Derivative of a function specified implicitly.
Derivative of a parametrically defined function

In this article we will look at two more typical tasks that are often found in tests in higher mathematics. In order to successfully master the material, you must be able to find derivatives at least at an intermediate level. You can learn to find derivatives practically from scratch in two basic lessons and Derivative of a complex function. If your differentiation skills are okay, then let's go.

Derivative of a function specified implicitly

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

Single variable function is a rule according to which 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 looked at functions defined in explicit form. What does it mean? Let's conduct a debriefing using specific examples.

Consider the function

We see that on the left we have a lone “player”, and on the right - only "X's". That is, the function explicitly expressed through the independent variable.

Let's look at another function:

This is where the variables are mixed up. Moreover impossible by any means express “Y” only through “X”. What are these methods? Transferring terms from part to part with a change of sign, moving them out of brackets, throwing factors according to the rule of proportion, etc. Rewrite the equality and try to express the “y” explicitly: . You can twist and turn the equation for hours, but you won’t succeed.

Let me introduce you: – example implicit function.

In the course of mathematical analysis it was proven that the implicit function exists(however, not always), it has a graph (just like a “normal” function). The implicit function is exactly the same exists first derivative, second derivative, etc. As they say, all rights of sexual minorities are respected.

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

Yes, and I’ll tell you the good news - the tasks discussed below are performed according to a fairly strict and clear algorithm without a stone in front of three tracks.

Example 1

1) At the first stage, we attach strokes to both parts:

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

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

- just to the point of 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 there is only one letter “y” - IS ITSELF A FUNCTION(see definition at the beginning of the lesson). Thus, sine is an external function and is an internal function. We use the rule for differentiating a complex function :

We differentiate the product according to the usual rule :

Please note that – is also a complex function, any “game with bells and whistles” is a complex function:

The solution itself should look something like this:

If there are brackets, then expand them:

4) On the left side we collect the terms that contain a “Y” with a prime. Move everything else to the right side:

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 using the algorithm just discussed. In fact, the phrases “implicit function” and “implicit function” differ in one semantic nuance. The phrase “implicitly specified function” is more general and correct, – this function is specified implicitly, but here you can express the “game” and present the function explicitly. The phrase “implicit function” refers to the “classical” implicit function when the “y” cannot be expressed.

Second solution

Attention! You can familiarize yourself with the second method only if you know how to confidently find partial derivatives. Calculus beginners and dummies, please don't read and skip this point, otherwise your head will be a complete mess.

Let's find the derivative of the implicit function using the second method.

We move all the terms to the left side:

And consider a function of two variables:

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

Thus:

The second solution allows you to perform a check. But it is not advisable for them to write out the final version of the assignment, since partial derivatives are mastered later, and a student studying the topic “Derivative of a function of one variable” should not yet know partial derivatives.

Let's look at a few more examples.

Example 2

Find the derivative of a function given implicitly

Add strokes to both parts:

We use linearity rules:

Finding derivatives:

Opening all the brackets:

We move 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 sample design at the end of the lesson.

It is not uncommon for fractions to arise after differentiation. In such cases, you need to get rid of fractions. Let's look at two more examples.

Example 4

Find the derivative of a function given implicitly

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

Differentiate using the rule for differentiating a complex function and the rule of differentiation of quotients :

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 contains . Multiply on . In detail, it will look like this:

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

On the left side we put it out of brackets:

Final answer:

Example 5

Find the derivative of a function given implicitly

This is an example for you to solve on your own. The only thing is that before you get 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

Let’s not stress, everything in this paragraph is also quite simple. You can write down the general formula for a parametrically defined function, but 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 brackets, 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 in human terms: “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 a “regular” function, for the American Indians of a parametrically defined function, all rights are also respected: you can build a graph, find derivatives, etc. By the way, if you need to plot a graph of a parametrically defined function, you can use my program.

