Determine the value of the derivative of the function at the point x0. Calculating the derivative of a function online. Equation of a tangent to a graph of a function
The calculator calculates the derivatives of all elementary functions, giving a detailed solution. The differentiation variable is determined automatically.
Derivative of a function- one of the most important concepts in mathematical analysis. Such problems as, for example, calculating the instantaneous velocity of a point at a time moment, if the path is known depending on time, the problem of finding the tangent to a function at a point, led to the appearance of the derivative.
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 how to differentiate functions, you need to learn and understand differentiation rules and learn to use derivative table.
Differentiation rules
Let and be arbitrary differentiable functions of a real variable, be some real constant. Then
- the rule of differentiation of the product of functions
- the rule of differentiation of private functions
0 "height =" 33 "width =" 370 "style =" vertical-align: -12px; "> - differentiation of function with variable exponent
- differentiation rule complex function
- the rule of differentiation of the power function
Derivative 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 to avoid long and tedious calculations. The online calculator will also be useful in the case when there is a need to check your solution for correctness, and if it is incorrect, quickly find the error.
Much theory has been written about the geometric meaning. I will not go into the output of the function increment, I will remind the main thing for performing tasks:
The derivative at the 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 take the task from the exam at once and begin to understand it:
Task number 1. The figure shows function graph 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) into the lying (horizontal) and you will be happy if you don’t forget about the sign (if the straight line decreases (→ ↓), then the answer should be with a minus, if the straight 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 proposed 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 it so?
If we draw tangents at other points x2, x1, etc. the tangents will be different.
Let's go back to grade 7 to build a straight line!
The straight line equation is given by the equation y = kx + b, where
k - tilt relative to the X-axis.
b is the distance between the y-intercept and the origin.
The derivative of the straight line is always the same: y "= k.
At whatever point on the straight line we take the derivative, it will be unchanged.
Therefore, it remains only to find tgα (as mentioned above: we divide the standing side by the recumbent side). We divide the opposite leg by the adjacent one, we get that k = 0.5. However, if the graph is decreasing, the coefficient is negative: k = −0.5.
I advise you to check yourself in the second way:
A straight line can be specified by two points. Let's find the coordinates of any two points. For example, (-2; -2) and (2; -4):
Substitute y = kx + b in the equation instead of y and x coordinates of points:
−2 = −2k + b
Solving this system, we get b = −3, k = −0.5
Conclusion: The second method is longer, but in it you will not forget about the sign.
Answer: - 0.5
Task number 2. The figure shows derivative plot function f (x). Eight points are marked on the abscissa: x1, x2, x3, ..., x8. How many of these points lie on the intervals of increasing function f (x)?
If the graph of the function decreases - the derivative is negative (and vice versa).
If the graph of the function increases, the derivative is positive (and vice versa).
These two phrases will help you solve most of the problems.
Watch carefully a drawing of a derivative or a 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 the sections of increasing: x4; x5; x6.
Answer: 3
Task number 3. The function f (x) is defined on the interval (-6; 4). The figure shows graph of its derivative... Find the abscissa of the point at which the function takes its greatest value.
I advise you to always build how the graph of the function goes with such arrows or schematically with signs (as in # 4 and # 5):
Obviously, if the graph increases to −2, then the maximum point is −2.
Answer: −2
Problem number 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 task is the opposite, the graph of the function is given, you need to schematically build how the graph of the derivative of the function will look, and calculate 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 tasks when asked about some terrible "extremes"? It will not be difficult for you to find what it is, but I will explain it for the graphs.
Problem number 5. The figure shows the 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 segment [-11; 5].
Let's mark the interval from -11 to 5!
Let's turn our bright eyes to the plate: the graph of the derivative of the function is given => then the extrema are the points of intersection with the X-axis.
Answer: 3
Problem number 6. The figure shows the 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 segment [-12; 5].
Let's note the interval from -12 to 5!
You can look at the plate 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
Problem number 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 0.
