Rules for differentiating implicitly specified functions. Differentiating complex and implicit functions of several variables
We will learn to find derivatives of functions specified implicitly, that is, specified by certain equations connecting variables x And y. Examples of functions specified implicitly:
,
Derivatives of functions specified implicitly, or derivatives of implicit functions, are found quite simply. Now let’s look at the corresponding rule and example, and then find out why this is needed at all.
In order to find the derivative of a function specified implicitly, you need to differentiate both sides of the equation with respect to x. Those terms in which only X is present will turn into the usual derivative of the function from X. And the terms with the game must be differentiated using the rule for differentiating a complex function, since the game is a function of X. To put it quite simply, the resulting derivative of the term with x should result in: the derivative of the function from the y multiplied by the derivative from the y. For example, the derivative of a term will be written as , the derivative of a term will be written as . Next, from all this, you need to express this “game stroke” and the desired derivative of the function specified implicitly will be obtained. Let's look at this with an example.
Example 1.
Solution. We differentiate both sides of the equation with respect to x, assuming that i is a function of x:
From here we get the derivative that is required in the task:
Now something about the ambiguous property of functions specified implicitly, and why special rules for their differentiation are needed. In some cases, you can make sure that substituting its expression in terms of x into a given equation (see examples above) instead of y leads to the fact that this equation turns into an identity. So. The above equation implicitly defines the following functions:
After substituting the expression for the squared game through x into the original equation, we obtain the identity:
.
The expressions that we substituted were obtained by solving the equation for the game.
If we were to differentiate the corresponding explicit function
then we would get the answer as in example 1 - from a function specified implicitly:
But not every function specified implicitly can be represented in the form y = f(x) . So, for example, the implicitly specified functions
are not expressed through elementary functions, that is, these equations cannot be resolved with respect to the game. Therefore, there is a rule for differentiating a function specified implicitly, which we have already studied and will further consistently apply in other examples.
Example 2. Find the derivative of a function given implicitly:
.
We express the prime and - at the output - the derivative of the function specified implicitly:
Example 3. Find the derivative of a function given implicitly:
.
Solution. We differentiate both sides of the equation with respect to x:
.
Example 4. Find the derivative of a function given implicitly:
.
Solution. We differentiate both sides of the equation with respect to x:
.
We express and obtain the derivative:
.
Example 5. Find the derivative of a function given implicitly:
Solution. We move the terms on the right side of the equation to the left side and leave zero on the right. We differentiate both sides of the equation with respect to x.
Higher order derivatives are found by successive differentiation of formula (1).
Example. Find and if (x ²+y ²)³-3(x ²+y ²)+1=0.
Solution. Denoting the left side given equation through f(x,y) find the partial derivatives
f"x(x,y)=3(x²+y²)²∙2x-3∙2x=6x[(x²+y²)-1],
f"y(x,y)=3(x²+y²)²∙2y-3∙2y=6y[(x²+y²)-1].
From here, applying formula (1), we obtain:
.
To find the second derivative, differentiate with respect to X the first derivative found, taking into account that at there is a function x:
.
2°. The case of several independent variables. Likewise, if the equation F(x, y, z)=0, Where F(x, y, z) - differentiable function of variables x, y And z, determines z as a function of independent variables X And at And Fz(x, y, z)≠ 0, then the partial derivatives of this implicitly given function, generally speaking, can be found using the formulas
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Another way to find the derivatives of the function z is as follows: by differentiating the equation F(x, y, z) = 0, we get:
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From here we can determine dz, and therefore .
Example. Find and if x ² - 2y²+3z² -yz +y =0.
1st method. Denoting the left side of this equation by F(x, y, z), let's find the partial derivatives F"x(x,y,z)=2x, F"y(x,y,z)=-4y-z+1, F"z(x,y,z)=6z-y.
Applying formulas (2), we obtain:
2nd method. Differentiating this equation, we get:
2xdx -4ydy +6zdz-ydz-zdy +dy =0
From here we determine dz, i.e. the total differential of the implicit function:
.
Comparing with formula , we see that
.
