Table of derivatives of elementary functions

Definition 1

The calculation of the derivative is called differentiation.

Denote the derivative $y"$ or $\frac(dy)(dx)$.

Remark 1

To find the derivative of a function, according to the basic rules, differentiation is converted into another function.

Consider the table of derivatives. Let us pay attention to the fact that functions after finding their derivatives are transformed into other functions.

The only exception is $y=e^x$, which turns into itself.

Derivative Differentiation Rules

Most often, when finding a derivative, it is required not only to look at the table of derivatives, but first to apply the rules of differentiation and the proof of the derivative of the product, and only then use the table of derivatives of elementary functions.

1. The constant is taken out of the sign of the derivative

$C$ is a constant (constant).

Example 1

Differentiate the function $y=7x^4$.

Decision.

Find $y"=(7x^4)"$. We take out the number $7$ for the sign of the derivative, we get:

$y"=(7x^4)"=7(x^4)"=$

using the table, you need to find the value of the derivative of the power function:

$=7 \cdot 4x^3=$

We transform the result to the form accepted in mathematics:

Answer:$28x^3$.

2. The derivative of the sum (difference) is equal to the sum (difference) of the derivatives:

$(u \pm v)"=u" \pm v"$.

Example 2

Differentiate the function $y=7+x-5x^3+4 \sin x-9\sqrt(x^2)+\frac(4)(x^4) -11\cot x$.

Decision.

$y"=(7+x-5x^5+4 \sin x-9\sqrt(x^2)+\frac(4)(x^4) -11\cot x)"=$

apply the rule of differentiation of the derivative sum and difference:

$=(7)"+(x)"-(5x^5)"+(4 \sin x)"-(9\sqrt(x^2))"+(\frac(4)(x^4) )"-(11\cot x)"=$

note that when differentiating, all powers and roots must be transformed to the form $x^(\frac(a)(b))$;

we take all the constants out of the sign of the derivative:

$=(7)"+(x)"-(5x^5)"+(4\sin x)"-(9x^(\frac(2)(5)))"+(4x^(-4) )"-(11\cot x)"=$

$=(7)"+(x)"-5(x^5)"+4(\sin x)"-9(x^(\frac(2)(5)))"+4(x^( -4))"-11(\cot x)"=$

having dealt with the rules of differentiation, some of them (for example, like the last two) are applied simultaneously in order to avoid rewriting a long expression;

we have obtained an expression from elementary functions under the sign of the derivative; Let's use the table of derivatives:

$=0+1-5 \cdot 5x^4+4\cos x-9 \cdot \frac(2)(5) x^(-\frac(3)(5))+12x^(-5)- 11 \cdot \frac(-1)(\sin^2 x)=$

transform to the form accepted in mathematics:

$=1-25x^4+4 \cos x-\frac(18)(5\sqrt(x^3))+\frac(12)(x^5) +\frac(11)(\sin^2 x)$

Note that when finding the result, it is customary to convert terms with fractional powers into roots, and with negative ones into fractions.

Answer: $1-25x^4+4 \cos x-\frac(18)(5\sqrt(x^3))+\frac(12)(x^5) +\frac(11)(\sin^2 x )$.

3. The formula for the derivative of the product of functions:

$(uv)"=u" v+uv"$.

Example 3

Differentiate the function $y=x^(11) \ln x$.

Decision.

First we apply the rule for calculating the derivative of the product of functions, and then we use the table of derivatives:

$y"=(x^(11) \ln x)"=(x^(11))" \ln x+x^(11) (\lnthx)"=11x^(10) \ln x+x^ (11) \cdot \frac(1)(x)=11x^(10) \ln x-\frac(x^(11))(x)=11x^(10) \ln x-x^(10)=x ^(10) (11 \ln x-1)$.

Answer: $x^(10) (11 \ln x-1)$.

4. The formula for the derivative of a private function:

$(\frac(u)(v))"=\frac(u" v-uv")(v^2)$.

Example 4

Differentiate the function $y=\frac(3x-8)(x^5-7)$.

Decision.

$y"=(\frac(3x-8)(x^5-7))"=$

according to the rules of precedence of mathematical operations, we first perform division, and then addition and subtraction, so we first apply the rule for calculating the derivative of the quotient:

$=\frac((3x-8)" (x^5-7)-(3x-8) (x^5-7)")((x^5-7)^2) =$

apply the rules of derivatives of the sum and difference, open the brackets and simplify the expression:

$=\frac(3(x^5-7)-5x^4 (3x-8))((x^5-7)^2) =\frac(3x^5-21-15x^5+40x^ 4)((x^5-7)^2) =\frac(-12x^5+40x^4-21)((x^5-7)^2)$ .

Answer:$\frac(-12x^5+40x^4-21)((x^5-7)^2)$.

Example 5

Let us differentiate the function $y=\frac(x^7-2x+3)(x)$.

Decision.

The function y is a quotient of two functions, so we can apply the rule for calculating the derivative of a quotient, but in this case we get a cumbersome function. To simplify this function, you can divide the numerator by the denominator term by term:

$y=\frac(x^7-13x+9)(x)=x^6-13+\frac(9)(x)$.

Let us apply to the simplified function the rule of differentiation of the sum and difference of functions:

$y"=(x^6-13+\frac(9)(x))"=(x^6)"+(-13)"+9(x^(-1))"=6x^5+ 0+9 \cdot (-x^(-2))=$

$=6x^5-\frac(9)(x^2)$.

Answer: $6x^5-\frac(9)(x^2)$.

