Finding the slope of the tangent. How to find the slope of a tangent to the graph of a function

The topic “The angular coefficient of a tangent as the tangent of the angle of inclination” in the certification exam is given several tasks at once. Depending on their condition, the graduate may be required to provide either a full answer or a short answer. In preparation for passing the Unified State Exam in mathematics, the student should definitely repeat problems in which it is required to calculate slope tangent.

It will help you do this educational portal"Shkolkovo". Our experts prepared and presented theoretical and practical material as accessible as possible. Having become familiar with it, graduates with any level of training will be able to successfully solve problems related to derivatives in which it is necessary to find the tangent of the tangent angle.

Highlights

To find the correct and rational decision similar tasks in the Unified State Exam must be remembered basic definition: The derivative represents the rate of change of a function; it is equal to the tangent of the tangent angle drawn to the graph of the function at a certain point. It is equally important to complete the drawing. It will allow you to find the right decision Unified State Exam problems on the derivative, in which it is necessary to calculate the tangent of the tangent angle. For clarity, it is best to plot the graph on the OXY plane.

If you have already read base material on the topic of derivatives and are ready to begin solving problems on calculating the tangent of the tangent angle, similar Unified State Exam assignments, you can do this online. For each task, for example, problems on the topic “Relationship of a derivative with the speed and acceleration of a body,” we wrote down the correct answer and solution algorithm. At the same time, students can practice completing tasks various levels complexity. If necessary, the exercise can be saved in the “Favorites” section so that you can discuss the solution with the teacher later.

You are already familiar with the concept of a tangent to the graph of a function. The graph of the function f differentiable at the point x 0 near x 0 practically does not differ from the tangent segment, which means it is close to the secant segment l passing through the points (x 0 ; f (x 0)) and (x 0 +Δx; f ( x 0 + Δx)). Any of these secants passes through point A (x 0 ; f (x 0)) of the graph (Fig. 1). In order to uniquely define a straight line passing through this point A, it is enough to indicate its slope. The angular coefficient Δy/Δx of the secant as Δх→0 tends to the number f ‘(x 0) (we will take it as the angular coefficient of the tangent) They say that the tangent is the limiting position of the secant at Δх→0.

If f'(x 0) does not exist, then the tangent either does not exist (like the function y = |x| at the point (0; 0), see figure) or is vertical (like the graph of the function at the point (0 ; 0), Fig. 2).

So, the existence of a derivative of the function f at the point xo is equivalent to the existence of a (non-vertical) tangent at the point (x 0, f (x 0)) of the graph, while tangent slope is equal to f" (x 0). This is geometric meaning derivative

The tangent to the graph of a function f differentiable at the point xo is a straight line passing through the point (x 0 ; f (x 0)) and having an angular coefficient f ‘(x 0).

Let's draw tangents to the graph of the function f at points x 1, x 2, x 3 (Fig. 3) and note the angles they form with the abscissa axis. (This is the angle measured in the positive direction from the positive direction of the axis to the straight line.) We see that angle α 1 is acute, angle α 3 is obtuse, and angle α 2 equal to zero, since straight line l is parallel to the Ox axis. Tangent acute angle is positive, obtuse is negative, tg 0 = 0. Therefore

F"(x 1)>0, f’(x 2)=0, f’(x 3)
Constructing tangents at individual points allows you to more accurately sketch graphs. So, for example, to construct a sketch of a graph of the sine function, we first find that at points 0; π/2 and π derivative of sine is equal to 1; 0 and -1 respectively. Let's construct straight lines passing through the points (0; 0), (π/2,1) and (π, 0) with angular coefficients of 1, 0 and -1, respectively (Fig. 4). It remains to fit into the resulting trapezoid formed by these straight lines and straight line Ox, graph of the sine so that for x equal to 0, π/2 and π, it touches the corresponding straight lines.

Note that the graph of the sine in the vicinity of zero is practically indistinguishable from the straight line y = x. Let, for example, the scales along the axes be chosen so that a unit corresponds to a segment of 1 cm. We have sin 0.5 ≈ 0.479425, i.e. |sin 0.5 - 0.5| ≈ 0.02, and on the chosen scale this corresponds to a segment 0.2 mm long. Therefore, the graph of the function y = sin x in the interval (-0.5; 0.5) will deviate (in the vertical direction) from the straight line y = x by no more than 0.2 mm, which approximately corresponds to the thickness of the drawn line.

Learn to take derivatives of functions. The derivative characterizes the rate of change of a function at a certain point lying on the graph of this function. IN in this case The graph can be either a straight or curved line. That is, the derivative characterizes the rate of change of a function at a specific point in time. Remember general rules, by which derivatives are taken, and only then proceed to the next step.

• Read the article.
• How to take the simplest derivatives, for example, derivative exponential equation, described. The calculations presented in the following steps will be based on the methods described therein.

