The geometric meaning of the derivative is: Definition of derivative, its geometric meaning

Lesson objectives:

Students should know:

• what is called the slope of a line;
• the angle between the straight line and the Ox axis;
• what is the geometric meaning of the derivative;
• equation of the tangent to the graph of a function;
• a method for constructing a tangent to a parabola;
• be able to apply theoretical knowledge on practice.

Lesson objectives:

Educational: create conditions for students to master a system of knowledge, skills and abilities with the concepts of mechanical and geometric meaning of a derivative.

Educational: to form a scientific worldview in students.

Developmental: to develop students’ cognitive interest, creativity, will, memory, speech, attention, imagination, perception.

Methods of organizing educational and cognitive activities:

• visual;
• practical;
• By mental activity: inductive;
• according to the assimilation of material: partially search, reproductive;
• by degree of independence: laboratory work;
• stimulating: encouragement;
• control: oral frontal survey.

Lesson Plan

1. Oral exercises (find the derivative)
2. Student message on the topic “Causes of mathematical analysis”.
3. Learning new material
4. Phys. Just a minute.
5. Solving tasks.
6. Laboratory work.
7. Summing up the lesson.
8. Commenting homework.

Equipment: multimedia projector (presentation), cards ( laboratory work).

“A person only achieves something where he believes in his own strength”

L. Feuerbach

Organization of the class throughout the lesson, students' readiness for the lesson, order and discipline.

Setting learning goals for students, both for the entire lesson and for its individual stages.

Determine the significance of the material being studied both in this topic and in the entire course.

Verbal counting

1. Find derivatives:

" , ()" , (4sin x)", (cos2x)", (tg x)", "

2. Logic test.

a) Insert the missing expression.

The general direction of the development of science is ultimately determined by the requirements of the practice of human activity. The existence of ancient states with a complex hierarchical management system would have been impossible without the sufficient development of arithmetic and algebra, because collecting taxes, organizing army supplies, building palaces and pyramids, and creating irrigation systems required complex calculations. During the Renaissance, connections between different parts of the medieval world expanded, trade and crafts developed. A rapid rise in the technical level of production begins, and new sources of energy that are not associated with the muscular efforts of humans or animals are being used industrially. In the XI-XII centuries, fulling and weaving machines appeared, and in the middle of the XV - printing press. Due to the need for the rapid development of social production during this period, the essence of the natural sciences, which had been descriptive since ancient times, changed. The goal of natural science is an in-depth study of natural processes, not objects. Mathematics, which operated with constant quantities, corresponded to the descriptive natural science of antiquity. It was necessary to create a mathematical apparatus that would describe not the result of the process, but the nature of its flow and its inherent patterns. As a result, by the end of the 12th century, Newton in England and Leibniz in Germany completed the first stage of creating mathematical analysis. What is “mathematical analysis”? How can one characterize and predict the characteristics of any process? Use these features? To penetrate deeper into the essence of a particular phenomenon?

Let's follow the path of Newton and Leibniz and see how we can analyze the process, considering it as a function of time.

Let's introduce a few concepts that will help us further.

The graph of the linear function y=kx+ b is a straight line, the number k is called the slope of the straight line. k=tg, where is the angle of the straight line, that is, the angle between this straight line and the positive direction of the Ox axis.

Picture 1

Consider the graph of the function y=f(x). Let us draw a secant through any two points, for example, the secant AM. (Fig.2)

Angular coefficient of the secant k=tg. In a right triangle AMC<МАС = (объясните почему?). Тогда tg = = , что с точки зрения физики есть величина средней скорости протекания любого процесса на данном промежутке времени, например, скорости изменения расстояния в механике.

Figure 2

Figure 3

The term “speed” itself characterizes the dependence of a change in one quantity on a change in another, and the latter does not necessarily have to be time.

So, the tangent of the angle of inclination of the secant tg = .

We are interested in the dependence of changes in quantities over a shorter period of time. Let us direct the increment of the argument to zero. Then the right side of the formula is the derivative of the function at point A (explain why). If x -> 0, then point M moves along the graph to point A, which means straight line AM is approaching some straight line AB, which is tangent to the graph of the function y = f(x) at point A. (Fig.3)

The angle of inclination of the secant tends to the angle of inclination of the tangent.

Geometric meaning derivative is that the value of the derivative at a point is equal to slope tangent to the graph of a function at a point.