1. (f(h(x))) "= f" (h(x)) x ∙ h"(x)

2. (sin x) " = cos x

3. (cos x) " = - sin x

4. (tg x) " = 1/cos 2 x

5.(ctg x)" = 1/sin 2 x

6. (a x) " = a x ∙ ln a

7. (e x) " = e x

8. (lnx)" = 1/x

9. (log a x) " = 1/ x ∙ ln a a

10.(arcsinx)" = 1/

11. (arccos x) "= -1/

12. (arctg x) "= 1/ 1+x 2

13. (arcctg x) " = -1/1+x 2

Example. Calculate Derivative

y=sin3(1-x2)

y"= (sin 3 (1-x 2))"* (sin (1-x 2))"* (1-x 2)" = 3 sin 2 (1-x 2) * cos (1-x 2) ) * (-2x) =

6x * sin 2 (1-x 2) * cos (1-x 2)

Definition. Let the function y = f(x), x Є(a;b) be differentiable at some point x o Є(a;b), i.e. at the point x o there is a limit lim Δf(x o) / Δx = f"’ (x o)

From here we have Δ f(x o) / Δx = f’(x o) + α , where α is an infinitesimal value at Δ x→0, i.e. limα = 0

So Δ f(x o) = f"" (x o) ∙ Δx + α∙ Δx.

The second term is infinitely small as Δx→0, therefore d f(x o)= f "(x o)∙ Δx or

Example. Calculate differential of function y = x 2 + cos 3x - 5

Dy \u003d (x 2 + cos 3x - 5) "dx \u003d (2x - 3 sin 3x) dx.

Definition. A differential function f(x) defined on some interval x is called antiderivative for a function f(x) defined on the same interval if for all x from this interval F"(x) = f(x) or d F(x ) = f(x) * dx

Definition. The set of all antiderivatives for the function f(x) defined on some interval x is called the indefinite integral of the function f(x) on this interval and is denoted by the symbol

∫ f(x) dx = f(x) + C, where F(x) is the antiderivative

C is the derivative constant.

To calculate the indefinite integral, there is a table of basic integrals (see the textbook Mathematics for technical schools by I.I. Valuta), p.251).

Example. To find

1. ∫(4x 3 - 6x 2 + 2x + 3)dx = ∫4x 3 dx - ∫6x 2 dx + ∫2xdx + ∫3dx = 4 x 4 /4 - 6 x 3 /3 + 2 x 2 /2 +

2. ∫(5x 4 – 8/cos 2 x + 3√x + 1) dx = ∫ 5x 4 dx – ∫8/cos 2 x * dx + ∫3√x dx + ∫dx =

5 * x 5 / 5 - 8 * tg x + 3 x 3/2 / 3/2 + x + C = x 5 - 8 tg x + 2x√x + x + C.

3. ∫2 3x * 3 x dx = ∫(2 3 * 3) x dx = ∫ 24 x dx = 24 x / ln 24 + C.

Definition. The increments F(b) - F (a) of any of the antiderivative functions f(x) + C when the argument changes from x = a to x = b is called the definite integral from a to b of the function f(x), and is denoted by f(x) dx = F(x) = F(b) – F(a), and is called the Newton-Leibniz formula.

Example. Calculate

1. ∫ (x 2 - 3x + 7)dx = ( x 3 - 3/2 x 2 + 7x) | = (1/3 * 2 3 - 3/2 * 2 2 + 7*2) - (1/3 *(-1) 3 -

3/2 (-1) 2 + 7*(-1)) = 19,5

Definition. The figure bounded by the graph of the function y \u003d f (x), a segment and straight lines x \u003d a and x \u003d b is called a curvilinear trapezoid.

S= ∫ f(x) dx = F(b) – F(a)

Example. Calculate the area of ​​a bounded figure y = ½ x 2 + 1 y = 0 x = -2 x = 3

S= ∫ (1/2 x 2 + 1) dx = (1/6 x 3 + x) | = (1/6 * 3 3 +3) -

- (1/6 (-2) 3 – 2) = 10 5/6

Topic 1.2. Ordinary differential equations

The solution of various problems by the method of mathematical modeling is reduced to finding an unknown function from an equation containing an independent variable, the desired function and the derivatives of this function. Such an equation is called a differential equation.

Definition. A solution to a differential equation is any function that turns the given equation into an identity.

Symbolically, the differential equation is written as follows:

F(x, y, y" , y"", .....y (h)) = 0

2x + y – 3y"= 0 y" 2 – 4 = 0, sin y"= cos xy, y"" = 2x are differential equations.

Definition 2. The order of a differential equation is the greatest order of the derivatives included in the given equation.

xy" + y - 2 = 0 - first order equation

y"" + 7y"- 3y = 0 - third order equation

Definition 3. A first-order differential equation is an equation of the form F(x, y, y") = 0

y"= f(x, y) is a first-order equation solved with respect to the derivative.

Definition 4. Any individual solution of a differential equation is called its particular solution.

Definition 5. The function given by the formula y = (e (x,C) or y = y(x,C) - represents the general solution of the differential solution F(x, y, y") = 0 or

Cauchy problem. When solving specific problems, it is often necessary to single out from the entire set of solutions of a differential equation that particular solution that is the answer to the question posed. In order to single out a separate integral curve from the entire set of solutions, the so-called initial conditions are set.

In the case of first-order differential equations y" = f(x, y), the initial condition for its solution y = y(x) is understood as the conditions that y = y o at x = x o i.e. y (x o) \u003d y o, where x o and y o are given numbers (initial data), such that when x \u003d x o and y \u003d y o, the function f (x, y) makes sense, i.e. there exists f (x o, y about).

Definition 6. The problem of finding a particular solution of a differential equation that satisfies given initial conditions is called the Cauchy problem.

