24.1. The concept of a function differential

Let the function y=ƒ(x) have a non-zero derivative at the point x.

Then, according to the theorem on the connection of a function, its limit and an infinitely small function, we can write D y / D x \u003d ƒ "(x) + α, where α → 0 for ∆x → 0, or ∆y \u003d ƒ" (x) ∆х+α ∆х.

Thus, the increment of the function ∆у is the sum of two terms ƒ "(х) ∆х and a ∆х, which are infinitesimal at ∆x→0. In this case, the first term is an infinitely small function of the same order with ∆х, since and the second term is an infinitely small function of a higher order than ∆x:

Therefore, the first term ƒ "(x) ∆x is called the main part of the increment functions ∆у.

function differential y \u003d ƒ (x) at the point x is called the main part of its increment, equal to the product of the derivative of the function and the increment of the argument, and is denoted dу (or dƒ (x)):

dy \u003d ƒ "(x) ∆x. (24.1)

The differential dу is also called first order differential. Let us find the differential of the independent variable x, that is, the differential of the function y=x.

Since y"=x"=1, then, according to formula (24.1), we have dy=dx=∆x, i.e., the differential of the independent variable is equal to the increment of this variable: dx=∆x.

Therefore, formula (24.1) can be written as follows:

dy \u003d ƒ "(x) dx, (24.2)

in other words, the differential of a function is equal to the product of the derivative of this function and the differential of the independent variable.

From formula (24.2) the equality dy / dx \u003d ƒ "(x) follows. Now the designation

the derivative dy/dx can be viewed as the ratio of the differentials dy and dx.

<< Пример 24.1

Find the differential of the function ƒ(x)=3x 2 -sin(l+2x).

Solution: According to the formula dy \u003d ƒ "(x) dx we find

dy \u003d (3x 2 -sin (l + 2x)) "dx \u003d (6x-2cos (l + 2x)) dx.

<< Пример 24.2

Find the differential of a function

Calculate dy at x=0, dx=0.1.

Decision:

Substituting x=0 and dx=0.1, we get

24.2. The geometric meaning of the differential of a function

Let us find out the geometric meaning of the differential.

To do this, we draw the tangent MT to the graph of the function y \u003d ƒ (x) at the point M (x; y) and consider the ordinate of this tangent for the point x + ∆x (see Fig. 138). In Figure ½ AM½ =∆x, |AM 1 |=∆y. From the right triangle MAB we have:

But, according to the geometric meaning of the derivative, tga \u003d ƒ "(x). Therefore, AB \u003d ƒ" (x) ∆x.

Comparing the result obtained with formula (24.1), we obtain dy=AB, i.e., the differential of the function y=ƒ(x) at the point x is equal to the increment of the ordinate of the tangent to the graph of the function at this point, when x receives the increment ∆x.

This is the geometric meaning of the differential.

24.3 Fundamental differential theorems

The main theorems about differentials are easy to obtain using the relation between the differential and the derivative of the function (dy=f"(x)dx) and the corresponding theorems about derivatives.

For example, since the derivative of the function y \u003d c is equal to zero, then the differential of a constant value is equal to zero: dy \u003d c "dx \u003d 0 dx \u003d 0.

Theorem 24.1. The differential of the sum, product and quotient of two differentiable functions is defined by the following formulas:

Let us prove, for example, the second formula. By definition of the differential, we have:

d(uv)=(uv)" dx=(uv" +vu" )dx=vu" dx+uv" dx=udv+vdu

Theorem 24.2. The differential of a complex function is equal to the product of the derivative of this function with respect to an intermediate argument and the differential of this intermediate argument.

Let y=ƒ(u) and u=φ(x) be two differentiable functions forming a complex function y=ƒ(φ(x)). By the theorem on the derivative of a compound function, one can write

y" x = y" u u" x .

Multiplying both parts of this equality by dx, we learn y "x dx \u003d y" u u "x dx. But y" x dx \u003d dy and u "x dx \u003d du. Therefore, the last equality can be rewritten as follows:

dy=y" u du.

