We are not tightened, in this paragraph, too, everything is quite simple. You can record the general formula parametrical specified functionbut, in order to be clear, I'll immediately write specific example. In parametric form, the function is defined by two equations :. Frequently, the equations are recorded not under curly brackets, but sequentially: ,.

The variable is called the parameter and can take values \u200b\u200bfrom the "minus infinity" to the "plus infinity". Consider, for example, the value and substitute it in both equations: . Or humanly: "If X is equal to four, then the game is equal to one." On the coordinate plane, you can mark the point, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the "TE" parameter. As for the "ordinary" function, for American IndianParmetrically specified function, all rights are also observed: you can build a schedule, find derivatives, etc. By the way, if there is a need to build a chart of a parametrically specified function, download my geometric prog on the page Mathematical formulas and tables.

In the simplest cases, it is possible to present a function explicitly. Express the parameter from the first equation: - and substitute it in the second equation: . As a result, an ordinary cubic function was obtained.

In more "severe" cases, such a focus does not roll. But this is not a misfortune, because there is a formula for finding a derivative of a parametric function:

Find a derivative of "Games by Variable TE":

All differentiation rules and a table of derivatives are fair, naturally, and for the letter, so some novelty in the process of finding derivatives. Just mentally replace all the "Iks" table in the table on the letter "TE".

Find a derivative of "IKSA for the TE variable":

Now only remained to substitute the found derivatives in our formula:

Ready. The derivative, as well as the function itself, also depends on the parameter.

As for the designations, in the formula, instead of recording, it was possible to simply record without a substrate index, since this is "ordinary" derivative "on X". But the literature always meets the option, so I will not deviate from the standard.

Example 6.

We use the formula

In this case:

In this way:

A feature of finding a derivative of parametric function is the fact that at every step, the result is beneficial to simplify. So, in the considered example, I revealed brackets under the root (although I could not do this). A great chance is that during substitution and in the formula, many things will be well reduced. Although there are, of course, examples and with porey responses.

Example 7.

Find a derivative from the function specified parametric

This is an example for self-decide.

In the article Simplest typical tasks With derivative We considered examples in which it was necessary to find the second derivative function. For a parametrically specified function, you can also find the second derivative, and it is located according to the following formula :. It is quite obvious that in order to find the second derivative, you must first find the first derivative.

Example 8.

Find the first and second derivatives from the function specified parametric

We will find the first derivative.
We use the formula

In this case:

Substitutes found derivatives in the formula. For simplification purposes, we use the trigonometric formula:

I noticed that in the task of finding a derivative of parametric function, it is quite often for the purpose of simplification. trigonometric formulas . Remember them or keep your hand, and do not skip the ability to simplify each intermediate result and answers. What for? Now we have to take a derivative from, and it is clearly better than finding a derivative from.

We find the second derivative.
We use the formula :.

Let's look at our formula. The denominator has already been found in the previous step. It remains to find a numerator - a derivative from the first derivative according to the "TE" variable:

It remains to take advantage of the formula:

To secure the material, I suggest a couple of examples for an independent solution.

Example 9.

Example 10.

Find and for a function specified parametric

I wish you success!

I hope this occupation was useful, and now you can easily find derivatives from functions defined implicitly and from parametric functions

Solutions and answers:

Example 3: Solution:






In this way:

Logarithmic differentiation

Derivatives elementary functions

Basic Differentiation Rules

Differential function

the main linear part Enrage function A.D. x. In determining the differentiability of the function

D. f \u003d F.(x.) - F.(x. 0)\u003d A.(x - X. 0)+ O.(x - X. 0), X®X. 0

called differential function f.(x.) at point x. 0 and denotes

df.(x. 0)\u003d F ¢(x. 0) D. x \u003d A.D. x.

Differential depends on the point x. 0 and from increment d x.On D. x.at the same time look like on an independent variable, so at each point, the differential is a linear function from increment D x.

If you consider as a function f.(x.)\u003d X., I get dX \u003d.D. x, dy \u003d adx. This is consistent with the labent designation.

Geometric interpretation of differential as the increments of ordinate tangent.

Fig. 4.3.

