Find the value of the derivative of the function at point x0. Calculate the derivative of a function online

Example 1

Reference: The following ways of notating a function are equivalent: In some tasks it is convenient to designate the function as “game”, and in others as “ef from x”.

First we find the derivative:

Example 2

Calculate the derivative of a function at a point

, , full function study and etc.

Example 3

Calculate the derivative of the function at the point. First, let's find the derivative:

Well, that's a completely different matter. Let's calculate the value of the derivative at the point:

If you do not understand how the derivative was found, return to the first two lessons of the topic. If you have any difficulties (misunderstanding) with the arctangent and its meanings, Necessarily study methodological material Graphs and properties elementary functions – the very last paragraph. Because there are still enough arctangents for the student age.

Example 4

Calculate the derivative of the function at the point.

Equation of the tangent to the graph of a function

To consolidate the previous paragraph, consider the problem of finding the tangent to function graph at this point. We encountered this task at school, and it also appears in the course of higher mathematics.

Let's look at the simplest “demonstration” example.

Write an equation for the tangent to the graph of the function at the abscissa point. I'll bring it right away graphic solution tasks (in practice this is not necessary in most cases):

A strict definition of a tangent is given using definition of the derivative of a function, but for now we will master the technical part of the issue. Surely almost everyone intuitively understands what a tangent is. If you explain it “on your fingers”, then the tangent to the graph of a function is straight, which concerns the graph of the function in the only one point. In this case, all nearby points of the line are located as close as possible to the graph of the function.

As applied to our case: at the tangent (standard notation) touches the graph of the function at a single point.

And our task is to find the equation of the line.

Derivative of a function at a point

How to find the derivative of a function at a point? Two obvious points of this task follow from the wording:

1) It is necessary to find the derivative.

2) It is necessary to calculate the value of the derivative at a given point.

Example 1

Calculate the derivative of a function at a point

Help: The following ways of notating a function are equivalent:


In some tasks it is convenient to designate the function as “game”, and in others as “ef from x”.

First we find the derivative:

I hope many have already become accustomed to finding such derivatives orally.

In the second step, we calculate the value of the derivative at the point:

A small warm-up example for solving it yourself:

Example 2

Calculate the derivative of a function at a point

Full solution and answer at the end of the lesson.

The need to find the derivative at a point arises in the following tasks: constructing a tangent to the graph of a function (next paragraph), study of a function for an extremum , study of a function for the inflection of a graph , full function study and etc.

But the task in question occurs in tests and by itself. And, as a rule, in such cases the function given is quite complex. In this regard, let's look at two more examples.

Example 3

Calculate the derivative of a function at point .
First, let's find the derivative:

The derivative, in principle, has been found, and you can substitute the required value. But I don’t really want to do anything. The expression is very long, and the meaning of “x” is fractional. Therefore, we try to simplify our derivative as much as possible. In this case, let’s try to bring the last three terms to a common denominator: at point .

This is an example for you to solve on your own.

How to find the value of the derivative of the function F(x) at the point Xo? How do you even solve this?

If the formula is given, then find the derivative and substitute X-zero instead of X. Calculate
If we're talking about o b-8 Unified State Exam, graph, then you need to find the tangent of the angle (acute or obtuse) that the tangent to the X axis forms (using the mental construction of a right triangle and determining the tangent of the angle)

Timur Adilkhodzhaev

First, you need to decide on the sign. If point x0 is at the bottom coordinate plane, then the sign in the answer will be minus, and if higher, then +.
Secondly, you need to know what tange is in a rectangle. And this is the ratio of the opposite side (leg) to the adjacent side (also leg). There are usually a few black marks on the painting. From these marks you make right triangle and you find tanges.

How to find the value of the derivative of the function f x at point x0?

no specific question posed - 3 years ago

In the general case, in order to find the value of the derivative of a function with respect to some variable at some point, you need to differentiate the given function with respect to this variable. In your case, by variable X. In the resulting expression, instead of X, put the value of X at the point for which you need to find the value of the derivative, i.e. in your case, substitute zero X and calculate the resulting expression.

Well, your desire to understand this issue, in my opinion, undoubtedly deserves a +, which I give with a clear conscience.

This formulation of the problem of finding the derivative is often posed to consolidate the material on geometric meaning derivative. A graph of a certain function is proposed, completely arbitrary and not given by the equation and you need to find the value of the derivative (not the derivative itself, mind you!) at the specified point X0. To do this, a tangent to a given function is constructed and the points of its intersection with the coordinate axes are found. Then the equation of this tangent is drawn up in the form y=кx+b.

