Definition. Let the function \(y = f(x)\) be defined in a certain interval containing the point \(x_0\). Let's give the argument an increment \(\Delta x \) such that it does not leave this interval. Let's find the corresponding increment of the function \(\Delta y \) (when moving from the point \(x_0 \) to the point \(x_0 + \Delta x \)) and compose the relation \(\frac(\Delta y)(\Delta x) \). If there is a limit to this ratio at \(\Delta x \rightarrow 0\), then the specified limit is called derivative of a function\(y=f(x) \) at the point \(x_0 \) and denote \(f"(x_0) \).
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x_0) $$
The symbol y is often used to denote the derivative." Note that y" = f(x) is new feature, but naturally associated with the function y = f(x), defined at all points x at which the above limit exists. This function is called like this: derivative of the function y = f(x).
Geometric meaning of derivative is as follows. If it is possible to draw a tangent to the graph of the function y = f(x) at the point with abscissa x=a, which is not parallel to the y-axis, then f(a) expresses the slope of the tangent:
\(k = f"(a)\)
Since \(k = tg(a) \), then the equality \(f"(a) = tan(a) \) is true.
Now let’s interpret the definition of derivative from the point of view of approximate equalities. Let the function \(y = f(x)\) have a derivative at a specific point \(x\):
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x) $$
This means that near the point x the approximate equality \(\frac(\Delta y)(\Delta x) \approx f"(x)\), i.e. \(\Delta y \approx f"(x) \cdot\Delta x\). The meaningful meaning of the resulting approximate equality is as follows: the increment of the function is “almost proportional” to the increment of the argument, and the coefficient of proportionality is the value of the derivative in given point X. For example, for the function \(y = x^2\) the approximate equality \(\Delta y \approx 2x \cdot \Delta x \) is valid. If we carefully analyze the definition of a derivative, we will find that it contains an algorithm for finding it.
Let's formulate it.
How to find the derivative of the function y = f(x)?
1. Fix the value of \(x\), find \(f(x)\)
2. Give the argument \(x\) an increment \(\Delta x\), go to a new point \(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the increment of the function: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Create the relation \(\frac(\Delta y)(\Delta x) \)
5. Calculate $$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) $$
This limit is the derivative of the function at point x.
If a function y = f(x) has a derivative at a point x, then it is called differentiable at a point x. The procedure for finding the derivative of the function y = f(x) is called differentiation functions y = f(x).
Let us discuss the following question: how are continuity and differentiability of a function at a point related to each other?
Let the function y = f(x) be differentiable at the point x. Then a tangent can be drawn to the graph of the function at point M(x; f(x)), and, recall, the angular coefficient of the tangent is equal to f "(x). Such a graph cannot “break” at point M, i.e. the function must be continuous at point x.
These were “hands-on” arguments. Let us give a more rigorous reasoning. If the function y = f(x) is differentiable at the point x, then the approximate equality \(\Delta y \approx f"(x) \cdot \Delta x \) holds. If in this equality \(\Delta x \) tends to zero, then \(\Delta y \) will tend to zero, and this is the condition for the continuity of the function at a point.
So, if a function is differentiable at a point x, then it is continuous at that point.
The reverse statement is not true. For example: function y = |x| is continuous everywhere, in particular at the point x = 0, but the tangent to the graph of the function at the “junction point” (0; 0) does not exist. If at some point a tangent cannot be drawn to the graph of a function, then the derivative does not exist at that point.
One more example. The function \(y=\sqrt(x)\) is continuous on the entire number line, including at the point x = 0. And the tangent to the graph of the function exists at any point, including at the point x = 0. But at this point the tangent coincides with the y-axis, i.e., it is perpendicular to the abscissa axis, its equation has the form x = 0. Such a straight line does not have an angle coefficient, which means that \(f"(0)\) does not exist.
So, we got acquainted with a new property of a function - differentiability. How can one conclude from the graph of a function that it is differentiable?
