Calculation of the derivative given parametrically online. Functions defined parametrically

Do not strain, in this paragraph, too, everything is quite simple. One can write the general formula parametrically given function, but, in order to make it clear, I will immediately write down specific example. In parametric form, the function is given by two equations: . Often, equations are written not under curly braces, but sequentially:,.

A variable is called a parameter and can take values ​​from "minus infinity" to "plus infinity". Consider, for example, the value and substitute it into both equations: . Or humanly: "if x is equal to four, then y is equal to one." You can mark a point on the coordinate plane, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the parameter "te". As for the "ordinary" function, for the American Indians of a parametrically given function, all rights are also respected: you can plot a graph, find derivatives, and so on. By the way, if there is a need to build a graph of a parametrically given function, download my geometric program on the page Mathematical Formulas and tables.

In the simplest cases, it is possible to represent the function explicitly. We express the parameter from the first equation: and substitute it into the second equation: . The result is an ordinary cubic function.

In more "severe" cases, such a trick does not work. But this does not matter, because there is a formula to find the derivative of a parametric function:

We find the derivative of "the player with respect to the variable te":

All the rules of differentiation and the table of derivatives are valid, of course, for the letter , thus, there is no novelty in the process of finding derivatives. Just mentally replace all the "x"s in the table with the letter "te".

We find the derivative of "x with respect to the variable te":

Now it only remains to substitute the found derivatives into our formula:

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

As for the notation, instead of writing in the formula, one could simply write it without a subscript, since this is the “ordinary” derivative “by x”. But there is always a variant in the literature, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

In this way:

A feature of finding the derivative of a parametric function is the fact that at each step, it is advantageous to simplify the result as much as possible. So, in the considered example, when finding, I opened the brackets under the root (although I might not have done this). There is a great chance that when substituting and into the formula, many things will be well reduced. Although there are, of course, examples with clumsy answers.

Example 7

Find the derivative of a function given parametrically

This is an example for independent decision.

In the article Protozoa typical tasks with derivative we considered examples in which it was required to find the second derivative of a function. For a parametrically given function, you can also find the second derivative, and it is found by the following formula: . It is quite obvious that in order to find the second derivative, one must first find the first derivative.

Example 8

Find the first and second derivatives of a function given parametrically

Let's find the first derivative first.
We use the formula

In this case:

Substitutes the found derivatives into the formula. For the sake of simplicity, we use the trigonometric formula:

I noticed that in the problem of finding the derivative of a parametric function, quite often, in order to simplify, one has to use trigonometric formulas . Remember them or keep them handy, and don't miss the opportunity to simplify each intermediate result and answers. What for? Now we have to take the derivative of , and this is clearly better than finding the derivative of .

Let's 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 the numerator - the derivative of the first derivative with respect to the variable "te":

It remains to use the formula:

To consolidate the material, I offer a couple more examples for an independent solution.

Example 9

Example 10

Find and for a function defined parametrically

Wish you success!

I hope this lesson was useful, and now you can easily find derivatives of implicit functions and parametric functions

Solutions and answers:

Example 3: Solution:






In this way:

The derivative of a function given implicitly.
Derivative of a parametrically defined function

In this article, we will consider two more typical tasks that are often found in control work in higher mathematics. In order to successfully master the material, it is necessary to be able to find derivatives at least at an average level. You can learn how to find derivatives practically from scratch in two basic lessons and Derivative of a compound function. If everything is in order with differentiation skills, then let's go.

Derivative of a function defined implicitly

Or, in short, the derivative of an implicit function. What is an implicit function? Let's first recall the very definition of a function of one variable:

Function of one variable is the rule that each value of the independent variable corresponds to one and only one value of the function.

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

So far, we have considered functions defined in explicit form. What does it mean? Let's arrange a debriefing on specific examples.

Consider the function

We see that on the left we have a lone “y”, and on the right - only x's. That is, the function explicitly expressed in terms of the independent variable .

Let's consider another function:

Here the variables and are located "mixed". And impossible in any way express "Y" only through "X". What are these methods? Transferring terms from part to part with a change of sign, bracketing, throwing factors according to the rule of proportion, etc. Rewrite the equality and try to express “y” explicitly:. You can twist and turn the equation for hours, but you will not succeed.

Allow me to introduce: - an example implicit function.

In the course of mathematical analysis, it was proved that the implicit function exists(but not always), it has a graph (just like a "normal" function). It's the same for an implicit function. exists first derivative, second derivative, etc. As they say, all the rights of sexual minorities are respected.

