Geometric meaning of derivative. What is a derivative? Definition and meaning of a derivative function
Derivative of a function.
1. Definition of a derivative, its geometric meaning.
2. Derivative of a complex function.
3. Derivative of the inverse function.
4. Higher order derivatives.
5. Parametrically defined functions and implicitly.
6. Differentiation of functions specified parametrically and implicitly.
Introduction.
The origins of differential calculus were two questions raised by the demands of science and technology in the 17th century.
1) Question about calculating speed for an arbitrarily given law of motion.
2) The question of finding (using calculations) a tangent to an arbitrarily given curve.
The problem of drawing a tangent to some curves was solved by the ancient Greek scientist Archimedes (287-212 BC), using the drawing method.
But only in the 17th and 18th centuries, in connection with the progress of natural science and technology, these issues received due development.
One of the important questions when studying any physical phenomenon Usually the question is about the speed, the speed of the occurring phenomenon.
The speed at which an airplane or car moves is always the most important indicator of its performance. The rate of population growth of a particular state is one of the main characteristics of its social development.
The original idea of speed is clear to everyone. However, this general idea is not enough to solve most practical problems. It is necessary to have such a quantitative definition of this quantity, which we call speed. The need for such precise quantification has historically been one of the main motivations for the creation of mathematical analysis Yu. An entire section of mathematical analysis is devoted to solving this basic problem and drawing conclusions from this solution. We move on to studying this section.
Definition of derivative, its geometric meaning.
Let a function be given that is defined in a certain interval (a,c) and continuous in it.
1. Let's give the argument X increment , then the function will get
increment:
2. Let's create a relation 
.
3. Passing to the limit at and, assuming that the limit
exists, we obtain a quantity called
derivative of a function with respect to argument X.
Definition. The derivative of a function at a point is the limit of the ratio of the increment of the function to the increment of the argument when →0.
The value of the derivative obviously depends on the point X, in which it is found, therefore the derivative of the function is, in turn, some function of X. Denoted by .
By definition we have
or (3)
Example. Find the derivative of the function.
1. ; 
In the coordinate plane xOy consider the graph of the function y=f(x). Let's fix the point M(x 0 ; f (x 0)). Let's add an abscissa x 0 increment Δx. We will get a new abscissa x 0 +Δx. This is the abscissa of the point N, and the ordinate will be equal f (x 0 +Δx). The change in the abscissa entailed a change in the ordinate. This change is called the function increment and is denoted Δy.
Δy=f (x 0 +Δx) - f (x 0). Through dots M And N let's draw a secant MN, which forms an angle φ with positive axis direction Oh. Let's determine the tangent of the angle φ from a right triangle MPN.
Let Δx tends to zero. Then the secant MN will tend to take a tangent position MT, and the angle φ will become an angle α . So, the tangent of the angle α is the limiting value of the tangent of the angle φ :
The limit of the ratio of the increment of a function to the increment of the argument, when the latter tends to zero, is called the derivative of the function at a given point:
Geometric meaning of derivative lies in the fact that the numerical derivative of the function at a given point is equal to the tangent of the angle formed by the tangent drawn through this point to the given curve and the positive direction of the axis Oh:
Examples.
1. Find the increment of the argument and the increment of the function y= x 2, If initial value the argument was equal 4 , and new - 4,01 .
Solution.
New argument value x=x 0 +Δx. Let's substitute the data: 4.01=4+Δх, hence the increment of the argument Δx=4.01-4=0.01. The increment of a function, by definition, is equal to the difference between the new and previous values of the function, i.e. Δy=f (x 0 +Δx) - f (x 0). Since we have a function y=x2, That Δу=(x 0 +Δx) 2 - (x 0) 2 =(x 0) 2 +2x 0 · Δx+(Δx) 2 - (x 0) 2 =2x 0 · Δx+(Δx) 2 =
2 · 4 · 0,01+(0,01) 2 =0,08+0,0001=0,0801.
Answer: argument increment Δx=0.01; function increment Δу=0,0801.
The function increment could be found differently: Δy=y (x 0 +Δx) -y (x 0)=y(4.01) -y(4)=4.01 2 -4 2 =16.0801-16=0.0801.
2. Find the angle of inclination of the tangent to the graph of the function y=f(x) at the point x 0, If f "(x 0) = 1.
Solution.
The value of the derivative at the point of tangency x 0 and is the value of the tangent of the tangent angle (the geometric meaning of the derivative). We have: f "(x 0) = tanα = 1 → α = 45°, because tg45°=1.
Answer: the tangent to the graph of this function forms an angle with the positive direction of the Ox axis equal to 45°.
3. Derive the formula for the derivative of the function y=x n.
Differentiation is the action of finding the derivative of a function.
When finding derivatives, use formulas that were derived based on the definition of a derivative, in the same way as we derived the formula for the derivative degree: (x n)" = nx n-1.
These are the formulas.