In this equation, the coefficient k and will be the value of the derivative. All that remains is to find the value of the coefficient b. To do this, we find the value of y at x = o, let it be equal to 3 - this is the value of the coefficient b. We substitute the values ​​of X0 and Y0 into the original equation and find k - our value of the derivative at this point.

Problem B9 gives a graph of a function or derivative from which you need to determine one of the following quantities:

1. The value of the derivative at some point x 0,
2. Maximum or minimum points (extremum points),
3. Intervals of increasing and decreasing functions (intervals of monotonicity).

The functions and derivatives presented in this problem are always continuous, making the solution much easier. Despite the fact that the task belongs to the section of mathematical analysis, even the weakest students can do it, since no deep theoretical knowledge is required here.

To find the value of the derivative, extremum points and monotonicity intervals, there are simple and universal algorithms - all of them will be discussed below.

Read the conditions of problem B9 carefully to avoid making stupid mistakes: sometimes you come across quite lengthy texts, but there are few important conditions that affect the course of the solution.

Calculation of the derivative value. Two point method

If the problem is given a graph of a function f(x), tangent to this graph at some point x 0, and it is required to find the value of the derivative at this point, the following algorithm is applied:

1. Find two “adequate” points on the tangent graph: their coordinates must be integer. Let's denote these points A (x 1 ; y 1) and B (x 2 ; y 2). Write down the coordinates correctly - this is key moment solutions, and any mistake here results in an incorrect answer.
2. Knowing the coordinates, it is easy to calculate the increment of the argument Δx = x 2 − x 1 and the increment of the function Δy = y 2 − y 1 .
3. Finally, we find the value of the derivative D = Δy/Δx. In other words, you need to divide the increment of the function by the increment of the argument - and this will be the answer.

Let us note once again: points A and B must be looked for precisely on the tangent, and not on the graph of the function f(x), as often happens. The tangent line will necessarily contain at least two such points - otherwise the problem will not be composed correctly.

Consider points A (−3; 2) and B (−1; 6) and find the increments:
Δx = x 2 − x 1 = −1 − (−3) = 2; Δy = y 2 − y 1 = 6 − 2 = 4.

Let's find the value of the derivative: D = Δy/Δx = 4/2 = 2.

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

Consider points A (0; 3) and B (3; 0), find the increments:
Δx = x 2 − x 1 = 3 − 0 = 3; Δy = y 2 − y 1 = 0 − 3 = −3.

Now we find the value of the derivative: D = Δy/Δx = −3/3 = −1.

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

Consider points A (0; 2) and B (5; 2) and find the increments:
Δx = x 2 − x 1 = 5 − 0 = 5; Δy = y 2 − y 1 = 2 − 2 = 0.

It remains to find the value of the derivative: D = Δy/Δx = 0/5 = 0.

From the last example, we can formulate a rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of tangency is zero. In this case, you don’t even need to count anything - just look at the graph.

Calculation of maximum and minimum points

Sometimes, instead of a graph of a function, Problem B9 gives a graph of the derivative and requires finding the maximum or minimum point of the function. In this situation, the two-point method is useless, but there is another, even simpler algorithm. First, let's define the terminology:

1. The point x 0 is called the maximum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≥ f(x).
2. The point x 0 is called the minimum point of the function f(x) if in some neighborhood of this point the following inequality holds: f(x 0) ≤ f(x).

In order to find the maximum and minimum points from the derivative graph, just follow these steps:

1. Redraw the derivative graph, removing all unnecessary information. As practice shows, unnecessary data only interferes with the decision. Therefore, we mark the zeros of the derivative on the coordinate axis - and that’s it.
2. Find out the signs of the derivative on the intervals between zeros. If for some point x 0 it is known that f'(x 0) ≠ 0, then only two options are possible: f'(x 0) ≥ 0 or f'(x 0) ≤ 0. The sign of the derivative is easy to determine from the original drawing: if the derivative graph lies above the OX axis, then f'(x) ≥ 0. And vice versa, if the derivative graph lies below the OX axis, then f'(x) ≤ 0.
3. We check the zeros and signs of the derivative again. Where the sign changes from minus to plus is the minimum point. Conversely, if the sign of the derivative changes from plus to minus, this is the maximum point. Counting is always done from left to right.