Comparing the formulas dy=y "x dx and dy=y" u du, we see that the first differential of the function y=ƒ(x) is determined by the same formula, regardless of whether its argument is an independent variable or is a function of another argument.

This property of the differential is called the invariance (invariance) of the form of the first differential.

The formula dy \u003d y "x dx in appearance coincides with the formula dy \u003d y" u du, but there is a fundamental difference between them: in the first formula x is an independent variable, therefore, dx \u003d ∆x, in the second formula and there is a function of x , so, generally speaking, du≠∆u.

With the help of the definition of the differential and the fundamental theorems on differentials, it is easy to transform a table of derivatives into a table of differentials.

For example: d(cosu)=(cosu)" u du=-sinudu

24.4. Differential table

24.5. Applying the Differential to Approximate Calculations

As already known, the increment ∆у of the function y=ƒ(х) at the point x can be represented as ∆у=ƒ"(х) ∆х+α ∆х, where α→0 as ∆х→0, or dy+α ∆x Discarding the infinitesimal α ∆x of a higher order than ∆x, we obtain the approximate equality

∆у≈dy, (24.3)

moreover, this equality is the more accurate, the smaller ∆x.

This equality allows us to calculate approximately the increment of any differentiable function with great accuracy.

The differential is usually found much easier than the increment of a function, so formula (24.3) is widely used in computational practice.

<< Пример 24.3

Find the approximate value of the increment of the function y \u003d x 3 -2x + 1 for x \u003d 2 and ∆x \u003d 0.001.

Solution: We apply the formula (24.3): ∆у≈dy=(х 3 -2х+1)" ∆х=(3х 2 -2) ∆х.

So, ∆у» 0.01.

Let's see what error was made by calculating the differential of the function instead of its increment. To do this, we find ∆у:

∆y \u003d ((x + ∆x) 3 -2 (x + ∆x) + 1) - (x 3 -2x + 1) \u003d x 3 + 3x 2 ∆x + 3x (∆x) 2 + (∆x ) 3 -2x-2 ∆x + 1-x 3 + 2x-1 \u003d ∆x (3x 2 + 3x ∆x + (∆x) 2 -2);

The absolute approximation error is equal to

|∆у-dy|=|0.010006-0.011=0.000006.

Substituting into equality (24.3) the values ​​∆у and dy, we obtain

ƒ(х+∆х)-ƒ(х)≈ƒ"(х)∆х

ƒ(х+∆х)≈ƒ(х)+ƒ"(х) ∆х. (24.4)

Formula (24.4) is used to calculate approximate values ​​of functions.

<< Пример 24.4

Calculate approximately arctg(1.05).

Solution: Consider the function ƒ(х)=arctgx. According to formula (24.4) we have:

arctg(x+∆х)≈arctgx+(arctgx)" ∆х,

i.e.

Since x+∆x=1.05, then for x=1 and ∆x=0.05 we get:

It can be shown that the absolute error of formula (24.4) does not exceed the value M (∆x) 2, where M is the largest value of |ƒ"(x)| on the segment [x;x+∆x].

<< Пример 24.5

What distance will the body travel in free fall on the Moon in 10.04 s from the beginning of the fall. Body free fall equation

H \u003d g l t 2 /2, g l \u003d 1.6 m / s 2.

Solution: It is required to find H(10,04). We use the approximate formula (ΔH≈dH)

H(t+∆t)≈H(t)+H"(t) ∆t. At t=10 s and ∆t=dt=0.04 s, H"(t)=g l t, we find

Task (for independent solution). A body of mass m=20 kg moves with a speed ν=10.02 m/s. Calculate approximately the kinetic energy of the body

24.6. Higher order differentials

Let y=ƒ(x) be a differentiable function, and its argument x be independent variable. Then its first differential dy=ƒ"(x)dx is also a function of x; one can find the differential of this function.

The differential from the differential of the function y=ƒ(x) is called her second differential(or a second-order differential) and is denoted d 2 y or d 2 ƒ(x).

So, by definition d 2 y=d(dy). Let us find the expression for the second differential of the function y=ƒ(x).

