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Example 2. Find points of discontinuity of the function

f (x)=

3

 

 

.

 

1

 

 

2 + 4 x+1

Solution. The given function is determined on all the numerical axis except

for a point

x = −1. We find the left — and the right-sided limits at this point:

 

 

 

f (10) =

lim

 

 

 

3

 

 

 

 

=

 

 

 

3

 

 

= 3

;

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x→−10

 

 

 

 

 

 

2

+ 4−∞

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

+ 4x+1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f (1+ 0) =

lim

 

 

 

3

 

 

 

=

 

 

 

3

 

 

 

= 0.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

+ 4+∞

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x→−1+0

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

+ 4x+1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Thus,

x = −1 is a point of essential dis-

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

continuity (of the first kind).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

lim f (x) = lim

 

3

 

 

=

3

=

1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

2 + 40

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

x→±∞

x→±∞

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

+ 4

x+1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The graph of the function

f (x)

is rep-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

–2

–1

 

 

 

0 1

2

3

resented on Fig. 3.12.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

–1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example 3. Find points of discontinuity

 

 

 

 

 

 

 

 

Fig. 3.12

 

 

 

 

 

of the function and sketch its graph:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

x

if

0 x 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f (x)

 

 

4 2x, if

1 < x < 2,5

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2x 7,

if

2,5 x ≤ ∞

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution. A function f (x)

is determined for all values

 

x 0.

But it does

not mean that it is continuous for x 0,

as far as this function is not elementary.

It is given by three different formulas for different intervals of changing the argument x and can have discontinuities at points where its analytical expression varies, i.e. at x = 1, x = 2,5. At all other points of the domain of

definition the function f (x) is continuous so as any of three formulas represents

an elementary function, continuous in the interval of changing the argument x. Let us study the behavior of the function at the points x = 1 and x = 2,5 :

lim

f (x) = lim 2 x = 2 .

lim

f (x) = lim (4 2x) = 2 .

x10

x10

x1+0

x1+0

191

The value of the function f (x) for x = 1 is f (1) = 2 1 = 2 , i.e. at the point

x = 1 function f (x)

is continuous.

 

 

2

 

lim

f (x) =

lim (4 2x) = −1 ,

 

 

 

 

x2,50

 

x2,50

 

 

lim

f (x) =

lim (2x 7) = −2 ,

 

 

x2,5+0

 

x2,5+0

0

2

4

f (2,5) = −2 .

 

 

 

2

 

At the point x = 2,5 the function has essen-

 

tial discontinuity. The graph of the function is

 

Fig. 3.13

represented on Fig. 3.13.

Micromodule 16

CLASS AND HOME ASSIGNMENT

Find the points of discontinuity of functions, investigate their character, in case of removable discontinuity predetermine the function to be continuous. Sketch graphs of non-elementary functions:

1.

f (x) =

 

x 1

 

.

 

 

 

 

 

 

 

 

 

3x 2 x 2

4.

f (x) =

 

2x + 5

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2x + 5

 

 

7.

f (x) =

 

2x + 1

 

.

 

 

x2 (x + 1)

 

 

 

 

 

 

x2

10. f (x) = 5 x2 1 .

2.

f (x) =

 

 

x + 2

.

 

 

 

 

 

 

 

 

 

 

x3 + 8

 

 

f (x)=2

 

3

 

 

 

5.

x7

.

 

8.

f (x) = sin x .

 

 

 

 

 

 

x

 

11.

f (x) =

3

 

 

 

 

 

1

 

 

 

 

 

 

 

 

3.

f (x) =

 

 

4x + 1

 

 

 

.

4x2 7x + 2

6.

f (x) =

2x2 + 7x + 6

.

 

 

 

 

 

3x2 + 5x 2

9.

f (x) = cos x .

 

 

 

x 1

. 12.

f (x) =

2

.

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 + 4 2x

 

 

1 + 2 x+1

 

 

 

 

 

1

 

 

1

 

1 .

 

 

1

 

1

 

 

 

 

 

13.

f (x) = (x + 2)arctg

. 14.

f (x) = 4

x+3

 

15.

f (x) =

 

x

x + 1

 

.

