dx
	=	t	,		=		,
dx	=	xt′	,	dx2	=	xt′	,
dx		xt′		dx2		xt′

230

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