Since dx=∆x does not depend on x, we assume that dx is constant when differentiating:

d 2 y=d(dy)=d(f"(x)dx)=(ƒ"(x)dx)" dx=f"(x)dx dx=f"(x)(dx) 2 i.e. .

d 2 y \u003d ƒ "(x) dx 2. (24.5)

Here dx 2 stands for (dx) 2 .

The third-order differential is defined and found similarly

d 3 y \u003d d (d 2 y) \u003d d (ƒ "(x) dx 2) ≈ f" (x) (dx) 3.

And, in general, the differential of the nth order is the differential of the differential of the (n-1)th order: d n y=d(d n-l y)=f (n) (x)(dx) n .

Hence we find that, In particular, for n=1,2,3

respectively we get:

i.e., the derivative of a function can be viewed as the ratio of its differential of the corresponding order to the corresponding power of the differential of the independent variable.

Note that all the above formulas are valid only if x is an independent variable. If the function y \u003d ƒ (x), where x - function of some other independent variable, then the differentials of the second and higher orders do not have the form invariance property and are calculated using other formulas. Let us show this by the example of a second-order differential.

Using the product differential formula (d(uv)=vdu+udv), we get:

d 2 y \u003d d (f "(x) dx) \u003d d (ƒ "(x)) dx + ƒ" (x) d (dx) \u003d ƒ "(x) dx dx + ƒ" (x) d 2 x , i.e.

d 2 y \u003d ƒ "(x) dx 2 + ƒ" (x) d 2 x. (24.6)

Comparing formulas (24.5) and (24.6), we see that in the case of a complex function, the second-order differential formula changes: the second term appears ƒ "(x) d 2 x.

It is clear that if x is an independent variable, then

d 2 x=d(dx)=d(l dx)=dx d(l)=dx 0=0

and formula (24.6) goes over into formula (24.5).

<< Пример 24.6

Find d 2 y if y=e 3x and x is the independent variable.

Solution: Since y"=3e 3x, y"=9e 3x, then by formula (24.5) we have d 2 y=9e 3x dx 2 .

<< Пример 24.7

Find d 2 y if y=x 2 and x=t 3 +1 and t is the independent variable.

Solution: We use formula (24.6): since

y"=2x, y"=2, dx=3t 2 dt, d 2 x=6tdt 2,

then d 2 y=2dx 2 +2x 6tdt 2 =2(3t 2 dt) 2 +2(t 3 +1)6tdt 2 =18t 4 dt 2 +12t 4 dt 2 +12tdt 2 =(30t 4 +12t)dt 2

Another solution: y=x 2 , x=t 3 +1. Therefore, y \u003d (t 3 +1) 2. Then by formula (24.5)

d 2 y=y ¢¢ dt 2 ,

d 2 y=(30t 4 +12t)dt 2 .

Higher order differentials.

Let the function y= ¦(x) be defined in some interval X (for example, an interval) and have derivatives of all orders at each interior point. Then its differential is dy=y 1 dx. We will call it the first-order differential.

At each particular point, the differential of the function is a number. On the interval it is a function of x. Therefore, we can speak of a differential from the first differential.

Definition: The differential of the first-order differential of the function y \u003d ¦ (x) is called the second-order differential of this function and is symbolically written d (dy) \u003d d 2 y.

Generally: the differential of the n-th order of the function y \u003d ¦ (x) is called the differential of the differential (n-1) of the order of the function d n y \u003d d (d n-1 y).

The designations d¦(x) , d 2 ¦(x) , d n ¦(x) are also applicable

Differentials of order higher than the first are called differentials of higher orders.

When calculating differentials of higher orders, it must be taken into account that dx is an arbitrary number that does not depend on x, and when differentiating with respect to x, it must be considered a constant factor.

Therefore, dy \u003d y 1 dx, d 2 y \u003d d (dy) \u003d d (y 1 dx) \u003d dx d (y 1) \u003d dx (y 11 dx) \u003d y 11 (dx) 2. It is customary to write the degree of the differential without brackets (dx) 2 = dx 2.