This scheme only works for continuous functions - there are no others in problem B9.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−5; 5]. Find the minimum point of the function f(x) on this segment.

Let's get rid of unnecessary information and leave only the boundaries [−5; 5] and zeros of the derivative x = −3 and x = 2.5. We also note the signs:

Obviously, at the point x = −3 the sign of the derivative changes from minus to plus. This is the minimum point.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7]. Find the maximum point of the function f(x) on this segment.

Let's redraw the graph, leaving only the boundaries [−3; 7] and zeros of the derivative x = −1.7 and x = 5. Let us note the signs of the derivative on the resulting graph. We have:

Obviously, at the point x = 5 the sign of the derivative changes from plus to minus - this is the maximum point.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−6; 4]. Find the number of maximum points of the function f(x) belonging to the segment [−4; 3].

From the conditions of the problem it follows that it is enough to consider only the part of the graph limited by the segment [−4; 3]. Therefore, we build a new graph on which we mark only the boundaries [−4; 3] and zeros of the derivative inside it. Namely, points x = −3.5 and x = 2. We get:

On this graph there is only one maximum point x = 2. It is at this point that the sign of the derivative changes from plus to minus.

A small note about points with non-integer coordinates. For example, in the last problem the point x = −3.5 was considered, but with the same success we can take x = −3.4. If the problem is compiled correctly, such changes should not affect the answer, since the points “without a fixed place of residence” do not directly participate in solving the problem. Of course, this trick won’t work with integer points.

Finding intervals of increasing and decreasing function

In such a problem, like the points of maximum and minimum, it is proposed to use the graph of the derivative to find areas in which the function itself increases or decreases. First, let's define what increasing and decreasing are:

1. A function f(x) is said to be increasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≤ f(x 2). In other words, the larger the argument value, the larger the function value.
2. A function f(x) is said to be decreasing on a segment if for any two points x 1 and x 2 from this segment the following statement is true: x 1 ≤ x 2 ⇒ f(x 1) ≥ f(x 2). Those. A larger argument value corresponds to a smaller function value.

Let us formulate sufficient conditions for increasing and decreasing:

1. In order for a continuous function f(x) to increase on the segment , it is sufficient that its derivative inside the segment be positive, i.e. f’(x) ≥ 0.
2. In order for a continuous function f(x) to decrease on the segment , it is sufficient that its derivative inside the segment be negative, i.e. f’(x) ≤ 0.

Let us accept these statements without evidence. Thus, we obtain a scheme for finding intervals of increasing and decreasing, which is in many ways similar to the algorithm for calculating extremum points:

1. Remove all unnecessary information. In the original graph of the derivative, we are primarily interested in the zeros of the function, so we will leave only them.
2. Mark the signs of the derivative at the intervals between zeros. Where f’(x) ≥ 0, the function increases, and where f’(x) ≤ 0, it decreases. If the problem sets restrictions on the variable x, we additionally mark them on a new graph.
3. Now that we know the behavior of the function and the constraints, it remains to calculate the quantity required in the problem.

Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7.5]. Find the intervals of decrease of the function f(x). In your answer, indicate the sum of the integers included in these intervals.

As usual, let's redraw the graph and mark the boundaries [−3; 7.5], as well as zeros of the derivative x = −1.5 and x = 5.3. Then we note the signs of the derivative. We have:

Since the derivative is negative on the interval (− 1.5), this is the interval of decreasing function. It remains to sum all the integers that are inside this interval:
−1 + 0 + 1 + 2 + 3 + 4 + 5 = 14.

Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−10; 4]. Find the intervals of increasing function f(x). In your answer, indicate the length of the largest of them.

Let's get rid of unnecessary information. Let us leave only the boundaries [−10; 4] and zeros of the derivative, of which there were four this time: x = −8, x = −6, x = −3 and x = 2. Let’s mark the signs of the derivative and get the following picture:

We are interested in the intervals of increasing function, i.e. such where f’(x) ≥ 0. There are two such intervals on the graph: (−8; −6) and (−3; 2). Let's calculate their lengths:
l 1 = − 6 − (−8) = 2;
l 2 = 2 − (−3) = 5.

Since we need to find the length of the largest of the intervals, we write down the value l 2 = 5 as an answer.

The calculator calculates the derivatives of all elementary functions, giving detailed solution. The differentiation variable is determined automatically.