In the simplest cases, it is possible to represent the function explicitly. Let us 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, this trick does not work. But it doesn’t matter, because there is a formula for finding the derivative of a parametric function:

We find the derivative of the “game with respect to the variable te”:

All differentiation rules and the table of derivatives are valid, naturally, 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 all that remains is 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 it in the formula, one could simply write it without a subscript, since this is a “regular” derivative “with respect to X”. But in literature there is always an option, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

Thus:

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

Example 7

Find the derivative of a function specified parametrically

This is an example for you to solve on your own.

In the article The simplest typical problems with derivatives we looked at examples in which we needed to find the second derivative of a function. For a parametrically defined function, you can also find the second derivative, and it is found using 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 of a function given parametrically

First, let's find the first derivative.
We use the formula

In this case:

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

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 in human terms: “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 a “regular” function, for the American Indians of a parametrically defined function, all rights are also respected: you can build a graph, find derivatives, etc. By the way, if you need to plot a graph of a parametrically specified function, download my geometric program on the page Mathematical formulas and tables.

In more “severe” cases, this trick does not work. But it doesn’t matter, because there is a formula for finding the derivative of a parametric function:

We find the derivative of the “game with respect to the variable te”:

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

Now all that remains is to substitute the found derivatives into our formula:

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

Example 6

We use the formula

In this case:

Thus:

Example 7

Find the derivative of a function specified parametrically

This is an example for you to solve on your own.

Example 8

Find the first and second derivatives of a function given parametrically

First, let's find the first derivative.
We use the formula

In this case:

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

I noticed that in the problem of finding the derivative of a parametric function, quite often for the purpose of simplification it is necessary to use trigonometric formulas . Remember them or keep them handy, and don't miss the opportunity to simplify each intermediate result and answers. For what? 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 reinforce the material, I offer a couple more examples for you to solve on your own.

Example 9

Example 10

Find and for a function specified parametrically

I wish you success!

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

Solutions and answers:

Example 3: Solution:

Thus:

The function can be specified in several ways. It depends on the rule that is used to specify it. The explicit form of specifying the function is y = f (x). There are times when its description is impossible or inconvenient. If there are many 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 .

Definition of a parametric function

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

There are cases when, to study a function, it is necessary to search for the derivative with respect to x. Let's consider the formula for the derivative of a parametrically defined 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 defined 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 begin with, you should move from a parametric task to an explicit one. To do this, you need to obtain 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 obtain that y " x = ψ Θ (x) = 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 differentiation rule.

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, from here we obtain that φ " (t) = t 2 + 1 ", ψ " (t) = t " = 1. You must 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 h, the parameter t specifies the expression of the argument x through the same parameter t, so as not to lose the connection between the values of the derivative and the parametrically defined 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 the transformation

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

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

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

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

y x "" = - t sin (2 t) φ " t = - 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 specifying the 2nd order derivative using a 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 using another method. Then

φ " t = (cos (2 t)) " = - sin (2 t) 2 t " = - 2 sin (2 t) ⇒ φ "" t = - 2 sin (2 t) " = - 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

From here 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 = = sin (2 t) - 2 t cos (2 t) 2 s i n 3 (2 t)

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

Higher order derivatives with parametrically defined functions are found in a similar way.

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

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

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 using the rules for differentiating 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 us introduce the notation:
.
Then the previous formula takes the form:
.

Let's take advantage of the fact that the function has an inverse function in the neighborhood of the point.
Let us introduce the following notation:
; ;
; .
Divide the numerator and denominator of the fraction by:
.
At , . Then
.

The rule has been proven.

Higher order derivatives

To find derivatives of higher orders, it is necessary to perform differentiation several times. Let's say we need to find the second-order derivative of a function defined parametrically, of the following form:
(1)

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

Let us denote the first derivative by the variable:
.
Then, to find the second derivative of a 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 through the functions and . To do this, let’s substitute and apply the derivative fraction formula:
.
Then
.

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

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

Continuing the process, you can obtain derivatives of functions from a variable of third and higher orders.