And in this lesson we will learn how to find the derivative of a function given implicitly. It's not that hard! All differentiation rules, the table of derivatives of elementary functions remain in force. The difference is in one peculiar point, which we will consider right now.

Yes, I'll let you know good news- the tasks discussed below are performed according to a rather rigid and clear algorithm without a stone in front of three tracks.

Example 1

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

2) We use the rules of linearity of the derivative (the first two rules of the lesson How to find the derivative? Solution examples):

3) Direct differentiation.
How to differentiate and completely understandable. What to do where there are “games” under the strokes?

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

How to differentiate
Here we have complex function. Why? It seems that under the sine there is only one letter "Y". But, the fact is that only one letter "y" - IS A FUNCTION IN ITSELF(see the definition at the beginning of the lesson). So the sine is outer function, – internal function. We use the differentiation rule complex function :

The product is differentiable according to the usual rule :

Note that is also a complex function, any “twist toy” is a complex function:

The design of the solution itself should look something like this:


If there are brackets, then open them:

4) On the left side, we collect the terms in which there is a “y” with a stroke. On the right side - we transfer everything else:

5) On the left side, we take the derivative out of brackets:

6) And according to the rule of proportion, we drop these brackets into the denominator of the right side:

The derivative has been found. Ready.

It is interesting to note that any function can be rewritten implicitly. For example, the function can be rewritten like this: . And differentiate it according to the algorithm just considered. In fact, the phrases "implicit function" and "implicit function" differ in one semantic nuance. The phrase "implicitly defined function" is more general and correct, - this function is given implicitly, but here you can express "y" and present the function explicitly. The phrase "implicit function" means a "classical" implicit function, when "y" cannot be expressed.

The second way to solve

Attention! You can familiarize yourself with the second method only if you know how to confidently find partial derivatives. Beginners to study mathematical analysis and teapots please do not read and skip this paragraph, otherwise the head will be a complete mess.

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

We move all the terms to the left side:

And consider a function of two variables:

Then our derivative can be found by the formula
Let's find partial derivatives:

In this way:

The second solution allows you to perform a check. But it is undesirable to draw up a final version of the task for him, since partial derivatives are mastered later, and a student studying the topic “Derivative of a function of one variable” should not know partial derivatives.

Let's look at a few more examples.

Example 2

Find the derivative of a function given implicitly

We hang strokes on both parts:

We use the rules of linearity:

Finding derivatives:

Expanding all parentheses:

We transfer all the terms with to the left side, the rest - to the right side:

Final answer:

Example 3

Find the derivative of a function given implicitly

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

It is not uncommon for fractions to appear after differentiation. In such cases, fractions must be discarded. Let's look at two more examples.

Example 4

Find the derivative of a function given implicitly

We conclude both parts under strokes and use the linearity rule:

We differentiate using the rule of differentiation of a complex function and the rule of differentiation of the quotient :

Expanding the brackets:

Now we need to get rid of the fraction. This can be done later, but it is more rational to do it right away. The denominator of the fraction is . Multiply on the . In detail, it will look like this:

Sometimes after differentiation, 2-3 fractions appear. If we had one more fraction, for example, then the operation would have to be repeated - multiply each term of each part on the

On the left side, we put it out of brackets:

Final answer:

Example 5

Find the derivative of a function given implicitly

This is a do-it-yourself example. The only thing in it, before getting rid of the fraction, you will first need to get rid of the three-story structure of the fraction itself. Full solution and answer at the end of the lesson.

Derivative of a parametrically defined function

Do not strain, in this paragraph, too, everything is quite simple. You can write down the general formula of a parametrically given function, but, in order to make it clear, I will immediately write down a specific example. In parametric form, the function is given by two equations: . Often, equations are written not under curly braces, but sequentially:,.

The variable is called a parameter and can take values ​​from "minus infinity" to "plus infinity". Consider, for example, the value and substitute it into both equations: . Or humanly: "if x is equal to four, then y is equal to one." You can mark a point on the coordinate plane, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the parameter "te". As for the "ordinary" function, for the American Indians of a parametrically given function, all rights are also respected: you can plot a graph, find derivatives, and so on. By the way, if there is a need to build a graph of a parametrically given function, you can use my program.

In the simplest cases, it is possible to represent the function explicitly. We express the parameter from the first equation: and substitute it into the second equation: . The result is an ordinary cubic function.