Derivative of a function- one of the most important concepts in mathematical analysis. The emergence of the derivative was led to such problems as, for example, calculating the instantaneous speed of a point at a moment in time, if the path depending on time is known, the problem of finding the tangent to a function at a point.

Most often, the derivative of a function is defined as the limit of the ratio of the increment of the function to the increment of the argument, if it exists.

Definition. Let the function be defined in some neighborhood of the point. Then the derivative of the function at a point is called the limit, if it exists

How to calculate the derivative of a function?

In order to learn to differentiate functions, you need to learn and understand differentiation rules and learn to use table of derivatives.

Rules of differentiation

Let and be arbitrary differentiable functions of a real variable and be some real constant. Then

— rule for differentiating the product of functions

— rule for differentiating quotient functions

0" height="33" width="370" style="vertical-align: -12px;"> — differentiation of a function with a variable exponent

- differentiation rule complex function

— rule for differentiating a power function

Derivative of a function online

Our calculator will quickly and accurately calculate the derivative of any function online. The program will not make mistakes when calculating the derivative and will help you avoid long and tedious calculations. Online calculator It will also be useful in the case when there is a need to check the correctness of your solution, and if it is incorrect, quickly find the error.

A lot of theory has been written about geometric meaning. I won’t go into the derivation of the function increment, but let me remind you the basics for completing tasks:

The derivative at point x is equal to the slope of the tangent to the graph of the function y = f(x) at this point, that is, it is the tangent of the angle of inclination to the X axis.

Let’s immediately take the task from the Unified State Exam and begin to understand it:

Task No. 1. The picture shows graph of a function y = f(x) and the tangent to it at the point with the abscissa x0. Find the value of the derivative of the function f(x) at the point x0.
Who is in a hurry and does not want to understand the explanations: build up to any such triangle (as shown below) and divide the standing side (vertical) by the lying side (horizontal) and you will be lucky if you don’t forget about the sign (if the line is decreasing (→↓), then the answer should be minus, if the line is increases (→), then the answer must be positive!)

You need to find the angle between the tangent and the X axis, let's call it α: draw a straight line parallel to the X axis anywhere through the tangent to the graph, we get the same angle.

It is better not to take point x0, because You will need a large magnifying glass to determine the exact coordinates.

Taking any right triangle (3 options are suggested in the figure), we find tgα (the angles are then equal, as corresponding), i.e. we obtain the derivative of the function f(x) at the point x0. Why is this so?

If we draw tangents at other points x2, x1, etc. the tangents will be different.

Let's go back to 7th grade to build a line!

The equation of a straight line is given by the equation y = kx + b, where

k - inclination relative to the X axis.

b is the distance between the intersection point with the Y axis and the origin.

The derivative of a straight line is always the same: y" = k.

At whatever point on the line we take the derivative, it will be unchanged.

Therefore, all that remains is to find tgα (as mentioned above: divide the standing side by the lying side). We divide opposite leg to the adjacent one, we find that k = 0.5. However, if the graph is decreasing, the coefficient is negative: k = −0.5.

I advise you to check yourself second way:
You can define a straight line using two points. Let's find the coordinates of any two points. For example, (-2;-2) and (2;-4):

Let's substitute the coordinates of the points into the equation y = kx + b instead of y and x:

−2 = −2k + b

Solving this system, we obtain b = −3, k = −0.5

Conclusion: The second method takes longer, but in it you will not forget about the sign.

Answer: − 0.5

Task No. 2. The picture shows derivative graph functions f(x). Eight points are marked on the abscissa axis: x1, x2, x3, ..., x8. How many of these points lie on the intervals of increasing function f(x)?

If the graph of a function is decreasing - the derivative is negative (and vice versa is true).

If the graph of a function increases, the derivative is positive (and vice versa is true).

These two phrases will help you solve most problems.

Look carefully a drawing of a derivative or function is given to you, and then choose one of two phrases.

Let's construct a schematic graph of the function. Because We are given a graph of the derivative, then where it is negative, the graph of the function decreases, where it is positive, it increases!

It turns out that 3 points lie on increasing areas: x4; x5; x6.

Answer: 3

Task No. 3. The function f(x) is defined on the interval (-6; 4). The picture shows graph of its derivative. Find the abscissa of the point at which the function takes on its greatest value.

I advise you to always plot how the function graph goes, using arrows like this or schematically with signs (as in No. 4 and No. 5):

Obviously, if the graph increases to −2, then the maximum point is −2.