 

 

 

 

 

 

x

 

 

 

1

 

+ 1

 

 

 

1

 

1

 

 

 

 

 

 

 

 

 

 

4x+3

 

 

 

 

 

 

 

 

 

 

 

 

 

x 1

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

 

 

 

x +15 3

 

f (x) =

 

 

 

 

x 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

16.

f (x) = 3 x

4 .

 

17.

f (x) =

x2 36

. 18.

 

 

 

 

 

.

 

 

 

 

arctg(x

1)

192

19.

 

1 x,

f ( x) =

 

 

 

 

 

 

 

 

2x 1,

 

 

2x + 5,

21.

f ( x) =

1

 

 

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

2x ,

 

 

23.

f (x) = x 1,

 

 

 

 

 

x

2

+ 6x

 

 

 

 

25.

f (x) =

1 ln

1 + x

.

 

 

 

x

 

 

1 x

x≤ −1;

x> −1.

x< −1;

x≥ −1.

x1;

1< x < 2;

7, x 2.

 

 

1

 

2

 

20.

 

x

 

,

2

 

f ( x) =

 

 

 

 

 

 

 

 

 

 

x ,

 

 

 

x,

 

 

22.

f ( x) =

2

 

,

 

 

 

 

 

 

 

 

 

 

 

 

x 2

 

 

 

 

1 x2 ,

24.

f ( x) = (x 1)2 ,

 

 

 

4 x,

 

 

 

Micromodule 16

SELF-TEST ASSIGNMENTS

Investigate the function f (x)

for continuity.

 

 

 

 

16.1. а) f (x) =

2x2

x 1

;

b) f (x) =

3

 

.

3x2

x 2

 

1

 

 

 

 

 

 

 

 

 

2 + 5x2

 

 

 

 

 

 

 

 

 

16.2.а) f (x) = 5x2 3x 2 ;

x2 + x 2

16.3. а) f (x) =

5x2

x 4

 

;

x2 + 2x 3

 

 

16.4. а) f (x) =

3x2

+ 5x 8

;

x2

+ 4x 5

 

 

b)

f (x) = arctg

x 1

.

 

 

 

 

 

 

 

 

 

 

 

x + 1

 

 

2

 

 

 

 

 

 

b) f (x) = 7 x+3 .

 

 

 

 

 

 

5

 

 

 

 

 

 

b)

f (x) =

 

 

 

 

 

.

lg

 

x + 1

 

 

 

 

 

1

 

 

 

 

 

 

 

 

2x2

5x + 3

 

 

 

 

 

 

 

 

 

1

 

 

16.5. а) f (x) =

;

b)

f (x) =

5x1

.

 

2x2

+ x 3

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5x1 + 1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x2 4x + 3

 

 

 

 

 

1

 

1

 

 

 

 

 

16.6. а) f (x) =

;

b)

f (x) =

 

x

x + 1

 

.

 

 

2x2

+ 5x 7

 

 

 

 

 

 

1

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x 1

x

 

 

x 2;

x > 2.

x 0;

x > 0.

x < 1;

0 x 2; x > 2.

193

16.7.а) f (x) =

16.8.а) f (x) =

16.9.а) f (x) =

16.10.а) f (x) =

16.11.а) f (x) =

16.12.а) f (x) =

16.13.а) f (x) =

16.14.а) f (x) =

16.15.а) f (x) =

16.16.а) f (x) =

16.17.а) f (x) =

16.18.а) f (x) =

16.19.а) f (x) =

5x2 4x 1

;

 

x2 + 5x 6

 

 

 

 

2x2 + 7x 9

 

;

 

x2

+ 3x 4

 

 

 

4x2 3x 1

 

;

2x2

+ 3x 5

 

 

 

2x2 7x + 5

;

 

x2

3x + 2

 

 

 

 

3x2 7x + 4

;

 

x2

4x + 3

 

 

 

 

4x2 7x + 3

;

 

x2

5x + 4

 

 

 

 

5x2 7x + 2

;

 

x2

6x + 5

 

 

 

 

2x2 9x + 7

 

;

 

x2

7x + 6

 

 

 

 

 

3x2 5x + 2

;

 

x2

+ 3x 4

 

 

 

 

2x2 x 3

 

;

 

x2

x 2

 

 

 

 

2x2 7x 9

;

 

x2

3x 4

 

 

 

 

2x2 5x 7

;

 

x2

4x 5

 

 

 

 

3x2 4x 7

;

 

x2

2x 3

 

 

 

194

b) f (x) =

6

.