In more "severe" cases, such a trick does not work. But this does not matter, because there is a formula to find the derivative of a parametric function:

We find the derivative of "the player with respect to the variable te":

All the rules of differentiation and the table of derivatives are valid, of course, for the letter , thus, there is no novelty in the process of finding derivatives. Just mentally replace all the "x"s in the table with the letter "te".

We find the derivative of "x with respect to the variable te":

Now it only remains to substitute the found derivatives into our formula:

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

As for the notation, instead of writing in the formula, one could simply write it without a subscript, since this is the “ordinary” derivative “by x”. But there is always a variant in the literature, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

In this way:

A feature of finding the derivative of a parametric function is the fact that at each step, it is advantageous to simplify the result as much as possible. So, in the considered example, when finding, I opened the brackets under the root (although I might not have done this). There is a great chance that when substituting and into the formula, many things will be well reduced. Although there are, of course, examples with clumsy answers.

Example 7

Find the derivative of a function given parametrically

This is a do-it-yourself example.

In the article The simplest typical problems with a derivative we considered examples in which it was required to find the second derivative of a function. For a parametrically given function, you can also find the second derivative, and it is found by the following formula: . It is quite obvious that in order to find the second derivative, one must first find the first derivative.

Example 8

Find the first and second derivatives of a function given parametrically

Let's find the first derivative first.
We use the formula

In this case:

We substitute the found derivatives into the formula. For the sake of simplicity, we use the trigonometric formula:

The function can be defined in several ways. It depends on the rule that is used when setting it. The explicit form of the function definition is y = f (x) . There are cases when its description is impossible or inconvenient. If there is a set of pairs (x; y) that need to be calculated for the parameter t over the interval (a; b). To solve the system x = 3 cos t y = 3 sin t with 0 ≤ t< 2 π необходимо задавать окружность с центром координат с радиусом равным 3 .

Parametric function definition

Hence we have that x = φ (t) , y = ψ (t) are defined on for the value t ∈ (a ; b) and have an inverse function t = Θ (x) for x = φ (t) , then we are talking about setting a parametric equation of a function of the form y = ψ (Θ (x)) .

There are cases when, in order to study a function, it is required to search for the derivative with respect to x. Consider the formula for the derivative of a parametrically given function of the form y x " = ψ " (t) φ " (t) , let's talk about the derivative of the 2nd and nth order.

Derivation of the formula for the derivative of a parametrically given function

We have that x = φ (t) , y = ψ (t) , defined and differentiable for t ∈ a ; b , where x t " = φ " (t) ≠ 0 and x = φ (t) , then there is an inverse function of the form t = Θ (x) .

To get started, move from parametric task to the obvious. To do this, you need to get a complex function of the form y = ψ (t) = ψ (Θ (x)) , where there is an argument x .

Based on the rule for finding the derivative of a complex function, we get that y "x \u003d ψ Θ (x) \u003d ψ " Θ x Θ" x.

This shows that t = Θ (x) and x = φ (t) are inverse functions from the inverse function formula Θ "(x) = 1 φ" (t) , then y "x = ψ" Θ (x) Θ " (x) = ψ " (t) φ " (t) .

Let's move on to consider solving several examples using a table of derivatives according to the rule of differentiation.

Example 1

Find the derivative for the function x = t 2 + 1 y = t .

Solution

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

y "x = ψ" (t) φ "(t) = 1 2 t

Answer: y x " = 1 2 t x = t 2 + 1 .

When working with the derivative of a function, the parameter t specifies the expression of the argument x through the same parameter t in order not to lose the connection between the values ​​of the derivative and the parametrically specified function with the argument to which these values ​​correspond.

To determine the second-order derivative of a parametrically given function, you need to use the formula for the first-order derivative on the resulting function, then we get that

y""x = ψ"(t)φ"(t)"φ"(t) = ψ""(t) φ"(t) - ψ"(t) φ""(t)φ"( t) 2 φ "(t) = ψ "" (t) φ "(t) - ψ "(t) φ "" (t) φ "(t) 3 .

Example 2

Find the 2nd and 2nd order derivatives of the given function x = cos (2 t) y = t 2 .

Solution

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

Then after transformation

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 "= ψ" (t) φ "(t) = 2 t - 2 sin 2 t = - t sin (2 t) .

We get that the form of the derivative of the 1st order is x = cos (2 t) y x " = - t sin (2 t) .