Answer: −2

Task No. 4. The figure shows a graph of the function f(x) and twelve points on the abscissa axis: x1, x2, ..., x12. At how many of these points is the derivative of the function negative?

The problem is the opposite, given a graph of a function, you need to schematically plot what the graph of the derivative of the function will look like and count how many points will lie in the negative range.

Positive: x1, x6, x7, x12.

Negative: x2, x3, x4, x5, x9, x10, x11.

Answer: 7

Another type of task when asked about some terrible “extremes”? It won’t be difficult for you to find what it is, but I’ll explain it for the graphs.

Task No. 5. The figure shows a graph of the derivative of the function f(x), defined on the interval (-16; 6). Find the number of extremum points of the function f(x) on the interval [-11; 5].

Let's mark the interval from -11 to 5!

Let's turn our bright eyes to the sign: a graph of the derivative of the function is given => then the extrema are the points of intersection with the X axis.

Answer: 3

Task No. 6. The figure shows a graph of the derivative of the function f(x), defined on the interval (-13; 9). Find the number of maximum points of the function f(x) on the interval [-12; 5].

Let's mark the interval from -12 to 5!

You can look at the table with one eye; the maximum point is an extremum, such that before it the derivative is positive (the function increases), and after it the derivative is negative (the function decreases). Such points are circled.

The arrows show how the function graph behaves

Answer: 3

Task No. 7. The figure shows a graph of the function f(x) defined on the interval (-7; 5). Find the number of points at which the derivative of the function f(x) is equal to 0.

You can look at the table above (the derivative is zero, which means these are extremum points). And in this problem the graph of the function is given, which means you need to find number of inflection points!

Or you can, as usual: build a schematic graph of the derivative.

The derivative is zero when the graph of a function changes its direction (from increasing to decreasing and vice versa)

Answer: 8

Task No. 8. The picture shows derivative graph function f(x), defined on the interval (-2; 10). Find the intervals of increasing function f(x). In your answer, indicate the sum of integer points included in these intervals.

Let's construct a schematic graph of the function:

Where it increases, we get 4 integer points: 4 + 5 + 6 + 7 = 22.

Answer: 22

Task No. 9. The picture shows derivative graph function f(x), defined on the interval (-6; 6). Find the number of points f(x) at which the tangent to the graph of the function is parallel to or coincides with the line y = 2x + 13.

We are given a graph of the derivative! This means that our tangent also needs to be “translated” into a derivative.

Derivative of tangent: y" = 2.

Now let's construct both derivatives:

The tangents intersect at three points, which means our answer is 3.

Answer: 3

Task No. 10. The figure shows a graph of the function f(x), and the points -2, 1, 2, 3 are marked. At which of these points is the value of the derivative the smallest? Please indicate this point in your answer.

The task is somewhat similar to the first one: to find the value of the derivative, you need to construct a tangent to this graph at a point and find the coefficient k.

If the line is decreasing, k< 0.

If the line is increasing, k > 0.

Let's think about how the value of the coefficient will affect the slope of the line:

When k = 1 or k = − 1, the graph will be in the middle between the X and Y axes.

The closer the straight line is to the X-axis, the closer the k coefficient is to zero.

The closer the straight line is to the Y axis, the closer the coefficient k is to infinity.

At point -2 and 1 k<0, однако в точке 1 прямая убывает "быстрее" больше похоже на ось Y =>this is where the smallest value of the derivative will be

Answer: 1

Task No. 11.

The line is tangent y = 3x + 9 to the graph of the function y = x³ + x² + 2x + 8. Find the abscissa of the tangent point.

The straight line will be tangent to the graph when the graphs have a common point, as do their derivatives. Let's equate the graph equations and their derivatives:

Having solved the second equation, we get 2 points. To check which one is suitable, we substitute each of the x’s into the first equation. Only one will do.

I don’t want to solve a cubic equation at all, but I’d love to solve a quadratic equation.

But what should you write down in response if you get two “normal” answers?

When substituting x(x) into the original graphs y = 3x + 9 and y = x³ + x² + 2x + 8 you should get the same Y

y= 1³+1²+2×1+8=12

Answer: 1

Right! So x=1 will be the answer

Task No. 12.

The straight line y = − 5x − 6 is tangent to the graph of the function ax² + 5x − 5. Find a.

Let us similarly equate functions and their derivatives:

The task with derivatives is considered one of the most difficult in the first part of the Unified State Exam, however, with a little care and understanding of the question, you will succeed and you will increase the percentage of completion of this task!