The answer is actually given above. If at some point it is possible to draw a tangent to the graph of a function that is not perpendicular to the abscissa axis, then at this point the function is differentiable. If at some point the tangent to the graph of a function does not exist or it is perpendicular to the abscissa axis, then at this point the function is not differentiable.
Rules of differentiation
The operation of finding the derivative is called differentiation. When performing this operation, you often have to work with quotients, sums, products of functions, as well as “functions of functions,” that is, complex functions. Based on the definition of derivative, we can derive differentiation rules that make this work easier. If C is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$
Table of derivatives of some functions
$$ \left(\frac(1)(x) \right) " = -\frac(1)(x^2) $$ $$ (\sqrt(x)) " = \frac(1)(2\ sqrt(x)) $$ $$ \left(x^a \right) " = a x^(a-1) $$ $$ \left(a^x \right) " = a^x \cdot \ln a $$ $$ \left(e^x \right) " = e^x $$ $$ (\ln x)" = \frac(1)(x) $$ $$ (\log_a x)" = \frac (1)(x\ln a) $$ $$ (\sin x)" = \cos x $$ $$ (\cos x)" = -\sin x $$ $$ (\text(tg) x) " = \frac(1)(\cos^2 x) $$ $$ (\text(ctg) x)" = -\frac(1)(\sin^2 x) $$ $$ (\arcsin x) " = \frac(1)(\sqrt(1-x^2)) $$ $$ (\arccos x)" = \frac(-1)(\sqrt(1-x^2)) $$ $$ (\text(arctg) x)" = \frac(1)(1+x^2) $$ $$ (\text(arcctg) x)" = \frac(-1)(1+x^2) $ $Let the function be given implicitly as the equation 
. Differentiating this equation with respect to X and solving the resulting equation with respect to the derivative 
, let's find the first order derivative (first derivative). Differentiating by X the first derivative we obtain the second derivative of the implicit function. Substituting the already found value 
into the expression for the second derivative, we express 
through X And u. We proceed similarly to find the third-order derivative (and further).
Example.Find 
, If 
.
Solution: differentiate the equation with respect to X:
. From here we find 
. Further .
Derivatives of higher orders from functions specified parametrically.
Let the function 
given by parametric equations 
.
As is known, the first derivative 
is found by the formula 
. Let's find the second derivative 
, i.e. 
. Likewise 
.
Example. Find the second derivative 
.
Solution: find the first derivative 
. Finding the second derivative 
.
Function differential.
Let the function 
differentiable by 
. The derivative of this function at some point 
is determined by equality 
. Attitude 
at 
, therefore different from the derivative 
by the amount of b.m., i.e. can be written down 
(
). Let's multiply everything by 
, we get 
. Function increment 
consists of two terms. first term 
- the main part of the increment, there is a differential function.
Def. Function differential 
The product of the derivative and the increment of the argument is called. Designated 
.
The differential of the independent variable coincides with its increment 
.
(). Thus, the formula for the differential can be written 
. The differential of a function is equal to the product of its derivative and the differential of the independent variable. From this relation it follows that the derivative can be considered as a ratio of differentials 
.
The differential is used in approximate calculations. Since in the expression 
second term 
an infinitesimal quantity enjoys the approximate equality 
or in expanded form
Example: Calculate approximate value 
.
Function 
has a derivative 
.
According to the formula (*) : .
Example: find the differential of a function
Geometric meaning of differential.
To the graph of the function 
at point M( x;y) draw a tangent and consider the ordinate of this tangent for the point x+∆
x. In the figure AM=∆ X AM 1 =∆ at from ∆MAB 
, from here 
, but according to the geometric meaning of the tangent 
. That's why 
. Comparing this formula with the differential formula we get that 
, i.e. differential function 
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 gets increment ∆х.
Rules for calculating the differential.
Since the function differential 
differs from the derivative by a factor 
, then all the rules for calculating the derivative are used to calculate the differential (hence the term “differentiation”).
Let two differentiable functions be given 
And 
, then the differential is found according to the following rules:
1)
2)
With -const
3)
4)
(
)
5) for a complex function 
, Where 
(because 
).
The differential of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the differential of this intermediate argument.