1) f \u003d.const. f ¢ \u003d0, df \u003d.0d. x \u003d.0.

2) f \u003d u + v, f \u003d u ¢ + v ¢, df \u003d du + dv.

3) f \u003d uv, f \u003d u ¢ v + v ¢ u, df \u003d u dv + v du.

Corollary. (cF.(x.))¢ \u003d CF ¢(x.), (c. 1 f. 1 (x.)+ ... + C n f n(x.))¢ \u003d C. 1 f ¢ 1 (x.)+ ... + C n f ¢ n(x.)

4) f \u003d u / v, v(x. 0) ¹0 and the derivative exists, then f ¢ \u003d.(u ¢ V-V¢ u.)/v. 2 .

For brevity we will denote u \u003d U.(x.), U. 0 \u003d U.(x. 0), then

Turning to the limit when D x®0 we obtain the required equality.

5) a derivative of a complex function.

Theorem. If there are f ¢(x. 0), G ¢(x. 0) and X. 0 \u003d G.(t. 0)then in some surroundings t 0 The complex function f is defined(g.(t.)), it is differentiated at the point T 0 and

Evidence.

f.(x.) - F.(x. 0)\u003d F ¢(x. 0)(x-X. 0)+ a ( x.)(x-X. 0), X.Î U.(x. 0).

f.(g.(t.))- F.(g.(t. 0))\u003d F ¢(x. 0)( G.(t.)- G.(t. 0))+ a ( G.(t.))( G.(t.)- G.(t. 0)).

We share both parts of this equality on ( t - T. 0) and move to the limit when T®T. 0 .

6) Calculation of the reverse function derivative.

Theorem. Let F be continuous and strictly monotonne on[a, B.]. Let at the point x 0 Î( a, B.) There is f ¢(x. 0) ¹ 0 , then reverse function x \u003d f -1 (y.) has at the point y 0 derivative, equal

Evidence. Consider f.strictly monotonically increasing, then f. -1 (y.) continuous, monotonously increases on [ f.(a.), F.(b.)]. Put y. 0 \u003d F.(x. 0), y \u003d f(x.), X - X 0 \u003d D. x

y - Y. 0 \u003d D. y.. Due to the continuity of the reverse function D y.®0 þ D. x.®0, have

Turning to the limit, we obtain the desired equality.

7) Derivative even function odd, derivative odd function even

Indeed, if x® - X. 0 , then - x® X. 0 , so

For even functions for odd function

1) f \u003d.const F ¢.(x.)=0.

2) f.(x.)\u003d x, f ¢(x.)=1.

3) f.(x.)\u003d E X., f ¢(x.)\u003d E X. ,

4) f.(x.)\u003d a x,(a X.)¢ \u003d a xlN. a.

5) lN. a.

6) f.(x.) \u003d ln. x

Corollary. (different function derivative odd)

7) (X. M. )¢= m. x. M -1. , X.>0, X. M. \u003d E. M. LN. X. .

8) (sin X.)¢= cos. x

9) (COS X.)¢=- sin. x(COS. X.)¢= (sin ( x +.p / 2)) ¢= cOS ( x +.p / 2) \u003d - sin x.

10) (TG X.)¢= 1 / COS 2 x.

11) (CTG X.)¢= -1 / SIN 2 x.

16) sh xch X..

f (x), from where it follows that f ¢(x.)\u003d F.(x.) (LN. F.(x.))¢ .

The same formula can be obtained differently f.(x.)\u003d E. LN. F.(x.) f ¢ \u003d e LN. F.(x.) (LN. F.(x.))¢.

Example. Calculate the derivative function f \u003d x x.

\u003d x x \u003d x x \u003d x x \u003d x x(LN. X +.1).

Geometric location points on the plane

we will call the schedule of the function, set parametro. They also speak about the parametric task of the function.