3°. Implicit Function System. If a system of two equations
defines u And v as functions of the variables x and y and the Jacobian
,
then the differentials of these functions (and therefore their partial derivatives) can be found from the system of equations
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Example: Equations u+v=x+y, xu+yv=1 determine u And v as functions X And at; find .
Solution. 1st method. Differentiating both equations with respect to x, we get:
.
In a similar way we find:
.
2nd method. By differentiation we find two equations connecting the differentials of all four variables: du +dv =dx +dy,xdu +udx +ydv+vdy =0.
Solving this system for differentials du And dv, we get:
4°. Parametric function specification. If the function of r variables X And at is given parametrically by the equations x=x(u,v), y=y(u,v), z=z(u,v) And
,
then the differential of this function can be found from the system of equations
Knowing the differential dz=p dx+q dy, we find the partial derivatives and .
Example. Function z arguments X And at given by equations x=u+v, y=u²+v², z=u²+v² (u≠v).
Find and .
Solution. 1st method. By differentiation we find three equations connecting the differentials of all five variables:
From the first two equations we determine du And dv:
.
Let's substitute the found values into the third equation du And dv:
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2nd method. From the third given equation we can find:
Let us differentiate the first two equations with respect to X, then by at:
From the first system we find: .
From the second system we find: .
Substituting expressions and into formula (5), we obtain:
Replacing variables
When replacing variables in differential expressions, the derivatives included in them should be expressed in terms of other derivatives according to the rules for differentiating a complex function.
1°. Replacing variables in expressions containing ordinary derivatives.
,
believing .
at By X through derivatives of at By t. We have:
,
.
Substituting the found derivative expressions into this equation and replacing X through , we get:
Example. Convert Equation
,
taking it as an argument at, and for the function x.
Solution. Let us express the derivatives of at By X through derivatives of X By u.
.
Substituting these derivative expressions into this equation, we will have:
,
or, finally,
.
Example. Convert Equation
going to polar coordinates
x=r cos φ, y=r cos φ. |
Solution. Considering r as a function φ , from formulas (1) we obtain:
dх = сosφ dr – r sinφ dφ, dy=sinφ+r cosφ dφ,
Let the continuous function at from X is specified implicitly F(x, y) = 0, where F(x, y), F" x(x, y), F "y(x, y) are continuous functions in some domain D containing the point ( X, at), whose coordinates satisfy the relations F (x, y) = 0, F "y(x, y) ≠ 0. Then the function at from X has a derivative
Proof (see picture.). Let F "y(x, y) > 0. Since the derivative F "y(x, y) is continuous, then we can construct a square [ X 0 - δ" , X 0 + δ" , at 0 - δ" , at 0 + δ" ], so that for all its points there is F "y (x, y) > 0, that is F(x, y) is monotone in at at fixed X. Thus, all conditions of the existence theorem for the implicit function are satisfied at = f (x), such that F(x, f (x)) º 0.
Let's set the increment Δ X. New meaning X + Δ X will correspond at + Δ at = f (x + Δ x), such that these values satisfy the equation F (x + Δ x, y + Δ y) = 0. It is obvious that
Δ F = F(x + Δ x, y + Δ y) − F(x, y) = 0
and in this case
.
From (7) we have
.
Since the implicit function at = f (x) will be continuous, then Δ at→ 0 at Δ X→ 0, which means α → 0 and β → 0. Whence we finally have
.
Q.E.D.
Partial derivatives and differentials of higher orders.
Let the partial derivatives of the function z = f (x, y), defined in a neighborhood of a point M, exist at every point in this neighborhood. In this case, partial derivatives are functions of two variables X And at, defined in the indicated neighborhood of the point M. Let us call them partial derivatives of the first order. In turn, partial derivatives with respect to variables X And at of functions at point M, if they exist, are called second-order partial derivatives of the function f (M) at this point and are indicated by the following symbols
Second-order partial derivatives of the form , , are called mixed partial derivatives.