Mechanical meaning of derivative.

The tangent of the tangent angle is a value showing the instantaneous rate of change of the function at a given point, that is, a new characteristic of the process being studied. Leibniz called this quantity derivative, and Newton said that the derivative itself is called the instantaneous speed.

No. 91(1) page 91 – show on the board.

The angular coefficient of the tangent to the curve f(x) = x 3 at point x 0 – 1 is the value of the derivative of this function at x = 1. f’(1) = 3x 2 ; f’(1) = 3.

No. 91 (3.5) – dictation.

No. 92(1) – on the board if desired.

No. 92 (3) – independently with oral testing.

No. 92 (5) – at the board.

Answers: 45 0, 135 0, 1.5 e 2.

Goal: development of the concept “ mechanical sense derivative".

Applications of derivatives to mechanics.

The law has been set rectilinear motion points x = x(t), t.

1. Average speed of movement over a specified period of time;
2. Velocity and acceleration at time t 04
3. Moments of stopping; whether the point after the moment of stopping continues to move in the same direction or begins to move in the opposite direction;
4. Highest speed movements over a specified period of time.

The work is performed according to 12 options, the tasks are differentiated by level of difficulty (the first option is the lowest level of difficulty).

Before starting work, a conversation on the following questions:

1. What physical meaning derivative of displacement? (Speed).
2. Is it possible to find the derivative of speed?
3. Is this quantity used in physics? What's it called? (Acceleration). Instantaneous speed
4. equal to zero. What can be said about the movement of the body at this moment? (This is the moment of stopping).

What is the physical meaning of the following statements: the derivative of motion is equal to zero at point t 0; does the derivative change sign when passing through point t 0? (The body stops; the direction of movement changes to the opposite).

A sample of student work.

Figure 4

IN opposite direction.

Let's draw a schematic diagram of the speed. The highest speed is achieved at the point

t=10, v (10) =3· 10 2 -4· 10 =300-40=260

Figure 5

1) What is the geometric meaning of the derivative?
2) What is the mechanical meaning of a derivative?
3) Draw a conclusion about your work.

Page 90. No. 91(2,4,6), No.92(2,4,6,), p. 92 No. 112.

Used Books

• Textbook Algebra and beginnings of analysis.
  Authors: Yu.M. Kolyagin, M.V. Tkacheva, N.E. Fedorova, M.I. Shabunina.
  Edited by A. B. Zhizhchenko.
• Algebra 11th grade. Lesson plans according to the textbook by Sh. A. Alimov, Yu. M. Kolyagin, Yu. V. Sidorov. Part 1.
• Internet resources:

Abstract open lesson teacher of GBPOU " Teachers College No. 4 St. Petersburg"

Martusevich Tatyana Olegovna

Date: 12/29/2014.

Topic: Geometric meaning of derivatives.

Lesson type: learning new material.

Teaching methods: visual, partly search.

The purpose of the lesson.

Introduce the concept of a tangent to the graph of a function at a point, find out what the geometric meaning of the derivative is, derive the equation of the tangent and teach how to find it.

Educational objectives:

Achieve an understanding of the geometric meaning of the derivative; deriving the tangent equation; learn to solve basic problems;

provide repetition of material on the topic “Definition of a derivative”;

create conditions for control (self-control) of knowledge and skills.

Developmental tasks:

promote the formation of skills to apply techniques of comparison, generalization, and highlighting the main thing;

continue the development of mathematical horizons, thinking and speech, attention and memory.

Educational tasks:

promote interest in mathematics;

education of activity, mobility, communication skills.

Lesson type – a combined lesson using ICT.

Equipment – multimedia installation, presentationMicrosoftPowerPoint.

Lesson stage

Time

Teacher's activities

Student activity

1. Organizational moment.

State the topic and purpose of the lesson.

Topic: Geometric meaning of derivatives.

The purpose of the lesson.

Introduce the concept of a tangent to the graph of a function at a point, find out what the geometric meaning of the derivative is, derive the equation of the tangent and teach how to find it.

Preparing students for work in class.

Preparation for work in class.

Understanding the topic and purpose of the lesson.

Note-taking.

2. Preparation for learning new material through repetition and updating background knowledge.

Organization of repetition and updating of basic knowledge: definition of derivative and formulation of its physical meaning.