Table of derivatives It will be easier to memorize by pronouncing verbal formulations:
1. The derivative of a constant quantity is zero.
2. X prime is equal to one.
3. The constant factor can be taken out of the sign of the derivative.
4. The derivative of a degree is equal to the product of the exponent of this degree by a degree with the same base, but the exponent is one less.
5. The derivative of a root is equal to one divided by two equal roots.
6. The derivative of one divided by x is equal to minus one divided by x squared.
7. The derivative of the sine is equal to the cosine.
8. The derivative of the cosine is equal to minus sine.
9. The derivative of the tangent is equal to one divided by the square of the cosine.
10. The derivative of the cotangent is equal to minus one divided by the square of the sine.
We teach differentiation rules.
1.
The derivative of an algebraic sum is equal to the algebraic sum of the derivatives of the terms.
2. The derivative of a product is equal to the product of the derivative of the first factor and the second plus the product of the first factor and the derivative of the second.
3. The derivative of “y” divided by “ve” is equal to a fraction in which the numerator is “y prime multiplied by “ve” minus “y multiplied by ve prime”, and the denominator is “ve squared”.
4. A special case of the formula 3.
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 of basic 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 the derivative of a power function and elementary functions.
Finding the derivative of these functions using formulas.
Repetition of the properties of a 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, answers to teacher 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 answers to questions posed 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 a 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),
Message homework 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 how 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.
Lecture: The concept of the derivative of a function, the geometric meaning of the derivative
The concept of a derivative function
Let us consider some function f(x), which will be continuous over the entire interval of consideration. On the interval under consideration, we select the point x 0, as well as the value of the function at this point.
So, let's look at the graph on which we mark our point x 0, as well as the point (x 0 + ∆x). Recall that ∆х is the distance (difference) between two selected points.
It is also worth understanding that each x corresponds to eigenvalue functions y.
The difference between the values of the function at the point x 0 and (x 0 + ∆x) is called the increment of this function: ∆у = f(x 0 + ∆x) - f(x 0).
Let's pay attention to Additional information, which is on the graph is a secant called KL, as well as the triangle that it forms with intervals KN and LN.
The angle at which the secant is located is called its angle of inclination and is denoted α. It can be easily determined that the degree measure of the angle LKN is also equal to α.
Now let's remember the relationships in a right triangle tgα = LN / KN = ∆у / ∆х.
That is, the tangent of the secant angle is equal to the ratio of the increment of the function to the increment of the argument.
At one time, the derivative is the limit of the ratio of the increment of a function to the increment of the argument on infinitesimal intervals.
The derivative determines the rate at which a function changes over a certain area.
Geometric meaning of derivative
If you find the derivative of any function at a certain point, you can determine the angle at which the tangent to the graph in a given current will be located, relative to the OX axis. Pay attention to the graph - the tangential slope angle is denoted by the letter φ and is determined by the coefficient k in the equation of the straight line: y = kx + b.
That is, we can conclude that geometric sense the derivative is the tangent of the tangent at some point of the function.
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 is less than zero.
Show solutionSolution
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 obtain a system of equations \begin(cases) -24x_0+b=3,\\-12x_0^2+bx_0-10=3x_0+2. \end(cases)
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.
Answer
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
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.
Show solutionSolution
The angular coefficient of the straight line to the graph of the function y=-x^2+5x-7 at an arbitrary point x_0 is equal to y"(x_0). But y"=-2x+5, which means y"(x_0)=-2x_0+5. Angular the coefficient of the line y=-3x+4 specified in the condition is equal to -3. Parallel lines have the same slope coefficients. Therefore, we find a value x_0 such that =-2x_0 +5=-3.
We get: x_0 = 4.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
Show solutionSolution
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. notice, that tg \alpha =\frac(AC)(CB)=\frac(2-1)(-1-(-6))=\frac15. From here, using the reduction formulas, we get: tg(\pi -\alpha) =-tg \alpha =-\frac15=-0.2.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
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 is greater than zero.
Show solutionSolution
Let x_0 be the abscissa of the point on the graph of the function y=16x^2+bx+12 through which
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)
Solving the 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 greater than zero, so x_0=1, then b=-2-32x_0=-34.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
The figure shows a graph of the function y=f(x), defined on the interval (-2; 8). Determine the number of points at which the tangent to the graph of the function is parallel to the straight line y=6.
Solution
The straight line y=6 is parallel to the Ox axis. Therefore, we find points at which the tangent to the graph of the function is parallel to the Ox axis. On this chart, such points are extremum points (maximum or minimum points). As you can see, there are 4 extremum points.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
The line y=4x-6 is parallel to the tangent to the graph of the function y=x^2-4x+9. Find the abscissa of the tangent point.
Show solutionSolution
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. We get: x_0 = 4.
Answer
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: Geometric meaning of derivatives. Tangent to the graph of a function
Condition
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.
Solution
From the figure we determine that the tangent passes through points A(1; 1) and B(5; 4). 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.