Higher order differentials
We will consider dx in the expression for dy as a constant factor. Then the function dy represents an argument-only function x and its differential at the point x has the form (when considering the differential from dy we will use new notations for differentials):
δ ( d y) = δ [ f " (x) d x] = [f " (x) d x] " δ x = f "" (x) d(x) δ x .
Differential δ ( d y) from the differential dy at the point x, taken at δ x = dx, is called the second order differential of the function f (x) at point x and is designated d 2 y, i.e.
d 2 y = f ""(x)·( dx) 2 .
In turn, the differential δ( d 2 y) from the differential d 2 y, taken at δ x = dx, is called the third order differential of the function f(x) and is denoted d 3 y etc. Differential δ( d n-1 y) from differential d n -1 f, taken at δ x = dx, is called differential n- th order (or n- m differential) functions f(x) and is denoted d n y.
Let us prove that for n-th differential of the function the following formula is valid:
d n y = y (n) ·( dx)n, n = 1, 2, … (3.1)
In the proof we will use the method mathematical induction. For n= 1 and n= 2 formula (3.1) is proven. Let it be true for differentials of order n - 1
d n −1 y=y( n−1) ·( dx)n −1 ,
and function y (n-1) (x) is differentiable at some point x. Then
Assuming δ x = dx, we get
Q.E.D.
For anyone n equality is true
or
those. n- i is the derivative of the function y= f (x) at point x equal to the ratio n- th differential of this function at the point x To n- th degree of the differential of the argument.
Directional derivative of functions of several variables.
The function and the unit vector are considered. Direct l via t. M 0 with guide vector
Definition 1. Derivative of a function u = u(x, y, z) by variable t called derivative in the direction l
Since on this straight line u – complex function one variable, then the derivative with respect to t equal to the total derivative with respect to t(§ 12).
It is denoted and equal to
Derivative of a function specified implicitly.
Derivative of a parametrically defined function
In this article we will look at two more typical tasks, which are often found in tests By 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
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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, derivatives table elementary functions remain in effect. 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 words “implicit function” more often mean “classical” implicit function, when the “game” cannot be expressed.
It should also be noted that an “implicit equation” can implicitly specify two or even more functions at once, for example, the equation of a circle implicitly defines the functions , , that define semicircles. But, within the framework of this article, we will not make a special distinction between the terms and nuances, it was just information for general development.
Second solution
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 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
Complete solution and a 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 of a parametrically defined function, but, in order to make it clear, I will immediately write concrete 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.” On coordinate plane you can mark a point, 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:
Derivative of a complex function. Total derivative
Let z=ƒ(x;y) be a function of two variables x and y, each of which is a function of an independent variable t: x = x(t), y = y(t). In this case, the function z = f(x(t);y(t)) is a complex function of one independent variable t; the variables x and y are intermediate variables.
If z = ƒ(x;y) is a function differentiable at the point M(x;y) є D and x = x(t) and y = y(t) are differentiable functions of the independent variable t, then the derivative of the complex function z(t ) = f(x(t);y(t)) is calculated using the formula
Let's give the independent variable t an increment Δt. Then the functions x = = x(t) and y = y(t) will receive increments Δx and Δy, respectively. They, in turn, will cause the function z to increment Az.
Since by condition the function z - ƒ(x;y) is differentiable at the point M(x;y), its total increment can be represented as
where а→0, β→0 at Δх→0, Δу→0 (see paragraph 44.3). Let's divide the expression Δz by Δt and go to the limit at Δt→0. Then Δх→0 and Δу→0 due to the continuity of the functions x = x(t) and y = y(t) (according to the conditions of the theorem, they are differentiable). We get:
Special case: z=ƒ(x;y), where y=y(x), i.e. z=ƒ(x;y(x)) is a complex function of one independent variable x. This case reduces to the previous one, and the role of the variable t is played by x. According to formula (44.8) we have:
Formula (44.9) is called the total derivative formula.
General case: z=ƒ(x;y), where x=x(u;v), y=y(u;v). Then z= f(x(u;v);y(u;v)) is a complex function of the independent variables u and v. Its partial derivatives can be found using formula (44.8) as follows. Having fixed v, we replace it with the corresponding partial derivatives