Note 1.If a X, Y.continuous on [a, B.] and X.(t.) strictly monotonne on the segment (for example, strictly monotonically increases), then on [ a, B.] , a \u003d x(a) , b \u003d x(b) function defined F.(x.)\u003d y.(t.(x.))where T.(x.) – return to X (T) function. The schedule of this function coincides with the schedule of the function

If the definition area parametrically specified function can be divided into a finite number of segments , k \u003d1,2,..., N,on each of which the function x.(t.) strictly monotonne, then a parametrically specified function disintegrates to a finite number of conventional functions. f K.(x.)\u003d y.(t. -1 (x.)) with definition areas [ x.(A. K.), X.(B. K.)] for plots of increasing x.(t.) and with the fields of determination [ x.(B. K.), X.(A. K.)] for descending areas of the function x.(t.). The functions thus obtained are called unambiguous branches of a parametrically specified function.

Figure shows a graph of a parametrically specified function.

With the selected parametrization, the definition area is divided into five sections of the strict monotony of the SIN function (2 t.), exactly: t.Î T.Î , T.Î , T.Î , and, accordingly, the schedule will split into five unambiguous branches corresponding to these sites.

Fig. 4.4.

Fig. 4.5.

You can choose another parametrization of the same geometric location

In this case, there will be only four such branches. They will correspond to strict monotony sites t.Î , T.Î T.Î , T.Î functions sIN (2. t.).

Fig. 4.6.

Four parts of the SIN function monotonicity (2 t.) On the length of the long.

Fig. 4.7.

The image of both graphs in one figure allows you to approximately depict a graph of a parametrically specified function using the monotony sections of both functions.

Consider for example the first branch corresponding to the segment t.Î . At the ends of this section function x \u003d.sIN (2. t.) takes values \u200b\u200b-1. and 1. , Therefore, this branch will be defined on [-1,1]. After that you need to look at the second function monotonicity y \u003d.cOS ( t.), she is on two plots of monotonality . This allows you to say that the first branch has two portions of monotony. Finding the terminal points of the graph can be combined with direct in order to designate the character of the chart monotony. Having done it with every branch, we obtain the plots of monotony of the unambiguous branches of the graph (in the figure they are highlighted in red)

Fig. 4.8.

First unambiguous branch f. 1 (x.)\u003d y.(t.(x.)) corresponding to the site will be defined for x.Î [-1,1] . First unambiguous branch t.Î , X.Î [-1,1].

All other three branches will have a definition area, too, set [-1,1] .

Fig. 4.9.

Second branch t.Î X.Î [-1,1].

Fig. 4.10.

Third branch t.Î X.Î [-1,1]

Fig. 4.11

Fourth branch t.Î X.Î [-1,1]

Fig. 4.12.

Comment2. The same function can have different parametric tasks. Differences may concern how the functions themselves X.(t.), y.(t.) , and the field of definition these functions.

An example of various parametric tasks of the same function

and t.Î [-1, 1] .

Note 3.If x, y are continuous on , X.(t) -strictly monotonne on the segment and there are derivatives y ¢(t. 0), x ¢(t. 0) ¹0, then there is f ¢(x. 0)= .

Indeed.

The latter statement applies to the unambiguous branches of the parametrically specified function.

4.2 Derivatives and Differentials of the highest order

Senior derivatives and differentials. Differentiation of functions specified parametrically. Formula Leibnitsa.

Until now, equations of lines on the plane connecting the current coordinates of these lines directly. However, another method of specifying a line is often used, in which the current coordinates are considered as the functions of the third variable value.

Let two functions of the variable

considered for the same values \u200b\u200bt. Then, any of these values \u200b\u200bT corresponds to a certain value and a certain value of y, and therefore a certain point. When the variable T runs out all values \u200b\u200bfrom the function of determining the functions (73), the point describes some line C in the plane of equation (73) is called parametric equations of this line, and the variable is a parameter.

Suppose that the function has a reverse function substituting this function into second from equations (73), we obtain the equation

expressing as a function

We agree to say that this function is set by parametric equations (73). The transition from these equations to equation (74) is called the exception of the parameter. When considering functions specified parametrically, the exception of the parameter is not only not necessary, but not always almost possible.

In many cases, it is much more convenient, setting various values The parameter is then calculated by the formulas (73) corresponding values \u200b\u200bof the argument and functions.

Consider examples.