Derivative applications.
Mean value theorems.
Rolle's theorem. If the function 
continuous on the segment 
and differentiable in the open interval 
and if it takes equal values at the ends of the segment 
, then in the interval 
there is at least one such point With, in which the derivative goes to zero, i.e. 
,
a<
c<
b.
Geometrically, Rolle's theorem means that on the graph of the function 
there is a point at which the tangent to the graph is parallel to the axis Oh.
Lagrange's theorem. If the function 
continuous on the segment 
and differentiable on the interval 
, then there is at least one point 
such that the equality .
The formula is called the Lagrange formula or the finite increment formula: the increment of a differentiable function on an interval 
is equal to the increment of the argument multiplied by the value of the derivative at some internal point of this segment.
Geometric meaning of Lagrange's theorem: functions on the graph 
there is a point C(s;f(c))
, in which the tangent to the graph of the function is parallel to the secant AB.
Cauchy's theorem. If the functions 
And 
continuous on the segment 
, differentiable on the interval 
, and 
For 
, then there is at least one point 
such that the equality holds 
.
Cauchy's theorem provides the basis for a new rule for calculating limits.
L'Hopital's rule.
Theorem:(L'Hopital's rule - disclosure of uncertainties of the form 
). Let the functions 
And 
continuous and differentiable in a neighborhood of a point X 0
and vanish at this point 
. Let it go 
in the vicinity of a point X 0
. if there is a limit 
, That 
.
Proof: Apply to functions 
And 
Cauchy's theorem for a segment 
Lying in the vicinity of a point X 0
. Then 
, Where x 0
<
c<
x. Because 
we get 
. Let's go to the limit at 
. Because 
, That 
, That's why 
.
So the limit of the ratio of two b.m. equal to the limit of the ratio of their derivatives, if the latter exists 
.
Theorem.(L'Hopital's rule for disclosing uncertainties of the form 
) Let the functions 
And 
continuous and differentiable in a neighborhood of a point X 0
(except maybe the point X 0
), in this vicinity 
,
. If there is a limit
, That 
.
Uncertainties of the form ( 
) are reduced to two main ( 
),
through identical transformations.
Example: 
Very often, when solving practical problems (for example, in higher geodesy or analytical photogrammetry), complex functions of several variables appear, i.e. arguments x, y, z one function f(x,y,z) ) are themselves functions of new variables U, V, W ).
This, for example, happens when moving from a fixed coordinate system Oxyz into the mobile system O 0 UVW and back. At the same time, it is important to know all the partial derivatives with respect to the “fixed” - “old” and “moving” - “new” variables, since these partial derivatives usually characterize the position of an object in these coordinate systems, and, in particular, affect the correspondence of aerial photographs to a real object . In such cases, the following formulas apply:
That is, a complex function is given T three "new" variables U, V, W through three "old" variables x, y, z, Then:
Comment. There may be variations in the number of variables. For example: if
In particular, if z = f(xy), y = y(x) , then we get the so-called “total derivative” formula:
The same formula for the “total derivative” in the case of:
will take the form:
Other variations of formulas (1.27) - (1.32) are also possible.
Note: the “total derivative” formula is used in the physics course, section “Hydrodynamics” when deriving the fundamental system of equations of fluid motion.
Example 1.10. Given:
According to (1.31):
§7 Partial derivatives of an implicitly given function of several variables
As is known, an implicitly specified function of one variable is defined as follows: the function of the independent variable x is called implicit if it is given by an equation that is not solved with respect to y :
Example 1.11.
The equation
implicitly specifies two functions:
And the equation
does not specify any function.
Theorem 1.2 (existence of an implicit function).
Let the function z =f(x,y) and its partial derivatives f" x And f" y defined and continuous in some neighborhood U M0 points M 0 (x 0 y 0 ) . Besides, f(x 0 ,y 0 )=0 And f"(x 0 ,y 0 )≠0 , then equation (1.33) defines in the neighborhood U M0 implicit function y=y(x) , continuous and differentiable in a certain interval D centered at a point x 0 , and y(x 0 )=y 0 .