In the case of a first-order differential equation, the Cauchy problem is formulated as follows: find a solution y = y(x) of the equation y" = f(x, y) that satisfies, for given initial data (x o, y o), the initial condition

y (x o) \u003d y o, or, in another notation, y x \u003d x0 \u003d y o, where x o, y o are given numbers.

Definition 7. A differential equation is called an equation with separable variables if it has the following form: y "= f 1 (x) f 2 (y) or

dy/f 2 (y) = f 1 (x) dx.

Theorem: If the integrals ∫dy/f 2 (y) and ∫ f 1 (x) dx exist, then the general integral of the separated variable equation is given by the equation

F 2 (y) = F 1 (x) + C, where F 2 (y) and F 1 (x) are some antiderivatives of the functions 1/f 2 (y) and f 1 (x), respectively.

When solving differential equations with separating variables, one can be guided by the following algorithm:

1) separate the variables (taking into account the conditions when this can be done);

2) integrating term by term obtained equations with separated variables, find its general integral;

3) find out if the equation has solutions that are not obtained from the general integral;

4) find a partial integral (or solution) that satisfies the initial conditions (if required).

Example. Find a particular solution to the equation 2yy" = 1-3x 2 if y o = 3 for x o =1

This is a separated variable equation. Let's represent it in differentials:

Hence 2y * dy = (1-3 x 2) dx

We integrate both parts of the last equality, we find ∫ 2y * dy = ∫ (1-3x 2) dx we get y 2 = x - x 3 + C. Substituting the initial values ​​y o = 3 x o =1 we find

C: 9 \u003d 1-1 + C i.e. C = 9.

Therefore, the desired partial integral will be y 2 \u003d x - x 3 + 9 or

x 3 + y 2 - x - 9 \u003d 0

Topic 1.4. Rows.

Definition 1. A number series is an expression of the form

а 1 + а 2 + …а n + ………., where а 1 , а 2 , ……а n – numbers belonging to some specific numerical system.

For the abbreviated notation of the series, the summation sign Σ is used, and

namely a 1 + a 2 + …a n + ……….= Σ a n

Definition 2. The numbers a 1, a 2, ... and n, ..... are called members of the series; and n is called the common member of the series.

Definition 3. A series is called convergent if the sequence of its partial sums S 1 , S 2 , S 3 .........S n , ...... converges, i.e. if there is a finite limit

The number S is called the sum of the series. If Lim S n does not exist or Lim S n = ∞, then the series

h →∞ h →∞

is called divergent and no numerical value is assigned to it.

Theorem 1. If the series converges, then its common term a n tends to zero.

If Lim a n ≠ 0 or this limit does not exist, then the series diverges.

Theorem 2. Let a series а 1 + а 2 + …а n + ………., with positive terms be given.

a n + 1 a n + 1

Assume that Lim exists and Lim = P

h →∞ a n h →∞ a n

1) if P<1, то ряд сходится

2) if Р>1, then the series diverges.

Definition 3. Series containing both positive and negative terms are called regular.

Definition 4. A regular series is called absolutely convergent if the series converges

|a 1 | + |а 2 | + …+ | and n | + ………., composed of the modules of its members.

Definition 5. Series а 1 + а 2 + …а n + ………., is called conditionally convergent if it converges, and the series |а 1 | + |а 2 | + …+ | and n | + ………., composed of the modules of its members, diverges.

Definition 6. A series is called alternating if the positive and negative terms follow each other in turn (a 1 + a 2 + a 3 - a 4 +…..+(-1) n +1 *

Theorem 3. An alternating series converges if:

1) its members decrease modulo,

a 1 ≥ a 2 ≥ … ≥ a n ≥ ……..

2) its common term tends to zero,

Moreover, the sum S of the series satisfies the inequality 0≤ S ≤a 1

Definition 7. Let u 1 (x), u 2 (x),.....u n (x) ... be some sequence of functions.

An expression of the form Σ u n (x) = u 1 (x), u 2 (x),.....u n (x) + is called a functional series.

Definition 8. A functional series is called convergent at a point x o if

number series Σ u n (x o) = u 1 (x o), u 2 (x o),.....u n (x o) + ......

obtained from the functional series by substitution x = x o , is a convergent series. This is called the point of convergence of the series.

Definition 9. A power series is a functional series of the form

Σ a n (x-x o) n = a o + a 1 (x-x o), a 2 (x-x o) 2 ,.....a n (x-x o) n + ......

where х is an independent variable, х o is a fixed number, а o , а 1 , а 2 , … а n ….. are constant coefficients.

Section 2.1. Fundamentals of discrete mathematics.

Topic 2.1. Sets and relations. Relationship properties. Operations on sets.

A set is the basic concept of set theory, which is introduced without definition. At the very least, what is known about a set is that it consists of elements.

The set A is called

is element B (Fig. 1)

picture 1

Ways to specify sets:

1. By enumeration, i.e. list of its elements.

2. A generative procedure that describes a method for obtaining elements of a set from already obtained elements or other objects. In this case, the elements of the set are all objects that can be constructed using such a procedure.

3. A description of the characteristic properties that its elements should have.

Specify in various ways the set N of all natural numbers 1, 2, 3…..

a) the set N cannot be specified as a list because of its infinity.

b) the generating procedure contains two rules:

1) 1 О N ; 2) if n н N, then n + 1 н N

c) description of the characteristic property of the elements of the set N:

N = (x; x is a positive integer)

Operations on sets.