1

3+ 2 x+ 4

b)f (x) = 1x ln 11+xx .

b)f (x) = (x + 1) arctg 1x .

b)

f (x) =

 

 

 

 

 

 

 

1

 

 

 

 

 

 

.

 

1

 

 

 

 

1

 

 

 

 

 

3x+ 2

+ 3x2

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

b) f (x) = 2 x7 .

b)

f (x) =

6

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

6 + 6 x6

b)

f (x) =

 

 

 

 

 

 

 

1

 

 

 

 

 

.

 

1

 

 

 

 

1

 

 

 

4 x+1 + 4 x1

x

b)f (x) = 2 x2 9 .

b)f (x) = (2x + 3) arctg 2x .

b) f (x) =

 

x

 

.

lg 1+ x

)

 

 

 

(

 

 

 

b)

f (x) =

2

 

 

.

 

1

 

 

 

 

 

 

 

 

 

 

 

3 + 4 x5

 

 

x

b)f (x) = 3x2 4 .

b)f (x) = (x + 4) arctg 1x .

16.20.а) f (x)

16.21.а) f (x)

16.22.а) f (x)

16.23.а) f (x)

16.24.а) f (x)

16.25.а) f (x)

16.26.а) f (x)

16.27.а) f (x)

16.28.а) f (x)

16.29.а) f (x)

16.30.а) f (x)

=

4x2

x 5

;

x

2 4x 5

 

 

=5x2 x 6 ;

x2 5x 6

=6x2 x 7 ;

x2 5x 6

=

4x2

3x 7

;

x2

6x 7

 

 

=4x2 5x 9 ;

x2 7x 8

=6x2 x 7 ;

x2 x 2

=7x2 x 8 ;

x2 7x 8

=4x2 7x 11 ; x2 3x 4

=4x2 + x 3 ;

x2 + 4x + 3

=3x2 + x 2 ;

x2 + 3x + 2

=5x2 + x 4 ;

x2 + 5x + 4

 

 

 

 

 

 

x2

 

 

 

 

 

 

 

 

 

b) f (x) = 5 x1 .

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

b)

f (x) =

4 x+3

1

.

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4 x+3 + 1

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

b)

f (x) =

3x+ 2

1

.

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3x+ 2 + 1

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

b)

f (x) =

 

 

 

 

7 x

 

 

 

 

.

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4 + 2 x2

 

b)

f (x) =

 

 

 

 

x

 

 

 

 

 

.

lg (2 x)

b)

f (x) =

3

 

 

 

 

.

 

 

 

 

 

 

 

 

1

 

2+ 42x

x1

b)f (x) = arctg (x 1) .

 

 

 

1

 

 

 

 

 

 

b)

f (x) =

 

3x

 

 

.

 

 

 

 

 

x

 

 

 

4 + 2 x5

 

 

 

 

 

 

1

 

 

 

 

 

b)

f (x) =

 

2 x2

 

.

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1+ 3x+1

 

 

 

3

b)f (x) = 6 x2 1 .

b)f (x) = 1 ln 2 x .

x2 + x

195

Micromodule 17

BASIC THEORETICAL INFORMATION

DERIVATIVE

The derivative, its geometrical and physical concept. Differentiability and continuity. The rules of differentiation. The derivatives of the elementary functions. The derivative of composite function (The chain rule for differentiation).

Literature: [2, chapter 4], [3, chapter 4, items 4.1—4.2], [4, part 5], [6, chapter 5, §§ 1, 2], [7, chapter 6, § 16], [9, chapter 4, § 2, 4, 5], [10, chapter 4, § 1], [11, chapter 3, §§ 1—15], [8].

17.1. Some problems leading to understanding of the derivative definition

1. The instantaneous velocity of non-uniform motion. Suppose that a body begins to move at the moment of time t = 0 along the straight line. Let the path

traveled by the body over the time t is defined by the formula S = f (t) . A

function S= (f )t is called the law of body motion. Let us consider a path

traveled by the body over the time [t;t +

t] ; it is equal to

S = f (t + t) f (t) .