Thus, d 2 y \u003d y ''dx 2, but this should not be confused with d (x 2) \u003d 2xdx

Similarly: d 3 y \u003d d (y 11 dx 2) \u003d dx 2 d (y 11) \u003d dx 2 (y 111 dx) \u003d y 111 dx 3; d 3 y \u003d y 111 dx 3.

Here again dx 3 \u003d dx dx dx, and not d (x 3) \u003d 3x 2 dx

d n y \u003d y n dx n

Here dx n = (dx) n as before.

From the general formula for the nth order differential, in particular, the formula for the nth order derivative follows.

Y (n) \u003d d n y / dx n, i.e. the derivative of the nth order is the quotient of the nth differential of the function and the nth degree of the diff. independent change.

We have seen that the form of the first differential dy=y 1 dx does not depend on whether x is an independent variable or x is itself a function of some variable t.

The form of the differential of order n=2 is no longer preserved in this case, it does not have invariance.

In the case of an independent variable x d 2 y \u003d y 11 dx 2 is a second-order differential. Let now х= , dу 1 =у 1 dх. But now dx is no longer an arbitrary constant, dx = dt, i.e. dx- is a function of t and therefore, when finding d 2 y, we cannot take dx out of the differential sign.

d 2 y \u003d d (y 1 dx) \u003d d (y 1) dx + y 1 d (dx) \u003d y 11 dx 2 + y 1 d 2 x, i.e.

d 2 y \u003d y 11 dx 2 + y 1 d 2 x - the form of the differential has changed, the term y 1 d 2 x has been added. Moreover, the form d n y is not preserved. Hence, in the case when x is not an independent variable, the designation y (n) = d p y / dx p should be understood as a single symbol, and not as a ratio of differentials.

Partial derivatives of functions of two variables.
Concept and examples of solutions

In this lesson, we will continue our acquaintance with the function of two variables and consider, perhaps, the most common thematic task - finding partial derivatives of the first and second order, as well as the total differential of the function. Part-time students, as a rule, face partial derivatives in the 1st year in the 2nd semester. Moreover, according to my observations, the task of finding partial derivatives is almost always found in the exam.

In order to effectively study the following material, you necessary be able to more or less confidently find the "usual" derivatives of a function of one variable. You can learn how to handle derivatives correctly in the lessons How to find the derivative? and Derivative of a compound function. We also need a table of derivatives of elementary functions and differentiation rules, it is most convenient if it is at hand in printed form. You can find reference material on the page Mathematical formulas and tables.

Let's quickly repeat the concept of a function of two variables, I will try to limit myself to the bare minimum. A function of two variables is usually written as , with the variables being called independent variables or arguments.

Example: - a function of two variables.

Sometimes the notation is used. There are also tasks where the letter is used instead of a letter.

From a geometric point of view, a function of two variables is most often a surface of three-dimensional space (a plane, a cylinder, a ball, a paraboloid, a hyperboloid, etc.). But, in fact, this is already more analytical geometry, and we have mathematical analysis on the agenda, which my university teacher never let me write off is my “horse”.

We turn to the question of finding partial derivatives of the first and second orders. I have some good news for those of you who have had a few cups of coffee and are in the mood for unimaginably difficult material: partial derivatives are almost the same as the "ordinary" derivatives of a function of one variable.

For partial derivatives, all the rules of differentiation and the table of derivatives of elementary functions are valid. There are only a couple of small differences that we will get to know right now:

... yes, by the way, for this topic I did create small pdf book, which will allow you to "fill your hand" in just a couple of hours. But, using the site, you, of course, will also get the result - just maybe a little slower:

Example 1

Find partial derivatives of the first and second order of a function

First, we find the partial derivatives of the first order. There are two of them.

Notation:
or - partial derivative with respect to "x"
or - partial derivative with respect to "y"

Let's start with . When we find the partial derivative with respect to "x", then the variable is considered a constant (constant number).

Comments on the actions taken:

(1) The first thing we do when finding the partial derivative is to conclude all function in parentheses under the dash with subscript.

Attention important! Subscripts DO NOT LOSE in the course of the solution. In this case, if you draw a “stroke” somewhere without, then the teacher, at least, can put it next to the task (immediately bite off part of the score for inattention).