Formulating the definition of a derivative and formulating its physical meaning. Repetition, updating and consolidation of basic knowledge.

Organization of repetition and development of the skill of finding a derivative power function and elementary functions.

Finding the derivative of these functions using formulas.

Repetition of properties linear function.

Repetition, perception of drawings and teacher’s statements

3. Working with new material: explanation.

Explanation of the meaning of the relationship between function increment and argument increment

Explanation of the geometric meaning of the derivative.

Introduction of new material through verbal explanations using images and visual aids: multimedia presentation with animation.

Perception of explanation, understanding, answering teacher's questions.

Formulating a question to the teacher in case of difficulty.

Perception new information, its primary understanding and comprehension.

Formulation of questions to the teacher in case of difficulty.

Creating a note.

Formulation of the geometric meaning of the derivative.

Consideration of three cases.

Taking notes, making drawings.

4. Working with new material.

Primary comprehension and application of the studied material, its consolidation.

At what points is the derivative positive?

Negative?

Equal to zero?

Training in finding an algorithm for answering questions according to a schedule.

Understanding, making sense of, and applying new information to solve a problem.

5. Primary comprehension and application of the studied material, its consolidation.

Message of the task conditions.

Recording the conditions of the task.

Formulating a question to the teacher in case of difficulty

6. Application of knowledge: independent work of a teaching nature.

Solve the problem yourself:

Application of acquired knowledge.

Independent work on solving the problem of finding the derivative from a drawing. Discussion and verification of answers in pairs, formulation of a question to the teacher in case of difficulty.

7. Working with new material: explanation.

Deriving the equation of a tangent to the graph of a function at a point.

Detailed explanation deriving the equation of a tangent to the graph of a function at a point using a multimedia presentation for clarity, answering student questions.

Derivation of the tangent equation together with the teacher. Answers to the teacher's questions.

Taking notes, creating a drawing.

8. Working with new material: explanation.

In a dialogue with students, the derivation of an algorithm for finding the equation of the tangent to the graph of a given function at a given point.

In a dialogue with the teacher, derive an algorithm for finding the equation of the tangent to the graph of a given function at a given point.

Note-taking.

Message of the task conditions.

Training in the application of acquired knowledge.

Organizing the search for ways to solve a problem and their implementation. detailed analysis solutions with explanation.

Recording the conditions of the task.

Making assumptions about possible ways solving the problem when implementing each point of the action plan. Solving the problem together with the teacher.

Recording the solution to the problem and the answer.

9. Application of knowledge: independent work of a teaching nature.

Individual control. Consulting and assistance to students as needed.

Check and explain the solution using a presentation.

Application of acquired knowledge.

Independent work on solving the problem of finding the derivative from a drawing. Discussion and verification of answers in pairs, formulation of a question to the teacher in case of difficulty

10. Homework.

§48, problems 1 and 3, understand the solution and write it down in a notebook, with drawings.

№ 860 (2,4,6,8),

Homework message with comments.

Recording homework.

11. Summing up.

We repeated the definition of the derivative; physical meaning of derivative; properties of a linear function.

We learned what the geometric meaning of a derivative is.

We learned to derive the equation of a tangent to the graph of a given function at a given point.

Correction and clarification of lesson results.

Listing the results of the lesson.

12. Reflection.

1. You found the lesson: a) easy; b) usually; c) difficult.

a) have mastered it completely, I can apply it;

b) have learned it, but find it difficult to apply;

c) didn’t understand.

3. Multimedia presentation in class:

a) helped to master the material; b) did not help master the material;

c) interfered with the assimilation of the material.

Conducting reflection.

Job type: 7

Condition

The straight line y=3x+2 is tangent to the graph of the function y=-12x^2+bx-10. Find b, given that the abscissa of the tangent point.

Show solution

Solution

Let x_0 be the abscissa of the point on the graph of the function y=-12x^2+bx-10 through which the tangent to this graph passes. The value of the derivative at point x_0 is equal to the slope of the tangent, that is, y"(x_0)=-24x_0+b=3. On the other hand, the point of tangency belongs simultaneously to both the graph of the function and the tangent, that is, -12x_0^2+bx_0-10= 3x_0+2. We get a system of equations

Solving this system, we get x_0^2=1, which means either x_0=-1 or x_0=1.

According to the abscissa condition, the tangent points are less than zero, so x_0=-1, then b=3+24x_0=-21.