1. The union of sets A and B is called

the set of all those elements

that belong to at least one of the sets

A, B. (Figure 2)

Figure 2

2. The intersection of sets A and B is called

a set consisting of all those and only those elements

which belong to both A and B. (Figure 3)

Figure 3

3. The difference of sets A and B is the set

all those and only those elements of A that

Figure 4

4. The complement (up to B) of a set A is called B

BUT

the set of all elements not belonging to A (Fig. 5)

Figure 5

Perform operations on the sets A = (a, b, c, d) and B = (c, d, f.g, h)

A U B =(a, b, c, d, e, f.g,h)

A ∩ B = (c, d)

Complement operations on sets A and B cannot be performed i.e. the universal set is not defined.

Relationships are one of the ways to specify relationships between elements of a set. The most studied and most often used are the so-called down- and bi-pair relations.

Relationships can be set:

list;

Matrix.

Relationship properties.

Let R be a relation on the set M, R ≤ M x M, then:

1. R is reflexive if a R a holds for any a Î M.

2. R is antireflexive if neither for each a Î M does a R a hold.

3. R ​​is symmetrical if a R b implies bRа.

4. R is antisemmetric if aRb and bRa imply a=b, i.e. for any distinct elements a and b (a≠b) is not both aRb and bRa .

5. R is transitive if aRb and bRa imply aRc.

Topic 2.2 Basic concepts of graph theory

Graphical representations in a broad sense are any visual representations of the system, process, phenomenon under study on a plane. These may include drawings, drawings, dependency graphs of characteristics, plan maps of areas, flowcharts of processes, diagrams, etc.

Graphical representations are a convenient way to illustrate the content of various concepts related to other ways of formalized representations.

A powerful and most studied class of objects related to graphical representations are the so-called graphs.

Graph theory has huge applications, since its language, on the one hand, is clear and understandable, and on the other hand, it is convenient in formal research.

Graphical representations in the narrow sense are a description of the system, process, phenomenon under study by means of graph theory in the form of a set of two classes of objects: vertices and lines connecting them - edges or arcs.

Definition: a graph D is a collection of two sets: vertices V and edges E, between the elements of which the incidence relation is defined - each edge e E is incidentally equal to two vertices v", v"" V, which it connects.

Also about the theory of graphs, about the elements of graphs, get acquainted with the types of graphs and consider operations on them, you can study section 3 "Graph Theory", pp. 195-214 in the textbook for the XXI century, edited by G.I. Moskinov "Discrete Mathematics ".

For independent study of the topic 3.1. Fundamentals of probability theory and mathematical statistics. Probability. Theorems of addition and multiplication of probabilities. Topics 3.2. Random variable, its distribution function. Topics 3.3. Mathematical expectation and variance of a random variable. You can use the following literature: V.S. Shchipachev "Fundamentals of Higher Mathematics", as well as I.P. Natanson. A short course in higher mathematics or NV Bogomolov Practical lesson in mathematics.

Let the function y = f(x) be defined in the interval X. derivative function y \u003d f (x) at the point x o is called the limit

= .

If this limit finite, then the function f(x) is called differentiable at the point x o; moreover, it turns out to be necessarily and continuous at this point.

If the considered limit is equal to  (or - ), then provided that the function at the point X o is continuous, we will say that the function f(x) has at a point X o infinite derivative.

The derivative is denoted by the symbols

y , f (x o), , .

Finding the derivative is called differentiation functions. The geometric meaning of the derivative is that the derivative is the slope of the tangent to the curve y=f(x) at a given point X o ; physical sense - in that the time derivative of the path is the instantaneous speed of the moving point in rectilinear motion s = s(t) at the moment t o .

If a with is a constant number, and u = u(x), v = v(x) are some differentiable functions, then the following differentiation rules hold:

1) (c) " = 0, (cu) " = cu";

2) (u+v)" = u"+v";

3) (uv)" \u003d u "v + v" u;

4) (u / v) "= (u" v-v "u) / v 2;

5) if y = f(u), u = (x), i.e. y = f((x)) - complex function, or superposition, composed of differentiable functions  and f, then , or

6) if for the function y = f(x) there exists an inverse differentiable function x = g(y), and  0, then .

Based on the definition of the derivative and the rules of differentiation, one can compile a list of tabular derivatives of the basic elementary functions.

1. (u )" =  u  1 u" (  R).

2. (a u)" = a u lna u".

3. (e u)" = e u u".

4. (log a u)" = u"/(u ln a).

5. (ln u)" = u"/u.

6. (sin u)" = cos u u".

7. (cos u)" = - sin u u".

8. (tg u)" = 1/ cos 2 u u".

9. (ctg u)" = - u" / sin 2 u.

10. (arcsin u)" = u" / .

11. (arccos u)" = - u" / .

12. (arctg u)" = u"/(1 + u 2).

13. (arcctg u)" = - u"/(1 + u 2).

Let us calculate the derivative of the exponential expression y=u v , (u>0), where u and v essence of the function X having derivatives at a given point u",v".

Taking the logarithm of the equality y=u v , we obtain ln y = v ln u.

Equating derivatives with respect to X from both parts of the obtained equality using rules 3, 5 and the formula for the derivative of the logarithmic function, we will have:

y"/y = vu"/u + v" ln u, whence y" = y (vu"/u + v" ln u).

(u v)"=u v (vu"/u+v" log u), u > 0.

For example, if y \u003d x sin x, then y" \u003d x sin x (sin x / x + cos x ln x).

If the function y = f(x) is differentiable at a point x, i.e. has a finite derivative at this point y", then = y "+, where 0 at х 0; hence  y = y" х +  x.

The main part of the function increment, linear with respect to x, is called differential functions and is denoted by dy: dy \u003d y "x. If we put y \u003d x in this formula, then we get dx \u003d x" x \u003d 1x \u003d x, therefore dy \u003d y "dx, i.e. a symbol for the notation for the derivative can be thought of as a fraction.