No proof.
From Theorem 1.2 it follows that on this interval D :
that is, there is an identity in
where the “total” derivative is found according to (1.31)
That is, (1.35) gives the formula for finding the derivative implicitly given function one variable x .
An implicit function of two or more variables is defined similarly.
For example, if in some area V space Oxyz the following equation holds:
then under some conditions on the function F it implicitly defines a function
Moreover, by analogy with (1.35), its partial derivatives are found as follows.
Consider the function y(x), which is written implicitly in general view$ F(x,y(x)) = 0 $. The derivative of an implicit function is found in two ways:
- By differentiating both sides of the equation
- By using the ready-made formula $ y" = - \frac(F"_x)(F"_y) $
How to find?
Method 1
There is no need to cast the function explicitly. You need to immediately start differentiating the left and right sides of the equation with respect to $ x $. It is worth noting that the derivative $ y" $ is calculated according to the rule of differentiation of a complex function. For example, $ (y^2)"_x = 2yy" $. After finding the derivative, it is necessary to express $ y" $ from the resulting equation and place $ y" $ on the left side.
Method 2
You can use a formula that uses the partial derivatives of the implicit function $ F(x,y(x)) = 0 $ in the numerator and denominator. To find the numerator, take the derivative with respect to $ x $, and for the denominator, take the derivative with respect to $ y $.
The second derivative of the implicit function can be found by repeatedly differentiating the first derivative of the implicit function.
Examples of solutions
Let's look at practical examples of solutions for calculating the derivative of an implicitly specified function.
| Example 1 |
|
Find the derivative of the implicit function $ 3x^2y^2 -5x = 3y - 1 $ |
| Solution |
|
Let's use method No. 1. Namely, we differentiate the left and right sides of the equation: $$ (3x^2y^2 -5x)"_x = (3y - 1)"_x $$ When differentiating, don’t forget to use the formula for the derivative of a product of functions: $$ (3x^2)"_x y^2 + 3x^2 (y^2)"_x - (5x)"_x = (3y)"_x - (1)"_x $$ $$ 6x y^2 + 3x^2 2yy" - 5 = 3y" $$ $$ 6x y^2 - 5 = 3y" - 6x^2 yy" $$ $$ 6x y^2 - 5 = y"(3-6x^2 y) $$ $$ y" = \frac(6x y^2 - 5)(3 - 6x^2y ) $$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
| $$ y" = \frac(6x y^2 - 5)(3 - 6x^2y ) $$ |
| Example 2 |
|
The function is given implicitly, find the derivative $ 3x^4 y^5 + e^(7x-4y) -4x^5 -2y^4 = 0 $ |
| Solution |
|
Let's use method No. 2. Finding partial derivatives of the function $ F(x,y) = 0 $ Let $ y $ be constant and differentiate with respect to $ x $: $$ F"_x = 12x^3 y^5 + e^(7x-4y) \cdot 7 - 20x^4 $$ $$ F"_x = 12x^3 y^5 + 7e^(7x-4y) - 20x^4 $$ Now we consider $ x $ a constant and differentiate with respect to $ y $: $$ F"_y = 15x^4 y^4 + e^(7x-4y) \cdot (-4) - 8y^3 $$ $$ F"_y = 15x^4 y^4 - 4e^(7x-4y) - 8y^3 $$ Now we substitute $ y" = -\frac(F"_y)(F"_x) $ into the formula and get: $$ y" = -\frac(12x^3 y^5 + 7e^(7x-4y) - 20x^4)(15x^4 y^4 - 4e^(7x-4y) - 8y^3) $$ |
| Answer |
| $$ y" = -\frac(12x^3 y^5 + 7e^(7x-4y) - 20x^4)(15x^4 y^4 - 4e^(7x-4y) - 8y^3) $$ |
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, derivatives table elementary functions remain in effect. The difference is in one peculiar moment, which we will look at right now.
Yes, I'll let you know 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, - 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. 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 transfer all terms to 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 general formula parametrically defined 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 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 the 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: 