From the course in geometry and mathematics, schoolchildren are accustomed to the concept of a derivative being conveyed to them through the area of ​​a figure, differentials, limits of functions, and also limits. Let's try to look at the concept of a derivative from a different angle, and determine how the derivative and trigonometric functions can be linked.

So, consider some arbitrary curve, which is described by an abstract function y = f (x).

Let's imagine that the graph is a map of a tourist route. The increment ∆x (delta x) in the figure is a certain distance of the path, and ∆y is the change in the height of the trail above sea level.
Then it turns out that the ratio ∆x / ∆y will characterize the complex route on each segment of the path. Having learned this value, it is possible to say with confidence whether the ascent / descent is steep, whether climbing equipment is needed and whether tourists need a certain physical training... But this indicator will be valid only for one small interval ∆x.

If the organizer of the trip takes values ​​for the starting and ending points of the trail, that is, ∆x is equal to the length of the route, then he will not be able to obtain objective data on the degree of difficulty of the trip. Therefore, it is necessary to build another graph that will characterize the speed and "quality" of changes in the path, in other words, determine the ratio ∆x / ∆y for each “meter” of the route.

This graph will be a visual derivative for a specific trail and will objectively describe its changes at each interval of interest. It is very easy to verify this, the value of ∆x / ∆y is nothing more than a differential taken for a specific value of x and y. Let us apply differentiation not to certain coordinates, but to the function as a whole:

Derivative and trigonometric functions

Trigonometric functions are inextricably linked with the derivative. You can understand this from the following drawing. The figure of the coordinate axis shows the function Y = f (x) - blue curve.

K (x0; f (x0)) is an arbitrary point, x0 + ∆x is an increment along the OX axis, and f (x0 + ∆x) is an increment along the OY axis at some point L.

Draw a straight line through points K and L and construct right triangle KLN. If you mentally move the segment LN along the graph Y = f (x), then the points L and N will tend to the values ​​K (x0; f (x0)). Let's call this point the conditional beginning of the chart - the limit, but if the function is infinite, at least on one of the intervals, this aspiration will also be infinite, and its limit value is close to 0.

The nature of this tendency can be described by the tangent to the selected point y = kx + b or by the graph of the derivative of the original function dy - the green line.

But where is trigonometry here ?! Everything is very simple, consider the right-angled triangle KLN. The value of the differential for a particular point K is the tangent of the angle α or ∠K:

Thus, you can describe the geometric meaning of the derivative and its relationship with trigonometric functions.

Derivative formulas for trigonometric functions

Sine, cosine, tangent and cotangent transformations when determining the derivative must be memorized.

The last two formulas are not an error, the point is that there is a difference between the definition of a derivative of a simple argument and a function in the same capacity.

Consider a comparative table with formulas for derivatives of sinus, cosine, tangent and cotangent:

Formulas for the derivatives of the arcsine, arccosine, arctangent and arccotangent are also derived, although they are rarely used:

It should be noted that the above formulas are clearly not enough for a successful solution typical assignments Unified State Exam, which will be demonstrated when solving a specific example of finding the derivative of a trigonometric expression.

Exercise: It is necessary to find the derivative of the function and find its value for π / 4:

Solution: To find y ’it is necessary to recall the basic formulas for converting the original function into a derivative, namely.

The x-derivative of the cotangent of x is minus one divided by the sine squared of x:
(ctg x) ′ =.

Derivation of the cotangent derivative formula

To derive the formula for the derivative of the cotangent, we will use the following mathematical facts:
1) Expression of the cotangent in terms of cosine and sine:
(1) ;
2) The value of the derivative of the cosine:
(2) ;
3) The value of the sine derivative:
(3) ;
4) The formula for the derivative of the quotient:
(4) ;
5) Trigonometric formula:
(5) .

We apply these formulas and rules to the derivative of the cotangent.

.

Thus, we have obtained the formula for the derivative of the cotangent.

The formula for the derivative of a fraction (4) is valid for those values ​​of the variable x for which the derivatives of the functions exist and for which the denominator of the fraction does not vanish:
.
In our case
,. Since the derivatives of cosine and sine are defined for all values ​​of the variable x, the formula for the derivative of the cotangent is valid for all x, except for the points at which the sine is equal to zero. That is, apart from the points
,
where is an integer.
The function y = ctg x defined for all x except points
.
That's why the derivative of the cotangent is defined over the entire domain of the cotangent function.

Higher order derivatives

A simple formula for the nth order derivative of the cotangent y = ctg x, No. However, the calculation of derivatives of higher orders can be simplified. You can reduce the process itself to differentiation of the polynomial.

To do this, we express the derivative of the cotangent through the cotangent itself:
.
So we found:
(6) .

Let us find the derivatives of the left and right sides of equation (6) and apply the rule of differentiation of a complex function. We get second order derivative:
.
Substitute (6):
(7) .

Find the third-order derivative... For this, we differentiate equation (7), apply the differentiation rule complex function and use expression (6) for the first derivative:
.

In a similar way we find derivatives of the fourth and fifth orders:

;

.

V general view, nth order derivative, in the variable x of the cotangent function,, can be represented as a polynomial in powers of the cotangent:
.
The coefficients of this polynomial are related by the recurrence relation:
,
where
; ;
.

General formula

Let's represent the process of differentiation by one formula. To do this, note that
.
Then the nth derivative of the cotangent has the following form:
,
where .

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