Function increment  y is the increment of the ordinate of the curve, and the differential d y is the increment of the ordinate of the tangent.

Let us find for the function y=f(x) its derivative y = f (x). The derivative of this derivative is called second order derivative functions f(x), or second derivative, and denoted .

The following are defined and denoted in the same way:

third order derivative - ,

fourth order derivative -

and generally speaking nth order derivative - .

Example 3.15. Calculate the derivative of the function y=(3x 3 -2x+1)sin x.

Decision. By rule 3, y"=(3x 3 -2x+1)"sin x + (3x 3 -2x+1)(sin x)" = = (9x 2 -2)sin x + (3x 3 -2x +1) cos x.

Example 3.16 . Find y", y = tg x + .

Decision. Using the rules for differentiating the sum and the quotient, we get: y"=(tgx + )" = (tgx)" + ()" = + = .

Example 3.17. Find the derivative of a complex function y= , u=x 4 +1.

Decision. According to the rule of differentiation of a complex function, we get: y "x \u003d y " u u" x \u003d () " u (x 4 +1)" x \u003d (2u +. Since u \u003d x 4 +1, then (2 x 4 + 2+ .

Derivative, rules and formulas of differentiation

Let the function y = f(x) be defined in the interval X. derivative function y \u003d f (x) at the point x o is called the limit

= .

If this limit finite, then the function f(x) is called differentiable at the point x o; moreover, it turns out to be necessarily and continuous at this point.

If the considered limit is equal to ¥ (or - ¥), then provided that the function at the point x o is continuous, we will say that the function f(x) has at a point x o infinite derivative.

The derivative is denoted by the symbols

y ¢, f ¢(x o), , .

Finding the derivative is called differentiation functions. The geometric meaning of the derivative is that the derivative is the slope of the tangent to the curve y=f(x) at a given point x o; physical sense - in that the derivative of the path with respect to time is the instantaneous speed of the moving point during rectilinear motion s = s(t) at the moment t o .

If a with is a constant number, and u = u(x), v = v(x) are some differentiable functions, then the following differentiation rules hold:

1) (c) " = 0, (cu) " = cu";

2) (u+v)" = u"+v";

3) (uv)" \u003d u "v + v" u;

4) (u / v) "= (u" v-v "u) / v 2;

5) if y = f(u), u = j(x), i.e. y = f(j(x)) - complex function, or superposition, composed of differentiable functions j and f, then , or

6) if for a function y = f(x) there exists an inverse differentiable function x = g(y), and ¹ 0, then .

Based on the definition of the derivative and the rules of differentiation, one can compile a list of tabular derivatives of the basic elementary functions.

1. (u m)" = m u m- 1 u" (m О R).

2. (a u)" = a u lna × u".

3. (e u)" = e u u".

4. (log a u)" = u"/(u ln a).

5. (ln u)" = u"/u.

6. (sin u)" = cos u × u".

7. (cos u)" = - sin u × u".

8. (tg u)" = 1/ cos 2 u × u".

9. (ctg u)" = - u" / sin 2 u.

10. (arcsin u)" = u" / .

11. (arccos u)" = - u" / .

12. (arctg u)" = u"/(1 + u 2).

13. (arcctg u)" = - u"/(1 + u 2).

Let us calculate the derivative of the exponential expression y=u v , (u>0), where u and v essence of the function X having derivatives at a given point u",v".

Taking the logarithm of the equality y=u v , we obtain ln y = v ln u.

Equating derivatives with respect to X from both parts of the obtained equality using rules 3, 5 and the formula for the derivative of the logarithmic function, we will have:

y"/y = vu"/u + v" ln u, whence y" = y (vu"/u + v" ln u).

(u v)"=u v (vu"/u+v" log u), u > 0.

For example, if y \u003d x sin x, then y" \u003d x sin x (sin x / x + cos x × ln x).

If the function y = f(x) is differentiable at a point x, i.e. has a finite derivative at this point y", then \u003d y "+a, where a®0 at Dx® 0; hence D y \u003d y" Dx + a x.

The main part of the function increment, linear with respect to Dx, is called function differential and is denoted dy: dy \u003d y "Dx. If we put y \u003d x in this formula, then we get dx \u003d x" Dx \u003d 1 × Dx \u003d Dx, therefore dy \u003d y "dx, i.e. the symbol for denoting the derivative can be considered like a fraction.

D function increment y is the increment of the ordinate of the curve, and the differential d y is the increment of the ordinate of the tangent.

Let us find for the function y=f(x) its derivative y ¢= f ¢(x). The derivative of this derivative is called second order derivative functions f(x), or second derivative, and denoted .

The following are defined and denoted in the same way:

third order derivative - ,

fourth order derivative -

and generally speaking nth order derivative - .

Example 3.15. Calculate the derivative of the function y=(3x 3 -2x+1)×sin x.

Decision. By rule 3, y"=(3x 3 -2x+1)"×sin x + (3x 3 -2x+1)×(sin x)" =
= (9x 2 -2) sinx + (3x 3 -2x+1) cos x.

Example 3.16. Find y", y = tg x + .

Decision. Using the rules for differentiating the sum and the quotient, we get: y"=(tgx + )" = (tgx)" + ()" = + = .

Example 3.17. Find the derivative of a complex function y= ,
u=x 4 +1.

Decision. According to the rule of differentiation of a complex function, we get: y "x \u003d y " u u" x \u003d () " u (x 4 +1)" x \u003d (2u +. Since u \u003d x 4 +1, then
(2 x 4 +2+ .

Example 3.18.