Job type: 7
Answer

Condition

Topic: Geometric meaning of derivatives. Tangent to the graph of a function

Show solution

The straight line y=-3x+4 is parallel to the tangent to the graph of the function y=-x^2+5x-7. Find the abscissa of the tangent point. The slope of the straight line to the graph of the function y=-x^2+5x-7 in

arbitrary point

According to the abscissa condition, the tangent points are less than zero, so x_0=-1, then b=3+24x_0=-21.

x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. The slope of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same angular coefficients. Therefore, we find a value x_0 such that =-2x_0 +5=-3.

Job type: 7
Answer

Condition

Show solution

We get: x_0 = 4.

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova. From the figure we determine that the tangent passes through points A(-6; 2) and B(-1; 1). Let us denote by C(-6; 1) the point of intersection of the lines x=-6 and y=1, and by \alpha the angle ABC (you can see in the figure that it is acute). Then straight line AB forms an angle \pi -\alpha with the positive direction of the Ox axis, which is obtuse. As is known, tg(\pi -\alpha) will be the value of the derivative of the function f(x) at point x_0.

According to the abscissa condition, the tangent points are less than zero, so x_0=-1, then b=3+24x_0=-21.

x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. The slope of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same angular coefficients. Therefore, we find a value x_0 such that =-2x_0 +5=-3.

Job type: 7
Answer

Condition

notice, that tg \alpha =\frac(AC)(CB)=\frac(2-1)(-1-(-6))=\frac15..

Show solution

From here, using the reduction formulas, we get:

tg(\pi -\alpha) =-tg \alpha =-\frac15=-0.2.

The straight line y=-2x-4 is tangent to the graph of the function y=16x^2+bx+12. Find b, given that the abscissa of the tangent point

Above zero

According to the abscissa condition, the tangent points are less than zero, so x_0=-1, then b=3+24x_0=-21.

x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. The slope of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same angular coefficients. Therefore, we find a value x_0 such that =-2x_0 +5=-3.

Job type: 7
Answer

Condition

Let x_0 be the abscissa of the point on the graph of the function y=16x^2+bx+12 through which

Show solution

is tangent to this graph. The value of the derivative at point x_0 is equal to the slope of the tangent, that is, y"(x_0)=32x_0+b=-2. On the other hand, the point of tangency belongs simultaneously to both the graph of the function and the tangent, that is, 16x_0^2+bx_0+12=- 2x_0-4. We obtain a system of equations\begin(cases) 32x_0+b=-2,\\16x_0^2+bx_0+12=-2x_0-4. \end(cases)

According to the abscissa condition, the tangent points are less than zero, so x_0=-1, then b=3+24x_0=-21.

x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. The slope of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same angular coefficients. Therefore, we find a value x_0 such that =-2x_0 +5=-3.

Job type: 7
Answer

Condition

The line y=4x-6 is parallel to the tangent to the graph of the function y=x^2-4x+9.

Show solution

Find the abscissa of the tangent point.

According to the abscissa condition, the tangent points are less than zero, so x_0=-1, then b=3+24x_0=-21.

x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. The slope of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same angular coefficients. Therefore, we find a value x_0 such that =-2x_0 +5=-3.

Job type: 7
Answer

Condition

The slope of the tangent to the graph of the function y=x^2-4x+9 at an arbitrary point x_0 is equal to y"(x_0). But y"=2x-4, which means y"(x_0)=2x_0-4. The slope of the tangent y =4x-7 specified in the condition is equal to 4. Parallel lines have the same angular coefficients. Therefore, we find a value of x_0 such that 2x_0-4=4.

Show solution

The figure shows the graph of the function y=f(x) and the tangent to it at the point with the abscissa x_0.

Find the value of the derivative of the function f(x) at point x_0.

From the figure we determine that the tangent passes through points A(1; 1) and B(5; 4).

1. Let us denote by C(5; 1) the point of intersection of the lines x=5 and y=1, and by \alpha the angle BAC (you can see in the figure that it is acute). Then straight line AB forms an angle \alpha with the positive direction of the Ox axis. Subject. Derivative. Geometric and mechanical meaning of derivative

If this limit exists, then the function is said to be differentiable at a point. The derivative of a function is denoted by (formula 2).

Geometric meaning of derivative.