Example 1. Let the arbitrary point of the circle with the center at the beginning of the coordinates and the radius of R. Cartesian coordinates x and in this point are expressed through its polar radius and the polar angle, which we denote here by t, as follows (see ch. I, § 3, p. 3):

Equations (75) are called parametric circle equations. The parameter in them is the polar angle, which varies in the range from 0 to.

If equations (75) are raised to the square and folded, then, by virtue of identity, the parameter will be excluded and the circumference equation in the Cartesian coordinate system determines the two elementary functions:

Each of these functions is set by parametric equations (75), but the parameter change area for these functions is different. For the first of them; The graph of this function is the upper semicircle. For the second function, it is a lower part-friendly schedule.

Example 2. Consider simultaneously ellipse

and a circle with a center at the beginning of the coordinates and radius A (Fig. 138).

Each point of the ellipse is comparable to the point N of a circle having the same abscissa as the point M, and located with it one way from the axis oh. The position of the point N, and consequently, the points M is completely determined by the polar angle of T points at the same time for their general abscissa we obtain the following expression: x \u003d a. Operation at the point M we find from the Ellipse equation:

The sign is selected because the ordinate at the point M and the ordinate point N should have the same signs.

Thus, the following parametric equations were obtained for an ellipse:

Here the parameter T varies from 0 to.

Example 3. Consider a circle with a center at a point a) and a radius A, which obviously concerns the abscissa axis at the beginning of the coordinates (Fig. 139). Suppose it is this circle rolls without sliding along the abscissa axis. Then the point M of the circle, which coincided at the initial moment with the beginning of the coordinates, describes a line called cycloid.

We will withdraw parametric cycloid equations, adopting an Angle angle of MSV of the circumference when it moves its fixed point from the position of the M. Then for the coordinates and at the point M we will receive the following expressions:

Due to the fact that the circle is rolling along the axis without slip, the length of the segment of the OS is equal to the length of the arc VM. Since the length of the arc VM is equal to the product of the radius A on the central angle T, then. Therefore . But therefore

These equations are parametric cycloid equations. When the parameter is changed from 0 to the circumference, one full revolution will make it. Point M at the same time will describe one arch of cycloids.

The exclusion of parameter T leads here to bulky expressions and is almost inexpedient.

Parametric lines task is particularly often used in mechanics, and the role of the parameter plays time.

Example 4. We define the trajectory of the projectile, released from the gun with the initial speed at an angle and to the horizon. The resistance of the air and the dimensions of the projectile, considering it with a material point, neglect.

Select the coordinate system. For the beginning of the coordinates, we take a point of departure of the projectile from the blow. The axis Oh will send horizontally, and the OU axis is vertically, placing them in the same plane with the blowing point. If there was no earth's strength, then the projectile would move in a straight line, component of the angle and with the axis Oh and at the time of the time t would have passed the path of the projectile coordinate at time t would be equal to :. Due to the earth's pulp, the projectile must at this point vertically sink by magnitude, therefore, in reality, at the time T, the coordinates of the projectile are determined by the formulas:

In these equations - permanent values. When the T change, the coordinates will also be changed at the point of the projector's trajectory. Equations are parametric Equims of the projector's trajectory in which the time is the time

Expressing from the first equation and substituting it in

the second equation, we obtain the equation of the projectory of the projectile in the form of this - the equation of parabola.

The derivative function specified implicitly.
Derivative of parametrically specified function

In this article, we will consider two more typical tasks that are often found in test work by higher Mathematics. In order to successfully master the material, it is necessary to be able to find derivatives at least at the average level. Learn how to find derivatives with virtually zero can be on two basic lessons and Derivative complex function. If everything is in order with the differentiation skills, then drove.

The derivative function specified implicitly

Or shorter - derivative implicit function. What is an implicit function? Let's first remember the determination of the function of one variable:

The function of one variable - It is a rule by which each value of an independent variable corresponds to one and only one function value.

The variable is called independent variable or argument.
The variable is called dependent variable or function .

So far, we have considered the functions specified in applied form. What does it mean? We arrange the parsing of flights on specific examples.

Consider a function

We see that on the left we have a lonely "cheek", but on the right - only "Ikers". That is, a function explicitly expressed through an independent variable.