Note that we do not have to introduce a notation for the derivative. You can write it like this:
;
.

Example 1

Find the derivative of a function defined parametrically:

Solution

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

.
Here .

.
Here .

The required derivative:
.

Answer

Example 2

Find the derivative of the function expressed through the parameter:

Solution

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

Finding the derivative:

.

Finding 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 order derivatives of the function defined parametrically in Example 1:

Solution

In Example 1 we found the first order derivative:

Let us introduce the designation . Then the function is derivative with respect to . It is specified parametrically:

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

Let's differentiate by .
.
We found the derivative of in Example 1:
.
The second-order derivative with respect to is equal to the first-order derivative with respect to:
.

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

Now we find the third order derivative. Let us introduce the designation . Then we need to find the first-order derivative of the function, which is specified in a parametric way:

Find the derivative with respect to . To do this, we rewrite it in equivalent form:
.
From
.

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

Comment

You don’t have to enter the variables and , which are derivatives of and , respectively. Then you can write it like this:
;
;
;
;
;
;
;
;
.

Answer

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

Third order derivative.

Consider defining a line on a plane in which the variables x, y are functions of a third variable t (called a parameter):

For each value t certain values correspond to certain values from a certain interval x And y, a, therefore, a certain point M (x, y) of the plane. When t runs through all values from a given interval, then the point M (x, y) describes some line L. Equations (2.2) are called parametric line equations L.

If the function x = φ(t) has an inverse t = Ф(x), then substituting this expression into the equation y = g(t), we obtain y = g(Ф(x)), which specifies y as a function of x. In this case, we say that equations (2.2) define the function y parametrically.

Example 1. Let M(x,y)– arbitrary point on a circle of radius R and centered at the origin. Let t– angle between axis Ox and radius OM(see Fig. 2.3). Then x, y are expressed through t:

Equations (2.3) are parametric equations of a circle. Let us exclude the parameter t from equations (2.3). To do this, we square each equation and add it, we get: x 2 + y 2 = R 2 (cos 2 t + sin 2 t) or x 2 + y 2 = R 2 – the equation of a circle in the Cartesian coordinate system. It defines two functions: Each of these functions is given by parametric equations (2.3), but for the first function , and for the second .

Example 2. Parametric equations

define an ellipse with semi-axes a, b(Fig. 2.4). Excluding the parameter from the equations t, we obtain the canonical equation of the ellipse:

Example 3. A cycloid is a line described by a point lying on a circle if this circle rolls without sliding in a straight line (Fig. 2.5). Let us introduce the parametric equations of the cycloid. Let the radius of the rolling circle be a, point M, describing the cycloid, at the beginning of the movement coincided with the origin of coordinates.

Let's determine the coordinates x, y points M after the circle has rotated through an angle t
(Fig. 2.5), t = ÐMCB. Arc length M.B. equal to the length of the segment O.B. since the circle rolls without slipping, therefore

OB = at, AB = MD = asint, CD = acost, x = OB – AB = at – asint = a(t – sint),

y = AM = CB – CD = a – acost = a(1 – cost).

So, the parametric equations of the cycloid are obtained:

When changing a parameter t from 0 to 2π the circle rotates one revolution, and the point M describes one arc of a cycloid. Equations (2.5) give y as a function of x. Although the function x = a(t – sint) has an inverse function, but it is not expressed in terms of elementary functions, so the function y = f(x) is not expressed through elementary functions.

Let us consider the differentiation of a function defined parametrically by equations (2.2). The function x = φ(t) on a certain interval of change t has an inverse function t = Ф(x), Then y = g(Ф(x)). Let x = φ(t), y = g(t) have derivatives, and x"t≠0. According to the rule of differentiation of complex functions y"x=y"t×t"x. Based on the rule for differentiating the inverse function, therefore:

The resulting formula (2.6) allows one to find the derivative for a function specified parametrically.