To solve it, you need to apply the second-order derivative formula. We get an expression like

y x "" \u003d - t sin (2 t) φ "t \u003d - t " sin (2 t) - t (sin (2 t)) " sin 2 (2 t) - 2 sin (2 t) = = 1 sin (2 t) - t cos (2 t) (2 t) " 2 sin 3 (2 t) = sin (2 t) - 2 t cos (2 t) 2 sin 3 (2 t)

Then setting the 2nd order derivative using the parametric function

x = cos (2 t) y x "" = sin (2 t) - 2 t cos (2 t) 2 sin 3 (2 t)

A similar solution can be solved by 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) "= - 2 cos (2 t) (2 t)" = - 4 cos (2 t) ψ "(t) = (t 2)" = 2 t ⇒ ψ "" (t) = ( 2 t) " = 2

Hence we get that

y "" x = ψ "" (t) φ " (t) - ψ " (t) φ "" (t) φ " (t) 3 = 2 - 2 sin (2 t) - 2 t (- 4 cos (2 t)) - 2 sin 2 t 3 \u003d \u003d sin (2 t) - 2 t cos (2 t) 2 s i n 3 (2 t)

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

Similarly, higher order derivatives with parametrically specified functions are found.

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Let the function be given in a parametric way:
(1)
where is some variable called parameter. And let the functions and have derivatives at some value of the variable . Moreover, the function also has an inverse function in some neighborhood of the point . Then the function (1) has a derivative at the point, which, in a parametric form, is determined by the formulas:
(2)

Here and are derivatives of the functions and with respect to the variable (parameter) . They are often written in the following form:
;
.

Then system (2) can be written as follows:

Proof

By condition, the function has an inverse function. Let's denote it as
.
Then the original function can be represented as a complex function:
.
Let's find its derivative by applying the rules of differentiation of complex and inverse functions:
.

The rule has been proven.

Proof in the second way

Let's find the derivative in the second way, based on the definition of the derivative of the function at the point :
.
Let's introduce the notation:
.
Then the previous formula takes the form:
.

Let us use the fact that the function has an inverse function , in the vicinity of the point .
Let us introduce the notation:
; ;
; .
Divide the numerator and denominator of the fraction by:
.
At , . Then
.

The rule has been proven.

Derivatives of higher orders

To find derivatives of higher orders, it is necessary to perform differentiation several times. Suppose we need to find the second derivative of a function given in a parametric way, of the following form:
(1)

According to formula (2), we find the first derivative, which is also determined parametrically:
(2)

Denote the first derivative by means of a variable:
.
Then, to find the second derivative of the function with respect to the variable , you need to find the first derivative of the function with respect to the variable . The dependence of a variable on a variable is also specified in a parametric way:
(3)
Comparing (3) with formulas (1) and (2), we find:

Now let's express the result in terms of the functions and . To do this, we substitute and apply the formula for the derivative of a fraction:
.
Then
.

From here we obtain the second derivative of the function with respect to the variable:

It is also given in a parametric form. Note that the first line can also be written as follows:
.

Continuing the process, it is possible to obtain derivatives of functions from a variable of the third and higher orders.

Note that it is possible not to introduce the notation for the derivative . It can be written like this:
;
.

Example 1

Find the derivative of a function given in a parametric way:

Solution

We find derivatives of and with respect to .
From the table of derivatives we find:
;
.
We apply:

.
Here .

.
Here .

Desired derivative:
.

Answer

Example 2

Find the derivative of the function expressed through the parameter:

Solution

Let's open the brackets using formulas for power functions and roots:
.

We find the derivative:

.

We find the derivative. To do this, we introduce a variable and apply the formula for the derivative of a complex function.

.

We find the desired derivative:
.

Answer

Example 3

Find the second and third derivatives of the function given parametrically in example 1:

Solution

In example 1, we found the first order derivative:

Let's introduce the notation . Then the function is the derivative with respect to . It is set parametrically:

To find the second derivative with respect to , we need to find the first derivative with respect to .

We differentiate with respect to .
.
We found the derivative by in example 1:
.
The second order derivative with respect to is equal to the first order derivative with respect to:
.

So, we have found the second-order derivative with respect to the parametric form:

Now we find the derivative of the third order. Let's introduce the notation . Then we need to find the first derivative of the function , which is given in a parametric way:

We find the derivative with respect to . To do this, we rewrite in an equivalent form:
.
From
.

The third order derivative with respect to is equal to the first order derivative with respect to:
.

Comment

It is possible not to introduce variables and , which are derivatives of and , respectively. Then you can write it like this:
;
;
;
;
;
;
;
;
.

Answer

In the parametric representation, the second order derivative has the following form:

Derivative of the third order.