Consider another feature:

Here are variables and arranged "intention". Moreover no way impossible Express "Ix" only through "X". What kind of ways is it? The transfer of the components from the part to the part with the change of the sign, to the brackets, transferring multipliers to the rule of proportion and others. Rewrite the equality and try to express "Igarek" explicitly :. You can twist-thrust the hour equation, but you will not succeed.

Allow me to introduce: - Example implicit function.

In the course of mathematical analysis it is proved that an implicit function exists (However, not always), she has a schedule (just like the "normal" function). The implicit function in the same way exists The first derivative, the second derivative, etc. As they say, all the rights of sex minorities are observed.

And in this lesson, we will learn to find a derivative of the function specified implicitly. It's not so difficult! All differentiation rules, the table of derivative elementary functions remain in force. The difference in one kind of moment, which we consider right now.

Yes, and inform good news - The tasks discussed below are performed on a pretty hard and clear algorithm without stone in front of three tracks.

Example 1.

1) At the first stage, hanging strokes on both parts:

2) We use the linearity rules derivative (the first two rules of the lesson How to find a derivative? Examples of solutions):

3) direct differentiation.
How to differentiate and completely understandable. What to do where under strokes is "igraki"?

- just before disgrace the derivative of the function is equal to its derivative: .

How to differentiate
Here we have complex function. Why? It seems to be under the sinus only one letter "Igarek". But, the fact is that only one letter "Igarek" - In itself is a function (See definition at the beginning of the lesson). Thus, the sinus is an external function - an internal function. Use the differentiation rule of a complex function :

Work differentiating by the usual rule :

Please note that - also a complex function, any "Cheerk with Founds" - a complex function:

The decision itself should look something like this:


If there are brackets, then reveal them:

4) In the left side we collect the terms, in which there is "igrek" with a touche. IN right part - Transfer everything else:

5) In the left part, we carry out the derivative of the brackets:

6) And according to the rule of proportion, we discard these brackets to the denominator of the right part:

The derivative was found. Ready.

It is interesting to note that in an implicit form you can rewrite any function. For example, a function You can rewrite: . And differentiate it according to the algorithm just discussed. In fact, the phrase "function specified in implicit form" and "implicit function" are distinguished by one semantic nuance. Phrase "The function specified in implicit form" is more general and correct, - This function is set in an implicit form, but here you can express "igrek" and present the function explicitly. Under the phrase, the "implicit function" understand the "classic" implicit function, when "igrek" cannot be expressed.

The second way of solving

Attention! With the second way you can read only if you can confidently find private derivatives. Beginners to learn mathematical analysis and teapots, please do not read and skip this item.Otherwise there will be a full porridge in my head.

Find a derivative of an implicit function in the second way.

We carry all the components to the left:

And we consider the function of two variables:

Then our derivative can be found by the formula
We find private derivatives:

In this way:

The second solution method allows you to check. But it is undesirable to make it a finishing version of the task, since the private derivatives are mastered later, and the student learns the topic "derived function of one variable", to know the private derivatives as it should not yet.

Consider a few more examples.

Example 2.

Find a derivative from the function specified implicitly

Turn the touches on both parts:

We use linearity rules:

Find derivatives:

Reveal all brackets:

We transfer all the components with the left part, the rest - the right side:

Final answer:

Example 3.

Find a derivative from the function specified implicitly

Complete solution and sample design at the end of the lesson.

Not uncommon, when fractions arise after differentiation. In such cases, fractions need to get rid. Consider two more examples.

Example 4.

Find a derivative from the function specified implicitly

We conclude both parts for the touches and use the linearity rule:

Differentiating using a complex differentiation rule and the differentiation rule of private :

Reveal brackets:

Now we need to get rid of the fraraty. This can be done later, but more rational to do immediately. In denominator, the fraci is located. Multiply on the . If in detail then it will look like this:

Sometimes 2-3 fractions appear after differentiation. If we had another fraction, for example, the operation would need to repeat - multiply each part of each part on the

In the left part, we endure the bracket:

Final answer:

Example 5.