If the body is moving uniformly, the ratio of the covered path to the time

S =

f (t + t) f (t)

 

t

 

t

 

is velocity. It does not depend on t and t . In case of non-linear motion this

ratio depends both on the chosen time t and on increment

t and expresses the

average velocity of motion over the time interval [t; t + t]

. The less is the time

interval t , the more reasons we may have to consider the motion over the time t and t + t to be uniform.

The limit

lim

S

= lim

f (t +

t) f (t)

= v(t) ,

t

 

t

t0

t0

 

if it exists, is called the instantaneous velocity at the instant of time t .

2. The rate of the chemical reaction. Suppose the chemical reaction begins at an instant of time t . Let us designate the quantity of substance, which reacted

till the instant of time t is c(t) . The quantity of substance, which started to react during time interval [t; t + t] , equals c(t + t) c(t). The ratio

196

c(t + t) c(t)

 

 

 

t

 

 

 

defines the average rate of the reaction and the limit

 

lim

c(t +

t) c(t)

 

 

 

t

 

t0

 

is called the rate of the chemical reaction at an instant of time t .

17.2. The definition of the derivative

 

Let any function y = f (x) be defined at the interval

(a, b) . We take any

meaning of x within this interval and apply it to increment

x . The difference

y = f (x + x) f (x)

is called the increment of a function at point x. The increment of argument x 0 may gain both positive and negative meanings but the meaning x + x

does not go beyond the limits of the domain of the function f (x) .

Definition 3.19. The limit of the ratio of the

increment of a function

y = f (x + x) f (x) to the increment of argument

x , when the later tends

to zero, is called the derivative of a function y = f (x)

at point x , that is,

 

 

 

 

 

y(x) = lim

f (x + x) f (x)

 

 

 

 

 

x

 

 

x0

 

 

 

 

 

Definition 3.20. A function, which has the finite derivative at point x , is called differentiable at this point.

Notations for derivative: y(x), f (x) (Lagrange’s notation) or dydx , dfdx

(Leibniz’s notation). From the definition of the derivative it follows that the derivative y' (x) at point x is a number. However, if such a number exists for each inner point of the interval (a, b) , the derivative, then, can be considered as a function of a point x from the given interval.

If

lim

f (x +

x) f (x)

= ±∞ , the function f (x) has the infinite derivative

 

x 0

x

 

at point

x .

 

 

 

 

 

 

197

17.3. The geometrical, physical and mechanical concept of the derivative

Let us try to give the geometrical interpretation of the derivative. We’ll consider the graph of a function y = f (x) at the neighbourhood of point x0 (see

Fig. 3.14). Let P0 be a point of the curve with components (x0 ; f (x0 )) , and P be a point of the graph with components (x0 + x; f (x0 + x)). A straight line, drawn trough the points P0 and P , is called the secant line. If, at the indefinite approach of point P along the graph of a function y = f (x) to point P0 , the secant line P0 P approaches some boundary position ( straight line P0 K ), then this boundary position of the secant line is called the tangent line to the curve y = f (x) at the point P0 .

Let α be the angle, which the tangent line forms with the positive direction

of x-axis and β be the angle between the secant line

P0 P and x-axis. From the

right triangle P0QP it follows that

 

 

 

 

 

tgβ =

PQ

=

y

=

f (x0 + x) f (x0 )

.

 

 

x

 

 

P Q

 

 

x

 

 

 

0

 

 

 

 

 

 

 

There exists such a limit that

 

 

 

 

 

 

y(x0 ) = lim

 

f (x0

+ x) f (x)

= lim

tg β = tg α.

 

 

 

 

x

 

x0

 

 

( pp0 )

 

 

 

 

 

 

 

x0

 

 

The geometrical concept of the derivative is follows: the derivative of a function y = f (x) at pointі x0 equals the slope of the tangent line to the graph

of a given function at the respective point, that is,

 

 

 

f (x0 ) = tgα,

 

 

 

where

α is the angle, which is formed by the tangent line to the graph of the

function with positive direction of x-axis at point x0 .

 

The equation of the tangent line drawn to the graph of a function

y = f (x)

at point P0 (x0 , y0 ) looks as follows

 

 

 

 

 

 

 

 

y y0 = f (x0 )(x x0 ),

 

(3.8)

where

y0 = f (x0 ) .