1. Let's look at the graph of the function. From Fig. 1 it is clear that for any two points A and B of the graph of the function, formula 3 can be written). It contains the angle of inclination of the secant AB. Thus, the difference ratio is equal to the slope of the secant. If you fix point A and move point B towards it, then it decreases without limit and approaches 0, and the secant AB approaches the tangent AC. Therefore, the limit of the difference ratio is equal to the slope of the tangent at point A. This leads to the conclusion. The derivative of a function at a point is the slope of the tangent to the graph of this function at that point. This is the geometric meaning of the derivative. Tangent equation

. Let us derive the equation of the tangent to the graph of the function at a point. INgeneral case the equation of a straight line with an angular coefficient has the form: . To find b, we take advantage of the fact that the tangent passes through point A: . This implies: . Substituting this expression instead of b, we obtain the tangent equation (formula 4). Geometric meaning of derivative. The exam tasks related to this topic cause some difficulties for graduates. Most of them are actually very simple.

* Moreover, in these problems, at least two points through which this tangent passes are clearly marked on the sketch. What do you need to know to solve?

Let's build an arbitrary graph of a certain function y = f (x) on coordinate plane, construct a tangent at point x o, let us denote the angle between the straight line and the ox axis as α (alpha)

From the algebra course we know that the equation of a straight line has the form:

That is, the derivative of the functiony = f(x) at point x 0 equal to the slope of the tangent:

And the angular coefficient in turn equal to tangent angle α (alpha), that is:

Angle α (alpha) may be less than, greater than 90 degrees, or equal to zero.

Let us illustrate two cases:

1. The tangent angle is greater than 90 degrees (obtuse angle).

2. The angle of inclination of the tangent is zero degrees (the tangent is parallel to the axis Oh).

That is, problems in which a graph of a function is given, a tangent to this graph at a certain point, and it is required to find the derivative at the point of tangency, are reduced to finding the slope of the tangent (or the tangent of the angle of inclination of the tangent, which is the same thing).

Below we will consider solving such problems by finding the tangent of the angle between the tangent and the abscissa axis (axisOh), we will consider another solution method (finding the derivative through the angular coefficient) in the near future. We will also consider problems where knowledge of the properties of the derivative is required to read the graph of a function. Do not miss!

Please note that on the coordinate plane there are two points through which the tangent passes - this is very important point(one might say key in these tasks).

What else do you need?- this is knowledge for the tangent of an obtuse angle.

y = f(x) x 0 y = f(x) at the point x 0 .

The value of the derivative at the point of tangency is equal to the slope of the tangent, which in turn is equal to the tangent of the angle of inclination of this tangent to the abscissa axis. In order to find the tangent of this angle, we construct right triangle, where the segment bounded by two points on the graph will be the hypotenuse, and the legs are parallel to the axes. In this problem these are points (–5; –4), (1; 5).

Let me remind you: tangent acute angle in a right triangle is called the ratio opposite leg to the adjacent.

The legs are determined by the number of cells.

Angle of inclination of the tangent to the abscissa axis equal to angle BAC , Oh. Means

Answer: 1.5

y = f(x) x 0 y = f(x) at the point x 0 .

The task is similar to the previous one. We also construct a right triangle, where the segment bounded by two points on the graph will be the hypotenuse. In this problem these are points (–5; –7), (3; 3).

The legs are also determined by the number of cells.

The angle of inclination of the tangent to the x-axis is equal to the angle BAC , since the AC leg is parallel to the axis Oh. Means

Answer: 1.25

The figure shows the graph of the functiony = f(x) and the tangent to it at the abscissa pointx 0 . Find the value of the derivative of the functiony = f(x) at the point x 0 .

We construct a right triangle, where the segment bounded by two points on the graph will be the hypotenuse. In this problem these are points (–3; 3) and (5; 11). From point (5;11) we construct a continuation of the leg so that we get an external angle.

Since CD is parallel to the x-axis, angle ABD is equal to the angle of inclination of the tangent to the x-axis. Thus, we will calculate the tangent of angle ABD. Note that it is more than 90 degrees, so here you need to use the reduction formula for tangent:

Means

*The lengths of the legs are calculated by the number of cells.

Answer: -1.75

The figure shows the graph of the function y = f(x) and the tangent to it at the abscissa point x 0 . Find the value of the derivative of the function y = f(x) at the point x 0 . x 0

That's all! Good luck to you!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.