Learn to distinguish between problems in which the slope must be calculated through the derivative of a function. Problems do not always ask you to find the slope or derivative of a function. For example, you may be asked to find the rate of change of a function at point A(x,y). You may also be asked to find the slope of the tangent at point A(x,y). In both cases it is necessary to take the derivative of the function.

The straight line y = f(x) will be tangent to the graph shown in the figure at point x0 provided that it passes through this point with coordinates (x0; f(x0)) and has an angular coefficient f"(x0). Find this coefficient, taking into account the features of the tangent, is not difficult.

You will need

• - mathematical reference book;
• - notebook;
• - a simple pencil;
• - pen;
• - protractor;
• - compass.

Instructions

• Please note that the graph of the differentiable function f(x) at the point x0 is no different from the tangent segment. Therefore, it is quite close to the segment l, passing through the points (x0; f(x0)) and (x0+Δx; f(x0 + Δx)). To specify a straight line passing through point A with coefficients (x0; f(x0)), indicate its slope. Moreover, it is equal to Δy/Δx secant tangent (Δх→0), and also tends to the number f‘(x0).
• If there are no values ​​for f‘(x0), then perhaps there is no tangent, or perhaps it runs vertically. Based on this, the presence of the derivative of the function at the point x0 is explained by the existence of a non-vertical tangent, which is in contact with the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent is equal to f "(x0). The geometric meaning of the derivative becomes clear, that is, the calculation of the angular coefficient of the tangent.
• That is, in order to find the slope of the tangent, you need to find the value of the derivative of the function at the point of tangency. Example: find the angular coefficient of the tangent to the graph of the function y = x³ at the point with abscissa X0 = 1. Solution: Find the derivative of this function y΄(x) = 3x²; find the value of the derivative at the point X0 = 1. у΄(1) = 3 × 1² = 3. The angle coefficient of the tangent at the point X0 = 1 is 3.
• Draw additional tangents in the figure so that they touch the graph of the function at the following points: x1, x2 and x3. Mark the angles formed by these tangents with the abscissa axis (the angle is counted in the positive direction - from the axis to the tangent line). For example, the first angle α1 will be acute, the second (α2) will be obtuse, and the third (α3) will be equal to zero, since the tangent line drawn is parallel to the OX axis. In this case, tangent obtuse angle There is negative value, and the tangent of the acute angle is positive, with tg0 and the result is zero.

Let a function f be given, which at some point x 0 has a finite derivative f (x 0). Then the straight line passing through the point (x 0 ; f (x 0)), having an angular coefficient f ’(x 0), is called a tangent.

What happens if the derivative does not exist at the point x 0? There are two options:

1. There is no tangent to the graph either. Classic example- function y = |x | at point (0; 0).
2. The tangent becomes vertical. This is true, for example, for the function y = arcsin x at the point (1; π /2).

Tangent equation

Any non-vertical straight line is given by an equation of the form y = kx + b, where k is the slope. The tangent is no exception, and in order to create its equation at some point x 0, it is enough to know the value of the function and the derivative at this point.

So, let a function y = f (x) be given, which has a derivative y = f ’(x) on the segment. Then at any point x 0 ∈ (a ; b) a tangent can be drawn to the graph of this function, which is given by the equation:

y = f ’(x 0) (x − x 0) + f (x 0)

Here f ’(x 0) is the value of the derivative at point x 0, and f (x 0) is the value of the function itself.

Task. Given the function y = x 3 . Write an equation for the tangent to the graph of this function at the point x 0 = 2.

Tangent equation: y = f ’(x 0) · (x − x 0) + f (x 0). The point x 0 = 2 is given to us, but the values ​​f (x 0) and f ’(x 0) will have to be calculated.

First, let's find the value of the function. Everything is easy here: f (x 0) = f (2) = 2 3 = 8;
Now let’s find the derivative: f ’(x) = (x 3)’ = 3x 2;
We substitute x 0 = 2 into the derivative: f ’(x 0) = f ’(2) = 3 2 2 = 12;
In total we get: y = 12 · (x − 2) + 8 = 12x − 24 + 8 = 12x − 16.
This is the tangent equation.

Task. Write an equation for the tangent to the graph of the function f (x) = 2sin x + 5 at point x 0 = π /2.

This time we will not describe each action in detail - we will only indicate the key steps. We have:

f (x 0) = f (π /2) = 2sin (π /2) + 5 = 2 + 5 = 7;
f ’(x) = (2sin x + 5)’ = 2cos x;
f ’(x 0) = f ’(π /2) = 2cos (π /2) = 0;

Tangent equation:

y = 0 · (x − π /2) + 7 ⇒ y = 7

IN the latter case the straight line turned out to be horizontal, because its angular coefficient k = 0. There is nothing wrong with this - we just stumbled upon an extremum point.