 

198

 

 

 

 

 

 

The straight line that passes through the point of contact, perpendicular to the tangent line is called the normal line (it is the straight line P0 N in Fig. 3.14).

The equation of the normal line is

 

 

y y0

= −

1

(x x0 ).

 

(3.9)

 

 

f '(x0 )

 

 

 

 

 

 

 

 

 

If a function y = f (x)

describes some physical process, the derivative y,

then, is the rate of change of this process. This is the way

the physical concept

of derivative may be interpreted.

 

 

 

 

y = f (x) could express,

In other words, whatever dependence a function

 

the ratio

y can be considered as the average rate of change of a function y

 

x

x, and the derivative f (x) — as the instantaneous

with respect to the argument

rate of change of this function.

 

 

 

 

 

 

 

у

N

 

 

 

y = f(x)

 

 

 

 

P

 

f(x0 + x)

 

 

 

 

 

 

 

 

 

 

 

 

К

 

 

 

 

 

 

 

 

 

 

f(x0)

 

 

P0

 

Q

 

 

 

 

 

 

 

 

О

α

β

x0

x

 

 

х

 

 

 

x0 +

 

x

 

 

 

Fig. 3.14

 

 

 

 

The mechanical concept of the derivative. If S = S(t) is the law of motion of a material point (it means that the dependence of the traversed path S on t is given), the derivative S(t) , then, is velocity v of a point at moment t ; the second derivative S′′(t) is the instantaneous acceleration a of a point at moment t and the third derivative S ′′′(t) is a jerk of a point at moment t , that is,

v = S(t), a = S′′(t) = v(t), j = S′′′(t) = v′′(t) = a(t).

199

17.4. The main rules of differentiation

Let functions u(x), v(x), w(x) be differentiable at point хand С is a constant. Then

 

1. (u + v)′ = u′ + v.

 

2. (u v)′ = u′ − v.

 

 

 

 

 

 

3. (uv)′ = uv + uv.

 

4. (Cu)′ = Cu.

 

 

 

 

 

 

 

 

 

u

 

 

 

 

uv uv

 

6. (uvw)′ = uvw + uvw + uvw.

 

5.

 

 

=

 

 

 

 

 

 

 

 

 

 

 

 

(v ≠ 0) .

 

 

 

 

 

 

 

v

2

 

 

 

 

v

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

17.5. The table of derivatives

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(differentiation of the elementary functions table)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Table 17.1

 

1. (C)′ = 0

 

 

 

 

 

 

 

 

 

 

 

2. (xn )′ = nxn1

 

 

 

 

 

 

3. ( x)′ =

1

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

x

 

 

 

 

 

 

4.

= −

x2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

5. (ax )′ = ax ln a

 

 

 

 

 

 

6. (ex )′ = ex

 

 

 

 

 

 

 

 

 

 

7. (log

a

x)′ =

 

1

 

 

 

 

 

 

8. (ln x)′ =

1

 

 

 

 

 

 

 

 

 

 

x ln a

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

9. (sin x)′ = cos x

 

 

 

 

 

 

10. (cos x)′ = − sin x

 

11. (tg x)

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

= cos2 x

 

12. (ctg x)′ = − sin2 x

 

 

 

 

13. (arcsin x)′ =

 

1

 

 

 

14.

(arccos x)

= −

1

 

 

 

 

1x2

 

 

 

 

 

 

 

 

 

1x2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

15. (arctg x)

=

 

 

1

 

 

 

16.

(arcctg x)

= −

1

 

 

 

1+ x2

 

 

1+ x2

 

 

17. (sh x)′ = ch x

 

 

 

 

 

 

18. (ch x)′ = sh x

 

 

 

 

 

 

19. (th x)

=

 

1

 

 

 

 

 

 

 

 

 

(cth x)

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

20.

= −

 

 

 

ch2 x

 

 

 

 

 

 

 

 

 

 

 

 

 

sh2 x

 

 

 

 

 

 

 

 

 

17.6. The derivative of the composite function

 

 

 

(The chain rule for differentiation). Table of derivatives

 

If a function

 

 

y = f (u) has the derivative at point u , and a function u = g(x)

has the derivative at point x , a

composite

function

y = f (g(x)) then is

differentiable at point

x . In this case

 

 

 

 

 

 

 

 

 

 

 

 

200

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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