Example 4. Let the function y, depending on x, is specified parametrically:

Solution. .
Example 5. Find the slope k tangent to the cycloid at the point M 0 corresponding to the value of the parameter.
Solution. From the cycloid equations: y" t = asint, x" t = a(1 – cost), That's why

Tangent slope at a point M0 equal to the value at t 0 = π/4:

DIFFERENTIAL FUNCTION

Let the function at the point x 0 has a derivative. By definition:
therefore, according to the properties of the limit (Section 1.8), where a– infinitesimal at Δx → 0. From here

Δy = f "(x0)Δx + α×Δx. (2.7)

As Δx → 0, the second term in equality (2.7) is an infinitesimal of higher order, compared to , therefore Δy and f " (x 0)×Δx are equivalent, infinitesimal (for f "(x 0) ≠ 0).

Thus, the increment of the function Δy consists of two terms, of which the first f "(x 0)×Δx is main part increment Δy, linear with respect to Δx (for f "(x 0)≠ 0).

Differential function f(x) at point x 0 is called the main part of the increment of the function and is denoted: dy or df(x0). Hence,

df (x0) =f "(x0)×Δx. (2.8)

Example 1. Find the differential of a function dy and the increment of the function Δy for the function y = x 2 at:
1) arbitrary x and Δ x; 2) x 0 = 20, Δx = 0.1.

Solution

1) Δy = (x + Δx) 2 – x 2 = x 2 + 2xΔx + (Δx) 2 – x 2 = 2xΔx + (Δx) 2, dy = 2xΔx.

2) If x 0 = 20, Δx = 0.1, then Δy = 40×0.1 + (0.1) 2 = 4.01; dy = 40×0.1= 4.

Let us write equality (2.7) in the form:

Δy = dy + a×Δx. (2.9)

Increment Δy is different from differential dy to an infinitesimal of higher order, compared to Δx, therefore, in approximate calculations, the approximate equality Δy ≈ dy is used if Δx is small enough.

Considering that Δy = f(x 0 + Δx) – f(x 0), we obtain an approximate formula:

f(x 0 + Δx) ≈ f(x 0) + dy. (2.10)

Example 2. Calculate approximately.

Solution. Consider:

Using formula (2.10), we obtain:

So, ≈ 2.025.

Let us consider the geometric meaning of the differential df(x 0)(Fig. 2.6).

Let us draw a tangent to the graph of the function y = f(x) at the point M 0 (x0, f(x 0)), let φ be the angle between the tangent KM0 and the Ox axis, then f"(x 0) = tanφ. From ΔM0NP:
PN = tgφ×Δx = f "(x 0)×Δx = df(x 0). But PN is the increment of the tangent ordinate as x changes from x 0 to x 0 + Δx.

Consequently, the differential of the function f(x) at the point x 0 is equal to the increment of the ordinate of the tangent.

Let's find the differential of the function
y = x. Since (x)" = 1, then dx = 1×Δx = Δx. We will assume that the differential of the independent variable x is equal to its increment, i.e. dx = Δx.

If x is an arbitrary number, then from equality (2.8) we obtain df(x) = f "(x)dx, whence .
Thus, the derivative for a function y = f(x) is equal to the ratio of its differential to the differential of the argument.

Let's consider the properties of the differential of a function.

If u(x), v(x) are differentiable functions, then the following formulas are valid:

To prove these formulas, derivative formulas for the sum, product and quotient of a function are used. Let us prove, for example, formula (2.12):

d(u×v) = (u×v)"Δx = (u×v" + u"×v)Δx = u×v"Δx + u"Δx×v = u×dv + v×du.

Let's consider the differential of a complex function: y = f(x), x = φ(t), i.e. y = f(φ(t)).

Then dy = y" t dt, but y" t = y" x ×x" t, so dy =y" x x" t dt. Considering,

that x" t = dx, we get dy = y" x dx =f "(x)dx.

Thus, the differential of a complex function y = f(x), where x =φ(t), has the form dy = f "(x)dx, the same as in the case when x is an independent variable. This property is called invariance of the form of the differential A.