You can look at the above table (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 equal to zero when the graph of functions changes its direction (from increasing to decreasing and vice versa)
Answer: 8
Problem number 8. The figure shows derivative plot function f (x), defined on the interval (-2; 10). Find the intervals of increasing function f (x). In your answer, indicate the sum of whole 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
Problem number 9. The figure shows derivative plot 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 straight line y = 2x + 13.
We are given a graph of the derivative! This means that our tangent must also be "translated" into a derivative.
Derivative of the tangent line: y "= 2.
Now let's build both derivatives:
The tangents intersect at three points, so our answer is 3.
Answer: 3
Problem number 10. The figure shows the graph of the function f (x), and points -2, 1, 2, 3 are marked. At which of these points is the value of the derivative the smallest? Indicate this point in your answer.
The task is somewhat similar to the first: to find the value of the derivative, you need to build 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 straight line:
With 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 coefficient k 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 =>it is there that the least value of the derivative will be
Answer: 1
Task number 11. The line is tangent y = 3x + 9 to the graph of the function y = x³ + x² + 2x + 8. Find the abscissa of the touch point.
The straight line will be tangent to the graph when the graphs have a common point, like their derivatives. Let's equate the equations of the graphs and their derivatives:
Solving the second equation, we get 2 points. To check which one fits, 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 a square one for a sweet soul.
But what should you write down in response if you get two "normal" answers?
When substituting x (x) in 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
Right! So x = 1 and will be the answer
Answer: 1Task number 12. The straight line y = - 5x - 6 is tangent to the graph of the function ax² + 5x - 5. Find a.
Similarly, we equate functions and their derivatives:
Let's solve this system with respect to the variables a and x:
Answer: 25
The task with derivatives is considered one of the most difficult in the first part of the exam, however, with a small amount of attentiveness and understanding of the question, you will succeed, and you will increase the percentage of completion of this task!
Example 1
Reference: The following ways to denote a function are equivalent: In some tasks it is convenient to designate the function as "igrokom", and in some as "ff 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 a function at a 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:
In the event that you do not understand how the derivative was found, return to the first two lessons of the topic. If you have difficulty (misunderstanding) with the arctangent and its meanings, necessarily study the teaching material Graphs and properties of elementary functions- the most recent paragraph. Because there are still enough arctangents for the student age.
Example 4
Calculate the derivative of a function at a point.
Equation of a tangent to a graph of a function
To consolidate the previous section, 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.
Let's consider a "demo" simplest example.
Write the equation of 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 strict definition of a tangent is given by definition of the derivative of a function, but for now we will master the technical part of the question. 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 one 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 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 assignment 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 to denote a function are equivalent:
In some tasks it is convenient to designate the function as "igrokom", and in some as "ff from x".
First, we find the derivative:
I hope many have already got used to finding 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
Complete solution and answer at the end of the tutorial.
The need to find the derivative at a point arises in the following problems: construction of a tangent to the graph of a function (next paragraph), extremum function study , inflection of a graph function , full function study and etc.
But the task in question occurs in control works 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 the point.
First, let's find the derivative: 
The derivative, in principle, has been found, and the required value can be substituted. But I don't really want to do it. 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 bring the last three terms to a common denominator: 
at the point.
This is an example for a do-it-yourself solution.
How to find the value of the derivative of the function F (x) at the point Xo? How to solve this in general?
If the formula is given, then find the derivative and substitute X-zero instead of X. Calculate
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 with the X-axis (using the mental construction of a right-angled 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 a minus, and if it is higher, then +.
Secondly, you need to know what tanges is in a rectangular rectangle. And this is the ratio of the opposite side (leg) to the adjacent side (also leg). There are usually some black marks on the painting. From these marks you make a right-angled triangle and find tanges.
How to find the value of the derivative of the function f x at the point x0?
no specific question posed - 3 years agoIn general, in order to find the value of the derivative of any function with respect to some variable at any point, you need 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.
This 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, note!) At the specified point X0. For this, a tangent line to a given function is constructed and the point of its intersection with the coordinate axes is found. Then the equation of this tangent line 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 = 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.