(2) Use the rules of differentiation , . For a simple example like this one, both rules can be applied in the same step. Pay attention to the first term: since is considered a constant, and any constant can be taken out of the sign of the derivative, then we take it out of brackets. That is, in this situation, it is no better than a regular number. Now let's look at the third term: here, on the contrary, there is nothing to take out. Since it is a constant, it is also a constant, and in this sense it is no better than the last term - the “seven”.

(3) We use tabular derivatives and .

(4) We simplify, or, as I like to say, "combine" the answer.

Now . When we find the partial derivative with respect to "y", then the variableconsidered a constant (constant number).

(1) We use the same differentiation rules , . In the first term we take out the constant beyond the sign of the derivative, in the second term nothing can be taken out because it is already a constant.

(2) We use the table of derivatives of elementary functions. Mentally change in the table all "X" to "Y". That is, this table is equally valid for (and indeed for almost any letter). In particular, the formulas we use look like this: and .

What is the meaning of partial derivatives?

At their core, 1st order partial derivatives resemble "ordinary" derivative:

- This functions, which characterize rate of change functions in the direction of the axes and respectively. So, for example, the function characterizes the steepness of "climbs" and "slopes" surfaces in the direction of the abscissa axis, and the function tells us about the "relief" of the same surface in the direction of the ordinate axis.

! Note : here refers to directions that are parallel coordinate axes.

For the sake of better understanding, let's consider a specific point of the plane and calculate the value of the function (“height”) in it:
- and now imagine that you are here (ON THE VERY surface).

We calculate the partial derivative with respect to "x" at a given point:

The negative sign of the "X" derivative tells us about descending functions at a point in the direction of the x-axis. In other words, if we make a small-small (infinitesimal) step towards the tip of the axis (parallel to this axis), then go down the slope of the surface.

Now we find out the nature of the "terrain" in the direction of the y-axis:

The derivative with respect to "y" is positive, therefore, at a point along the axis, the function increases. If it’s quite simple, then here we are waiting for an uphill climb.

In addition, the partial derivative at a point characterizes rate of change functions in the relevant direction. The greater the resulting value modulo- the steeper the surface, and vice versa, the closer it is to zero, the flatter the surface. So, in our example, the "slope" in the direction of the abscissa is steeper than the "mountain" in the direction of the ordinate.

But those were two private paths. It is quite clear that from the point at which we are, (and in general from any point of the given surface) we can move in some other direction. Thus, there is an interest in compiling a general "navigation chart" that would tell us about the "landscape" of the surface. if possible at every point scope of this function in all available ways. I will talk about this and other interesting things in one of the next lessons, but for now let's get back to the technical side of the issue.

We systematize the elementary applied rules:

1) When we differentiate by , then the variable is considered a constant.

2) When differentiation is carried out according to, then is considered a constant.

3) The rules and the table of derivatives of elementary functions are valid and applicable for any variable (or any other) with respect to which differentiation is carried out.

Step two. We find partial derivatives of the second order. There are four of them.

Notation:
or - the second derivative with respect to "x"
or - the second derivative with respect to "y"
or - mixed derivative "x by y"
or - mixed derivative "Y with X"

There are no problems with the second derivative. In simple terms, the second derivative is the derivative of the first derivative.

For convenience, I will rewrite the first-order partial derivatives already found:

First we find the mixed derivatives:

As you can see, everything is simple: we take the partial derivative and differentiate it again, but in this case, already by “y”.

Similarly:

In practical examples, you can focus on the following equality:

Thus, through mixed derivatives of the second order, it is very convenient to check whether we have found the partial derivatives of the first order correctly.

We find the second derivative with respect to "x".
No inventions, we take and differentiate it by "X" again:

Similarly:

It should be noted that when finding , you need to show increased attention, since there are no miraculous equalities to test them.

The second derivatives also find wide practical application, in particular, they are used in the problem of finding extrema of a function of two variables. But everything has its time:

Example 2

Calculate the first order partial derivatives of the function at the point . Find derivatives of the second order.