Decision. Let us represent the function y= as a superposition of two functions: y = e u and u = x 2 . We have: y" x \u003d y " u u" x \u003d (e u)" u (x 2)" x \u003d e u ×2x. Substituting x2 instead of u, we get y=2x .

Example 3.19. Find the derivative of the function y=ln sin x.

Decision. Denote u=sin x, then the derivative of the complex function y=ln u is calculated by the formula y" = (ln u)" u (sin x)" x = .

Example 3.20. Find the derivative of the function y= .

Decision. The case of a complex function obtained as a result of several superpositions is exhausted by the successive application of Rule 5:

.

Example 3.21. Calculate derivative y=ln .

Decision. Taking logarithms and using the properties of logarithms, we get:

y=5/3ln(x 2 +4) +7/3ln(3x-1)-2/3ln(6x 3 +1)-1/3tg 5x.

Differentiating both parts of the last equality, we get:

Function extremum

The function y=f(x) is called increasing (waning) in some interval if for x 1< x 2 выполняется неравенство f(x 1) < f (x 2) (f(x 1) >f(x2)).

If a differentiable function y = f(x) on a segment increases (decreases), then its derivative on this segment f ¢(x) > 0 (f ¢(x)< 0).

Dot x o called local maximum point (minimum) of the function f(x) if there is a neighborhood of the point x o, for all points of which the inequality f(x) £ f(x o) (f(x) ³ f(x o)) is true.

The maximum and minimum points are called extremum points, and the values ​​of the function at these points are its extrema.

Necessary conditions for an extremum. If point x o is an extremum point of the function f(x), then either f ¢(x o) = 0, or f ¢(x o) does not exist. Such points are called critical, where the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

The first sufficient condition. Let be x o- critical point. If f ¢ (x) when passing through the point x o changes the plus sign to minus, then at the point x o the function has a maximum, otherwise it has a minimum. If the derivative does not change sign when passing through a critical point, then at the point x o there is no extremum.

The second sufficient condition. Let the function f(x) have a derivative
f ¢ (x) in a neighborhood of a point x o and the second derivative at the very point x o. If f ¢(x o) = 0, >0 (<0), то точка x o is a local minimum (maximum) point of the function f(x). If =0, then one must either use the first sufficient condition or involve higher derivatives.

On a segment, the function y = f(x) can reach its minimum or maximum value either at critical points or at the ends of the segment.

Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.

Decision. Since f ¢ (x) \u003d 6x 2 - 30x +36 \u003d 6 (x -2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. Since when passing through the point x 1 \u003d 2, the derivative changes sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Having calculated the values ​​of the function at the points x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.

Example 3.23. It is necessary to build a rectangular area near the stone wall so that it is fenced off with wire mesh on three sides, and adjoins the wall on the fourth side. For this there is a linear meters of the grid. At what aspect ratio will the site have the largest area?

Decision. Denote the sides of the site through x and y. The area of ​​the site is S = xy. Let be y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore, y = a - 2x and S = x(a - 2x), where 0 £ x £ a/2 (the length and width of the area cannot be negative). S ¢ = a - 4x, a - 4x = 0 for x = a/4, whence
y \u003d a - 2 × a / 4 \u003d a / 2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For x< a/4 S ¢ >0, and for x >a/4 S ¢<0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед).

Since S is continuous on and its values ​​at the ends of S(0) and S(a/2) are equal to zero, then the found value will be the largest value of the function. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is y = 2x.

Example 3.24. It is required to make a closed cylindrical tank with a capacity of V=16p » 50 m 3 . What should be the dimensions of the tank (radius R and height H) in order to use the least amount of material for its manufacture?

Decision. The total surface area of ​​the cylinder is S = 2pR(R+H). We know the volume of the cylinder V = pR 2 H Þ H = V/pR 2 =16p/ pR 2 = 16/ R 2 . Hence, S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S¢(R) = 2p(2R- 16/R 2) = 4p (R- 8/R 2). S ¢(R) = 0 for R 3 = 8, therefore,
R = 2, H = 16/4 = 4.

In all the formulas below, letters u and v differentiable functions of the independent variable are denoted x: , , but in letters a, c, n- permanent:

1.

3.

4.

5.

6.

The remaining formulas are written both for functions of an independent variable and for complex functions:

8.

9.

11.

12.

13.

14.

15.

16.

17.

7a.

8a.

9a.

11a.

12a.

13a.

16a.

17a.

When solving the examples below, detailed notes are made. However, one should learn to differentiate without intermediate entries.

Example 1 Find the derivative of a function .

Decision. This function is the algebraic sum of functions. We differentiate it using formulas 3, 5, 7 and 8:

Example 2 Find the derivative of a function

Decision. Applying formulas 6, 3, 7 and 1, we get

Example 3 Find the derivative of a function and calculate its value at

Decision. This is a complex function with an intermediate argument . Using formulas 7a and 10, we have

.

Example 4 Find the derivative of a function .

Decision. This is a complex function with an intermediate argument . Applying formulas 3, 5, 7a, 11, 16a, we get

Example 5 Find the derivative of a function .

Decision. We differentiate this function by formulas 6, 12, 3 and 1:

Example 6 Find the derivative of a function and calculate its value at .

Decision. First, we transform the function using the properties of logarithms:

Now we differentiate by formulas 3, 16a, 7 and 1:

.

Let us calculate the value of the derivative at .

Example 7 Find the derivative of the function and calculate its value at .

Decision. We use formulas 6, 3, 14a, 9a, 5 and 1:

.

Calculate the value of the derivative at :

.

The geometric meaning of the derivative.

The derivative of a function has a simple and important geometric interpretation.