Find a derivative from the function specified implicitly

This is an example for an independent solution. The only thing, in it, before getting rid of the fraction, will first need to get rid of the three-storey of the fraci itself. Complete solution and answer at the end of the lesson.

Derivative of parametrically specified function

We are not tightened, in this paragraph, too, everything is quite simple. You can record the general formula of a parametrically specified function, but, in order to be clear, I will immediately write a specific example. In parametric form, the function is defined by two equations :. Frequently, the equations are recorded not under curly brackets, but sequentially: ,.

The variable is called Parameter And it can take values \u200b\u200bfrom "minus infinity" to "plus infinity." Consider, for example, the value and substitute it in both equations: . Or humanly: "If X is equal to four, then the game is equal to one." On the coordinate plane, you can mark the point, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the "TE" parameter. As for the "ordinary" function, for American Indians of a parametrically specified function, all rights are also observed: you can build a schedule, find derivatives, etc. By the way, if there is a need to build a chart of a parametrically specified function, you can use my program.

In the simplest cases, it is possible to present a function explicitly. Express the parameter from the first equation: - and substitute it in the second equation: . As a result, an ordinary cubic function was obtained.

In more "severe" cases, such a focus does not roll. But this is not a misfortune, because there is a formula for finding a derivative of a parametric function:

Find a derivative of "Games by Variable TE":

All differentiation rules and a table of derivatives are fair, naturally, and for the letter, so some novelty in the process of finding derivatives. Just mentally replace all the "Iks" table in the table on the letter "TE".

Find a derivative of "IKSA for the TE variable":

Now only remained to substitute the found derivatives in our formula:

Ready. The derivative, as well as the function itself, also depends on the parameter.

As for the designations, in the formula, instead of recording, it was possible to simply record without a substrate index, since this is "ordinary" derivative "on X". But the literature always meets the option, so I will not deviate from the standard.

Example 6.

We use the formula

In this case:

In this way:

A feature of finding a derivative of parametric function is the fact that at every step, the result is beneficial to simplify. So, in the considered example, I revealed brackets under the root (although I could not do this). A great chance is that during substitution and in the formula, many things will be well reduced. Although there are, of course, examples and with porey responses.

Example 7.

Find a derivative from the function specified parametric

This is an example for an independent solution.

In the article The simplest typical tasks with derivative We considered examples in which it was necessary to find the second derivative function. For a parametrically specified function, you can also find the second derivative, and it is located according to the following formula :. It is quite obvious that in order to find the second derivative, you must first find the first derivative.

Example 8.

Find the first and second derivatives from the function specified parametric

We will find the first derivative.
We use the formula

In this case:

We substitute the derivatives found in the formula. For simplification purposes, we use the trigonometric formula:

The function can be set in several ways. It depends on the rule that is used in its task. An explicit type of function setting has the form y \u003d f (x). There are cases when its description is impossible or uncomfortable. If there are many pairs (x; y), which must be calculated for the parameter T at the interval (A; b). To solve the system x \u003d 3 · cos T y \u003d 3 · SIN T C 0 ≤ T< 2 π необходимо задавать окружность с центром координат с радиусом равным 3 .

Definition of parametric function

From here we have that x \u003d φ (t), y \u003d ψ (t) are defined by the value t ∈ (a; b) and have a reverse function t \u003d θ (x) for x \u003d φ (t), then this is speech On the task of the parametric equation of the function y \u003d ψ (θ (x)).

There are cases when the function of the function is required to search for a derivative by x. Consider the formula of the derivative of a parametrically specified function of the form y x "\u003d ψ" (t) φ "(t), let's talk about the derivative 2 and n-order.

Output of the formula for the derivative of the parametrically specified function

We have that x \u003d φ (t), y \u003d ψ (t), defined and differentiated with the value t ∈ A; b, where x t "\u003d φ" (t) ≠ 0 and x \u003d φ (t), then there is a reverse function of the form T \u003d θ (x).

To begin with, you should move from the parametric task to the apparent one. To do this, it is necessary to obtain a complex function of the form y \u003d (t) \u003d ψ (θ (x)), where there is an argument x.

