The derivative of the cotangent is equal. Derivatives of trigonometric functions: tangent, sine, cosine and others. Derivative formulas for trigonometric functions

The derivative with respect to the variable x from the cotangent of x is equal to 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 cotangent, we will use the following mathematical facts:
1) Expressing the cotangent in terms of cosine and sine:
(1) ;
2) The value of the cosine derivative:
(2) ;
3) The value of the sine derivative:
(3) ;
4) The quotient derivative formula:
(4) ;
5) Trigonometric formula:
(5) .

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

.

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

The derivative fraction formula (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 cosine and sine are defined for all values ​​of the variable x, the formula for the derivative of cotangent is valid for all x, except for points at which the sine is equal to zero. That is, except for the points
where is an integer. The function itself y = ctg x
.
defined for all x except points That's why.

the derivative of the cotangent is defined over the entire domain of definition of the cotangent function

Higher order derivatives The function itself y = A simple formula for the nth order derivative of the cotangent y = , No. But the calculation of higher order derivatives can be simplified. The process itself can be reduced to.

differentiating a polynomial
.
To do this, we express the derivative of the cotangent through the cotangent itself:
(6) .

So we found: Let's find the derivatives of the left and right sides of equation (6) and apply the rule for differentiating a complex function. We get:
.
second order derivative
(7) .

Let's substitute (6): Let's 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 view nth order derivative
.
, in the variable x of the cotangent function, , can be represented as a polynomial in powers of cotangent:
,
The coefficients of this polynomial are related by the recurrence relation:
; ;
.

Where

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

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

So, let's consider some arbitrary curve that is described by the abstract function y = f(x).

Let's imagine that the schedule 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 path above sea level.
Then it turns out that the ratio ∆x/∆y will characterize the complexity of the route on each segment of the route. Having learned this value, you can confidently say whether the ascent/descent is steep, whether climbing equipment will be needed, and whether tourists need certain physical training. But this indicator will be valid only for one small interval ∆x.

If the organizer of the trip takes the 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 construct 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 path and will objectively describe its changes at each interval of interest. It is very simple to verify this; the value ∆x/∆y is nothing more than a differential taken for a specific value of x and y. Let us apply differentiation not to specific coordinates, but to the function as a whole:

Derivative and trigonometric functions

Trigonometric functions are inextricably linked with derivatives. This can be understood 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 the increment along the OX axis, and f (x0 + ∆x) is the increment along the OY axis at a certain point L.

Let's 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 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 desire will also be infinite, and its limit value close to 0.

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

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

In this way, we can 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 when determining the derivative must be memorized.

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

Let's look at a comparative table with formulas for the derivatives of sinus, cosine, tangent and cotangent:

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

It is worth noting that the above formulas are clearly not enough to successfully solve typical USE tasks, 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.