Find the value of the derivative at the point x0. Find the value of the derivative of the function at the point x0

Example 1

Reference: The following ways of notating a function are equivalent: In some tasks, it can be convenient to designate the function as a “player”, and in some 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 . Let's find the derivative first:

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

In the event that you do not understand how the derivative was found, return to the first two lessons of the topic. If there are difficulties (misunderstanding) with the arc tangent and its meanings, necessarily study methodological material Graphs and properties of 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 .

The equation of the tangent to the graph of the function

To consolidate the previous paragraph, consider the problem of finding the tangent to function graphics at this point. We met this task at school, and it is also found in the course of higher mathematics.

Consider a "demonstration" elementary example.

Write an equation for the tangent to the graph of the function at the point with the abscissa. I will immediately give a ready-made graphical solution to the problem (in practice, this is not necessary in most cases):

A rigorous definition of a tangent is given by definitions 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 "on the fingers", then the tangent to the graph of the function is straight, which concerns the graph of the function in the only point. In this case, all nearby points of the straight 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 a straight 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 a “player”, and in some as “ef from x”.

First we find the derivative:

I hope that many have already adapted to find such derivatives orally.

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

A small warm-up example for an independent solution:

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 the function for the inflection of the graph , full function study and etc.

But the task under consideration is found in control papers and by itself. And, as a rule, in such cases, the function is given quite complex. In this regard, consider two more examples.

Example 3

Calculate the derivative of a function at point .
Let's find the derivative first:

The derivative, in principle, is found, and the required value can be substituted. But I don't really want to do anything. The expression is very long, and the value of "x" is fractional. Therefore, we try to simplify our derivative as much as possible. In this case, let's try to reduce the last three terms to a common denominator: at point .

This is a do-it-yourself example.

How to find the value of the derivative of the function F(x) at the Ho point? How to solve it in general?

If the formula is given, then find the derivative and substitute X-zero instead of X. count
If we are talking about b-8 USE, graph, then you need to find the tangent of the angle (acute or obtuse), which forms a tangent to the X axis (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 the point x0 is in the lower part of the coordinate plane, then the sign in the answer will be minus, and if it is higher, then +.
Secondly, you need to know what is tange in a rectangular 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 a right-angled triangle and find tange.

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

there is no specific question - 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 any point, it is necessary to differentiate the given function with respect to this variable. In your case, by the 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 +, which I put with a clear conscience.

Such a formulation of the problem of finding the derivative is often posed to fix the material on the geometric meaning of the derivative. A graph of a certain function is proposed, completely arbitrary and not given by an equation, and it is required to find the value of the derivative (not the derivative itself!) at the specified point X0. To do this, a tangent to the 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=kx+b.

In this equation, the coefficient k and will be the value of the derivative. it remains only to find the value of the coefficient b. To do this, we find the value of y at x \u003d 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.

In problem B9, a graph of a function or derivative is given, from which it is required to determine one of the following quantities:

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

The functions and derivatives presented in this problem are always continuous, which greatly simplifies the solution. Despite the fact that the task belongs to the section of mathematical analysis, it is quite within the power of even the weakest students, 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 condition of problem B9 carefully so as not to make stupid mistakes: sometimes quite voluminous texts come across, but there are few important conditions that affect the course of the solution.

Calculation of the value of the derivative. Two point method

If the problem is given a graph of the 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 as A (x 1 ; y 1) and B (x 2 ; y 2). Write down the coordinates correctly - this is the key point of the solution, and any mistake here leads to the wrong 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 function increment by the argument increment - and this will be the answer.

Once again, we note: points A and B must be sought precisely on the tangent, and not on the graph of the function f(x), as is often the case. The tangent will necessarily contain at least two such points, otherwise the problem is formulated incorrectly.

Consider the points A (−3; 2) and B (−1; 6) and find the increments:
Δx \u003d x 2 - x 1 \u003d -1 - (-3) \u003d 2; Δy \u003d y 2 - y 1 \u003d 6 - 2 \u003d 4.

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

Task. The figure shows the graph of the function y \u003d 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 the point x 0 .

Consider points A (0; 3) and B (3; 0), find increments:
Δx \u003d x 2 - x 1 \u003d 3 - 0 \u003d 3; Δy \u003d y 2 - y 1 \u003d 0 - 3 \u003d -3.

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

Task. The figure shows the graph of the function y \u003d 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 the point x 0 .

Consider points A (0; 2) and B (5; 2) and find increments:
Δx \u003d x 2 - x 1 \u003d 5 - 0 \u003d 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 the rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of contact is equal to zero. In this case, you don’t even need to calculate anything - just look at the graph.

Calculating High and Low Points

Sometimes instead of a graph of a function in problem B9, a derivative graph is given and it is required to find the maximum or minimum point of the function. In this scenario, 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 the following inequality holds in some neighborhood of this point: f(x 0) ≥ f(x).
2. The point x 0 is called the minimum point of the function f(x) if the following inequality holds in some neighborhood of this point: f(x 0) ≤ f(x).

In order to find the maximum and minimum points on the graph of the derivative, it is enough to perform the following steps:

1. Redraw the graph of the derivative, removing all unnecessary information. As practice shows, extra data only interfere 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. Conversely, if the derivative graph lies below the OX axis, then f'(x) ≤ 0.
3. We again check the zeros and signs of the derivative. Where the sign changes from minus to plus, there is a 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 works only 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 segment [−5; 5]. Find the minimum point of the function f(x) on this segment.

Let's get rid of unnecessary information - we will leave only the borders [−5; 5] and the zeros of the derivative x = −3 and x = 2.5. 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 segment [−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 the zeros of the derivative x = −1.7 and x = 5. 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 segment [−6; 4]. Find the number of maximum points of the function f(x) that belong to the interval [−4; 3].

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

On this graph, there is only one maximum point x = 2. It is in it 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 formulated correctly, such changes should not affect the answer, since the points "without a fixed place of residence" are not directly involved in solving the problem. Of course, with integer points such a trick will not work.

Finding intervals of increase and decrease of a function

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

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

We formulate sufficient conditions for increasing and decreasing:

1. 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. 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.

We accept these assertions without proof. Thus, we get a scheme for finding intervals of increase and decrease, which is in many ways similar to the algorithm for calculating extremum points:

1. Remove all redundant information. On the original graph of the derivative, we are primarily interested in the zeros of the function, so we 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 has restrictions on the variable x, we additionally mark them on the new chart.
3. Now that we know the behavior of the function and the constraint, it remains to calculate the required value 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 decreasing function f(x). In your answer, write the sum of integers included in these intervals.

As usual, we redraw the graph and mark the boundaries [−3; 7.5], as well as the zeros of the derivative x = −1.5 and x = 5.3. Then we mark 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 segment [−10; 4]. Find the intervals of increasing function f(x). In your answer, write the length of the largest of them.

Let's get rid of redundant information. We leave only the boundaries [−10; 4] and zeros of the derivative, which this time turned out to be four: x = −8, x = −6, x = −3 and x = 2. Note the signs of the derivative and get the following picture:

We are interested in the intervals of increasing function, i.e. 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 it is required to find the length of the largest of the intervals, we write the value l 2 = 5 in response.