Based on the rule of finding a derivative of a complex function, we obtain that y "x \u003d ψ θ (x) \u003d ψ" θ x · θ "x.

It can be seen that T \u003d θ (x) and x \u003d φ (t) are inverse functions from the feed function θ "(x) \u003d 1 φ" (t), then y "x \u003d ψ" θ (x) · θ "(x) \u003d ψ" (t) φ "(t).

Let us turn to the consideration of the solution of several examples using the derivatives table according to the range of differentiation.

Example 1.

Find a derivative for the function X \u003d T 2 + 1 Y \u003d T.

Decision

By condition, we have that φ (t) \u003d t 2 + 1, ψ (t) \u003d t, we obtain that φ "(t) \u003d t 2 + 1", ψ "(t) \u003d t" \u003d 1. It is necessary to use the derived formula and write down the answer in the form:

y "x \u003d ψ" (t) φ "(t) \u003d 1 2 t

Answer: y x "\u003d 1 2 t x \u003d t 2 + 1.

When working with the derivative of the function h, the T parameter indicates the expression of the X argument through the same parameter T so as not to lose the connection between the values \u200b\u200bof the derivative and parametrically specified function with the argument to which these values \u200b\u200bcorrespond.

To determine the derivative of the second order of the parametrically specified function, you need to use the formula of the first-order derivative on the result received, then we get that

y "" x \u003d ψ "(t) φ" (t) "φ" (t) \u003d ψ "" (t) · φ "(t) - ψ" (t) · φ "" (t) φ "( t) 2 φ "(t) \u003d ψ" "" (t) · φ "(t) - ψ" (t) · φ "" (t) φ "(t) 3.

Example 2.

Find the derivatives 2 and 2 of the order of the specified function X \u003d COS (2 T) Y \u003d T 2.

Decision

By condition, we obtain that φ (t) \u003d cos (2 t), ψ (t) \u003d t 2.

Then after the conversion

φ "(t) \u003d cos (2 t)" \u003d - sin (2 t) · 2 t "\u003d - 2 sin (2 t) ψ (t) \u003d t 2" \u003d 2 t

It follows that y x "\u003d ψ" (t) φ "(t) \u003d 2 t - 2 sin 2 t \u003d - t sin (2 t).

We obtain that the type of derivative of 1 of the order x \u003d cos (2 t) y x "\u003d - t sin (2 t).

To solve, you need to apply the formula for the second order derivative. We get the expression of the view

yx "" \u003d - t sin (2 t) φ "t \u003d - t" · sin (2 t) - t · (sin (2 t)) "sin 2 (2 t) - 2 sin (2 t) \u003d \u003d 1 · SIN (2 T) - T · COS (2 T) · (2 \u200b\u200bT) "2 SIN 3 (2 T) \u003d SIN (2 T) - 2 T COS (2 T) 2 SIN 3 (2 T)

Then the task of a 2-order derivative with a parametric function

x \u003d cos (2 t) y x "" \u003d sin (2 t) - 2 t cos (2 t) 2 sin 3 (2 t)

A similar solution is possible to solve another method. Then

φ "T \u003d (cos (2 t))" \u003d - sin (2 t) · 2 t "\u003d - 2 sin (2 t) ⇒ φ" "t \u003d - 2 sin (2 t)" \u003d - 2 · sin (2 t) "\u003d - 2 cos (2 t) · (2 \u200b\u200bt)" \u003d - 4 cos (2 t) ψ "(t) \u003d (t 2)" \u003d 2 t ⇒ ψ "" "(t) \u003d ( 2 t) "\u003d 2

From here we get that

y "" x \u003d ψ "" (t) · φ "(t) - ψ" (t) · φ "" "(t) φ" (t) 3 \u003d 2 · - 2 sin (2 t) - 2 t · (- 4 cos (2 t)) - 2 sin 2 t 3 \u003d sin (2 t) - 2 t · cos (2 t) 2 sin 3 (2 t)

Answer: y "" x \u003d sin (2 t) - 2 t · cos (2 t) 2 s i n 3 (2 t)

Similarly, there is a delicate of derivatives of higher orders with parametrically specified functions.

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