This is an example for self-solving (answers at the end of the lesson). If you have difficulty differentiating roots, go back to the lesson How to find the derivative? In general, pretty soon you will learn how to find similar derivatives on the fly.

We fill our hand with more complex examples:

Example 3

Check that . Write the total differential of the first order.

Solution: We find partial derivatives of the first order:

Pay attention to the subscript: next to the "X" it is not forbidden to write in brackets that it is a constant. This mark can be very useful for beginners to make it easier to navigate the solution.

Further comments:

(1) We take out all the constants outside the sign of the derivative. In this case, and , and, hence, their product is considered a constant number.

(2) Do not forget how to properly differentiate the roots.

(1) We take all the constants out of the sign of the derivative, in this case the constant is .

(2) Under the prime, we have the product of two functions, therefore, we need to use the product differentiation rule .

(3) Do not forget that is a complex function (although the simplest of the complex ones). We use the corresponding rule: .

Now we find mixed derivatives of the second order:

This means that all calculations are correct.

Let's write the total differential. In the context of the task under consideration, it makes no sense to tell what the total differential of a function of two variables is. It is important that this very differential very often needs to be written down in practical problems.

Total First Order Differential functions of two variables has the form:

In this case:

That is, in the formula you just need to stupidly just substitute the already found partial derivatives of the first order. Differential icons in this and similar situations, if possible, are best written in numerators:

And at the repeated request of readers, full differential of the second order.

It looks like this:

CAREFULLY find the "single-letter" derivatives of the 2nd order:

and write down the “monster”, carefully “attaching” the squares, the product and not forgetting to double the mixed derivative:

It's okay if something seemed difficult, you can always return to derivatives later, after you pick up the differentiation technique:

Example 4

Find first order partial derivatives of a function . Check that . Write the total differential of the first order.

Consider a series of examples with complex functions:

Example 5

Find partial derivatives of the first order of the function .

Decision:

Example 6

Find first order partial derivatives of a function .
Write down the total differential.

This is an example for self-solving (answer at the end of the lesson). I won't post the complete solution because it's quite simple.

Quite often, all of the above rules are applied in combination.

Example 7

Find first order partial derivatives of a function .

(1) We use the rule of differentiating the sum

(2) The first term in this case is considered a constant, since there is nothing in the expression that depends on "x" - only "y". You know, it's always nice when a fraction can be turned into zero). For the second term, we apply the product differentiation rule. By the way, in this sense, nothing would change if a function were given instead - it is important that here the product of two functions, EACH of which depends on "X", and therefore, you need to use the rule of differentiation of the product. For the third term, we apply the rule of differentiation of a complex function.

(1) In the first term, both the numerator and the denominator contain a “y”, therefore, you need to use the rule for differentiating the quotient: . The second term depends ONLY on "x", which means it is considered a constant and turns into zero. For the third term, we use the rule of differentiation of a complex function.

For those readers who courageously made it almost to the end of the lesson, I’ll tell you an old Mekhmatov anecdote for detente:

Once an evil derivative appeared in the space of functions and how it went to differentiate everyone. All functions scatter in all directions, no one wants to turn! And only one function does not escape anywhere. The derivative approaches it and asks:

"Why aren't you running away from me?"

- Ha. But I don't care, because I'm "e to the power of x", and you can't do anything to me!

To which the evil derivative with an insidious smile replies:

- This is where you are wrong, I will differentiate you by “y”, so be zero for you.

Who understood the joke, he mastered the derivatives, at least for the "troika").

Example 8

Find first order partial derivatives of a function .

This is a do-it-yourself example. A complete solution and a sample design of the problem are at the end of the lesson.

Well, that's almost all. Finally, I cannot help but please mathematicians with one more example. It's not even about amateurs, everyone has a different level of mathematical training - there are people (and not so rare) who like to compete with more difficult tasks. Although, the last example in this lesson is not so much complicated as cumbersome in terms of calculations.

Let y \u003d f (x) a differentiable function, and its argument x is an independent variable. Then its first differential dy = f ′ (x )dx is also a function otx ; find the differential of this function.