If the function differentiable at a point X, then the graph of this function has a tangent at the corresponding point, and the slope of the tangent is equal to the value of the derivative at the point under consideration.

The slope of the tangent drawn to the graph of the function at point ( X 0 , at 0), is equal to the value of the derivative of the function at x = x 0 , i.e. .

The equation for this tangent has the form

Example 8. Write an equation for a tangent to a function graph at point A (3.6).

Decision. To find the slope of the tangent, we find the derivative of this function:

X= 3:

The tangent equation has the form

, or , i.e.

Example 9 Compose the equation of the tangent drawn to the graph of the function at the point with the abscissa x=2.

Decision. First, find the ordinate of the touch point. Since point A lies on the curve, its coordinates satisfy the equation of the curve, i.e.

; .

The equation of the tangent drawn to the curve at the point has the form . To find the slope of the tangent, we find the derivative:

.

The slope of the tangent is equal to the value of the derivative of the function at X= 2:

The tangent equation is:

, , i.e.

The physical meaning of the derivative. If the body moves in a straight line according to the law s=s(t), then for a period of time (from the moment t until the moment ) it will go some way. Then there is the average speed of movement for a period of time .

speed body movements at a given time t is called the limit of the ratio of the path to the increment of time, when the increment of time tends to zero:

.

Therefore, the time derivative of the path s t equal to the speed of the rectilinear motion of the body at a given time:

.

The rate of physical, chemical and other processes is also expressed using the derivative.

Function derivative is equal to the rate of change of this function for a given value of the argument X:

Example 10 The law of movement of a point along a straight line is given by the formula (s - in meters, t - in seconds). Find the speed of the point at the end of the first second.

Decision. The speed of a point at a given time is equal to the derivative of the path s by time t:

,

So, the speed of the point at the end of the first second is 9 m/s.

Example 11. A body thrown vertically upward moves according to the law , where v 0 - initial speed, g is the free fall acceleration. Find the speed of this movement for any moment of time t. How long will the body rise and to what height will it rise if v0= 40 m/s?

Decision. The speed at which a point is moving at a given time t equal to the derivative of the path s by time t:

.

At the highest point of the ascent, the velocity of the body is zero:

, , , , with.

Over 40/ g seconds the body rises to a height

, m.

Second derivative.

Function derivative in general is a function of X. If we calculate the derivative of this function, then we get the second-order derivative or the second derivative of the function .

Second derivative functions is called the derivative of its first derivative .

The second derivative of a function is denoted by one of the symbols - , , . Thus, .

Derivatives of any order are defined and denoted in a similar way. For example, a third order derivative:

or ,

Example 12. .

Decision. First we find the first derivative

Example 13 Find the second derivative of a function and calculate its value at x=2.

Decision. First we find the first derivative:

Differentiating again, we find the second derivative:

Let us calculate the value of the second derivative at x=2; we have

The physical meaning of the second derivative.

If the body moves in a straight line according to the law s = s(t), then the second derivative of the path s by time t equal to the acceleration of the body at a given time t:

Thus, the first derivative characterizes the speed of some process, and the second derivative characterizes the acceleration of the same process.

Example 14 The point moves in a straight line according to the law . Find the speed and acceleration of the movement .

Decision. The speed of the body at a given time is equal to the derivative of the path s by time t, and acceleration is the second derivative of the path s by time t. We find:

; then ;

; then

Example 15 The speed of rectilinear motion is proportional to the square root of the path traveled (as, for example, in free fall). Prove that this motion occurs under the action of a constant force.

Decision. According to Newton's law, the force F causing the movement is proportional to the acceleration, i.e.

or

According to the condition . Differentiating this equality, we find

Therefore, the acting force .

Applications of the derivative to the study of a function.

1) The condition for the function to increase: A differentiable function y = f(x) monotonically increases on the interval X if and only if its derivative is greater than zero, i.e. y = f(x) f'(x) > 0. This condition geometrically means that the tangent to the graph of this function forms an acute angle with a positive direction to the x-axis.

2) The condition for the function to decrease: A differentiable function y = f(x) monotonically decreases on the interval X if and only if its derivative is less than zero, i.e.

y = f(x)↓ f'(x) This condition geometrically means that the tangent to the graph of this function forms an obtuse angle with the positive direction of the x-axis)

3) The condition of the constancy of the function: A differentiable function y = f(x) is constant on the interval X if and only if its derivative is equal to zero, i.e. y = f(x) - constant f'(x) = 0 . This condition geometrically means that the tangent to the graph of this function is parallel to the oX axis, i.e. α \u003d 0)

Function extremes.

Definition 1: The point x \u003d x 0 is called minimum point function y = f(x), if this point has a neighborhood, for all points of which (except for the point itself) the inequality f(x)> f(x 0)

Definition 2: The point x \u003d x 0 is called maximum point function y = f(x) if this point has a neighborhood for all points of which (except for the point itself) the inequality f(x)< f(x 0).

Definition 3: The minimum or maximum point of a function is called a point extremum. The value of the function at this point is called extreme.

Remarks: 1. The maximum (minimum) is not necessarily the maximum (smallest) value of the function;

2. A function can have several maximums or minimums;

3. A function defined on a segment can reach an extremum only at the interior points of this segment.

5) Necessary condition for an extremum: If the function y \u003d f (x) has an extremum at the point x \u003d x 0, then at this point the derivative is equal to zero or does not exist. These points are called critical points of the 1st kind.

