From the course of geometry and mathematics, schoolchildren are used to the fact that the concept of a derivative is conveyed to them through the area of ​​\u200b\u200bthe 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).

Imagine that the graph is a map 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 difficult route on each segment of the path. Knowing this value, you can say with confidence whether the ascent / descent is steep, whether climbing equipment is needed and whether tourists need certain physical fitness. But this indicator will be valid only for one small interval ∆x.

If the organizer of the hike takes the values ​​for the start and end points of the trail, that is, ∆x - will be 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 path changes, in other words, determine the ratio ∆x/∆y for each “meter” of the route.

This graph will be a visual derivative for a particular path 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 but the 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) - the 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 line through the 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 graph - 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 aspiration can be described by a tangent to the selected point y = kx + b or by the graph of the derivative of the original function dy - the green straight line.

But where is the trigonometry here ?! It's very simple to consider a right triangle KLN. The value of the differential for a particular point K is the tangent of the angle α or ∠K:

Thus, it is possible to describe the geometric meaning of the derivative and its relationship with trigonometric functions.

Derivative formulas for trigonometric functions

The transformations of sine, cosine, tangent and cotangent in determining the derivative must be memorized.

The last two formulas are not an error, the fact 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 sinis, cosine, tangent and cotangent:

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

It should be noted that the above formulas are clearly not enough for a successful solution. typical tasks USE, 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:

Decision: To find y’, you need to remember the basic formulas for converting the original function into a derivative, viz.

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The derivative with respect to x of the cotangent of x is minus one divided by the sine squared of x:
(ctgx)′ =.

Derivation of the formula for the derivative of the cotangent

To derive the formula for the derivative of the cotangent, we will use the following mathematical facts:
1) The expression of the cotangent in terms of cosine and sine:
(1) ;
2) The value of the cosine derivative:
(2) ;
3) The value of the derivative of the sine:
(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 there are derivatives of the functions and and for which the denominator of the fraction does not vanish:
.
In our case
, . Since the derivatives of the cosine and sine are defined for all values ​​of the variable x, then the cotangent derivative formula is valid for all x, except for points where the sine is zero. That is, apart from the dots
,
where is an integer.
The function itself y = ctg x defined for all x except points
.
So the derivative of the cotangent is defined on the entire domain of the definition of the cotangent function.

Derivatives of higher orders

A simple formula, for the nth derivative of the cotangent y = ctg x, No. But the calculation of higher order derivatives can be simplified. The process can be reduced to polynomial differentiation.

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

Let us find the derivatives of the left and right parts 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. To do 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 fourth and fifth order derivatives:

;

.

In general, derivative of the nth order, 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 recursive relation:
,
where
; ;
.

General formula

Let us represent the process of differentiation by one formula. For this, we note that
.
Then the nth derivative of the cotangent has the following form:
,
where .