The differential from the differential of the function y \u003d f (x) is called its second differential(or second order differential) and is denoted d 2 y or d 2 f (x):

d 2 y = f′′ (x) dx2

Here dx 2 stands for (dx )2 .

The third-order differential is defined and found similarly: d 3 y = d (d2 y) = d (f′′ (x) dx2) = f′′′ (x) dx3 .

In general, the n-th order differential is the differential of the (n-1)-th order differential: d n y \u003d d (d n - 1 y ) \u003d f (n) (x) (dx) n.

From here we find that f (n) (x) = d n y . In particular, for n = 1, 2, 3, respectively, we obtain: dx n

f′(x)=			f′′(x)=	d2y		f′′′(x ) =	d 3 y	Those. The derivative of a function can be viewed as

the ratio of its differential of the corresponding order to the corresponding degree of the differential of the independent variable.

Note that all the above formulas are valid only if x is an independent variable.

Example. Find d 2 y if y = e 3 x their is an independent variable. Solution: since y ′ = 3e 3 x ,y ′′ = 9e 3 x , then we have d 2 y = 9e 3 x dx 2 .

Rules of L'Hospital

L'Hopital's rules are used to reveal uncertainties of the form 0 0 and ∞ ∞ , which are called basic.

Theorem 3. (L'Hopital's rule for revealing uncertainties of the form 0 0 ).

Let the functions f (x) and g (x) be continuous and differentiable in the vicinity of the points 0 and

vanish at this point: f (x 0 ) =g (x 0 ) = 0. Let g ′ (x )≠ 0 in the vicinity of the point x 0 . If a

there is a limit

f'(x)

L , then

f(x)

f'(x)

g(x)

g (x)

x → x0

x → x0

x → x0

Example. Find lim1 − cos6 x .

x → 0

2x2

Solution: lim

1− cos 6x

p. L.

6sin 6x

p. L.

36 cos 6x

x → 0

x → 0

x → 0

Theorem 4. (L'Hopital's rule for revealing uncertainties of the form ∞ ∞ ).

Let the functions f (x) and g (x) be continuous and differentiable in a neighborhood of the points 0 (except,
maybe points x 0 ), in this neighborhood limf (x) = limg (x) = ∞,g ′ (x) ≠ 0. If there exists
	f'(x)			f(x)			f'(x)	x → x0	x → x0
limit lim
	g (x)			g(x)
x → x0			x → x0			x → x0	g (x)

tg 3x

Example. Find lim tg 5x

x → π 2

lim tan 3 x =

∞ =

Lim3cos

p. L.

p. L.

x→

tg5x

x→

x→

cos2 5x

lim − 10 cos 5 x sin 5 x

Lim sin10 x

lim 10cos10x

5 x →

− 6 cos 3x sin 3x

x→

sin6x

x→

6cos6x

Uncertainties of the form , [∞ − ∞ ], , [∞ 0 ], are reduced to two main ways of identical transformations.

Let f (x) → 0, and g (x) → 0 as x → x 0. Then the following transformations are obvious:

lim(f (x ) g (x )) =[ 0 ∞] = lim

f(x)

f(x)

∞ ).

x → x

x → x

x → x

g(x)

g(x)

Find lim tg

x

(2 − x ).

x → 2

2 − x

0=lim

−1

limtg π x (2− x ) = [ ∞ 0] = lim

p. L.

x → 2

x → 2

x

ctg 4

x → 2

2 x

Let f (x) → ∞, and g (x) → ∞ as x → x 0. Then you can do this:

lim (f (x ) −g (x )) =[ ∞ − ∞] =lim

g(x)

f(x)

x → x0

x → x0

x → x0

f(x)

g(x)

g(x)

f(x)

Let f (x) → 1, and g (x) → ∞, or f (x) → ∞, and g (x) → 0, or f (x) → 0, and g (x) → 0 with x → x 0.

To find the limit of the form lim f (x) g (x) we recall the property of the logarithm

x → x0

e lnf (x) g (x) \u003d f (x) g (x).

Example. Find lim x → 0 (cos2 x ) x 2 .