6) Sufficient conditions for the existence of the extremum of the function: Let the function y \u003d f (x) be continuous on the interval X and have inside this interval as a critical point of the 1st kind x \u003d x 0, then:

a) if this point has a neighborhood in which for x< х 0 f’(x) < 0, а при x>x 0 f’(x) > 0, then x = x 0 is a point minimum functions y = f(x);

b) if this point has a neighborhood in which for x< х 0 f’(x) >0, and for x> x 0

f'(x)< 0, то х = х 0 является точкой maximum functions y = f(x);

c) if this point has such a neighborhood that in it both to the right and to the left of the point x 0 the signs of the derivative are the same, then there is no extremum at the point x 0.

The intervals of decreasing or increasing functions are called intervals. monotony.

Definition1: The curve y = f(x) is called convex down on the interval a< х <в, если она лежит выше касательной в любой точке этого промежутка и кривая у = f(x) называется convex up on the interval a< х <в, если она лежит ниже касательной в любой точке этого промежутка.

Definition 2: The intervals in which the graph of the function is convex up or down are called swell at intervals function graph.

A sufficient condition for the curve to be convex. The graph of the differentiable function Y = f(x) is convex up on the interval a< х <в, если f”(x) < 0 и convex down, if f”(x) > 0.

Definition 1: The points at which the second derivative is zero or does not exist are called critical points of the second kind.

Definition 2: The point of the graph of the function Y = f(x), separating the intervals of the convexity of the opposite directions of this graph, is called the point inflection.

inflection point

Example: Given a function y \u003d x 3 - 2x 2 + 6x - 4. Investigate the function for intervals of monotonicity and extremum points. Determine the direction of the convexity and inflection points.

Solution: 1. Find the domain of the function: D(y) = ;

2. Find the first derivative: y’ = 3x 2 - 4x+ 6;

3. Let's solve the equation: y' = 0, 3x 2 - 4x+ 6 = 0, D 0, then this equation has no solution, therefore there are no extremum points. y' , then the function increases over the entire domain of definition.

4. Find the second derivative: y” = 6x - 4;

5. Solve the equation: y” = 0, 6x - 4 = 0, x =

Answer: ( ; - ) - inflection point, the function is convex upwards at x and convex upwards at x

Asymptotes.

1. Definition: The asymptote of a curve is a straight line to which the graph of a given function approaches indefinitely.

2. Types of asymptotes:

1) Vertical asymptotes. The graph of the function y = f(x) has a vertical asymptote if . The vertical asymptote equation has the form x = a

2) Horizontal asymptotes. The graph of the function y = f(x) has a horizontal asymptote if . The horizontal asymptote equation is y = b.

Example 1: For the function y = find the asymptotes.

3) Oblique asymptotes. The straight line y = kx + b is called the oblique asymptote of the graph of the function y = f(x) if . The values ​​of k and b are calculated by the formulas: k = ; b = .

Decision: , then y = 0 is the horizontal asymptote;

(since x - 3 ≠ 0, x ≠ 3), then x = 3 is the vertical asymptote. ,t. i.e. k = 0, then the curve has no oblique asymptote.

Example 2: For the function y = find the asymptotes.

Solution: x 2 - 25 ≠ 0 with x ≠ ± 5, then x \u003d 5 and x \u003d - 5 are horizontal asymptotes;

y = , then the curve has no vertical asymptote;

k = ; b = , i.e. y = 5x - oblique asymptote.

Examples of constructing function graphs.

Example 1 .

Investigate the function and build a graph of the function y \u003d x 3 - 6x 2 + 9x - 3

1. Find the domain of the function: D(y) = R

y (- x) \u003d (- x) 3 - 6 (- x) 2 + 9 (-x) - 3 \u003d - x 3 - 6x 2 - 9x - 3 \u003d - (x 3 + 6x 2 + 9x + 3), i.e.

(y \u003d x 5 - x 3 - odd, y \u003d x 4 + x 2 - even)

3. Is not periodic.

4. Find the points of intersection with the coordinate axes: if x \u003d 0, then y \u003d - 3 (0; - 3)

if Y = 0, x is hard to find.

5. Find the asymptotes of the graph of the function: There are no vertical asymptotes, because there are no x values ​​for which the function is indefinite; y = , i.e., there are no horizontal asymptotes;

k = , i.e., there are no oblique asymptotes.

6. We examine the function for intervals of monotonicity and its extrema: y’ = 3x 2 - 12x + 9,

y'= 0, 3x 2 - 12x + 9 = 0 x 1 = 1; x 2 = 3 - critical points of the 1st kind.

Let's determine the signs of the derivative: y'(0) = 9 > 0; y'(2) = - 3< 0; y’(4) = 9 > 0

y max = y(1) = 1, (1;1) - maximum point; y min \u003d y (3) \u003d - 3, (3; - 3) - minimum point, function y for x and y .

7. We examine the function for intervals of convexity and inflection points:

y” = (y’)’ = (3x 2 - 12x + 9)’ = 6x - 12, y” = 0, 6x - 12 = 0 x = 2 - critical point of the 1st kind.

Let's determine the signs of the second derivative: y”(0) = - 12< 0; y”(3) = 6 > 0

Y(2) = - 1 (2; - 1) - inflection point, the function is convex up at x and convex down at x.

8. Additional points:

X	- 1
at	- 19

9. Let's build a graph of the function:

Investigate the function and plot the function y =

1. Find the domain of the function: 1 - x ≠ 0, x ≠ 1, D(y) = .

2. Find out if the given function is even or odd: ,

y(- x) ≠ y(x) is not even and y(- x) ≠ - y(x) is not odd

3. Is not periodic.

4. Find the points of intersection with the coordinate axes: x \u003d 0, then y \u003d - 2; y = 0, then , i.e. (0; - 2); ().

5. Find the asymptotes of the graph of the function: since x ≠ 1, then the line x = 1 is the vertical asymptote;