Definition. Let the function \(y = f(x) \) be defined in some interval containing the point \(x_0 \) inside. Let's increment \(\Delta x \) to the argument so as not to leave this interval. Find the corresponding increment of the function \(\Delta y \) (when passing 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 of this relation at \(\Delta x \rightarrow 0 \), then the indicated limit is called derivative 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 a new function, 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 \u003d f (x).

The geometric meaning of the derivative consists of the following. If a tangent that is not parallel to the y axis can be drawn to the graph of the function y \u003d f (x) at a point with the abscissa x \u003d a, then f (a) expresses the slope of the tangent:
\(k = f"(a)\)

Since \(k = tg(a) \), the equality \(f"(a) = tg(a) \) is true.

And now we interpret the definition of the derivative in terms of approximate equalities. Let the function \(y = f(x) \) have a derivative at a particular 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. \(\Delta y \approx f"(x) \cdot\Deltax\). The meaningful meaning of the obtained 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 the 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 \u003d f (x) ?

1. Fix value \(x \), find \(f(x) \)
2. Increment \(x \) argument \(\Delta x \), go to new point\(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the function increment: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Compose 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 x.

If the function y = f(x) has a derivative at the point x, then it is called differentiable at the point x. The procedure for finding the derivative of the function y \u003d f (x) is called differentiation functions y = f(x).

Let us discuss the following question: how are the continuity and differentiability of a function at a point related?

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 the point M (x; f (x)) and, recall, the slope of the tangent is equal to f "(x). Such a graph cannot "break" at the point M, i.e., the function must be continuous at x.

It was reasoning "on the fingers". Let us present a more rigorous argument. 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. zero, then \(\Delta y \) will also 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 also continuous at that point.

The converse 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 “joint point” (0; 0) does not exist. If at some point it is impossible to draw a tangent to the function graph, then there is no derivative at this 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, that is, it is perpendicular to the abscissa axis, its equation has the form x \u003d 0. Slope there is no such line, which means that \(f"(0) \) does not exist either

So, we got acquainted with a new property of a function - differentiability. How can you tell if a function is differentiable from the graph of a function?

The answer is actually given above. If at some point a tangent can be drawn to the graph of a function that is not perpendicular to the x-axis, then at this point the function is differentiable. If at some point the tangent to the graph of the function does not exist or it is perpendicular to the x-axis, then at this point the function is not differentiable.

Differentiation rules

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 with "functions of functions", that is, complex functions. Based on the definition of the derivative, we can derive differentiation rules that facilitate this work. If C is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:

$$ C"=0 $$ $$ x"=1 $$ $$ (f+g)"=f"+g" $$ $$ (fg)"=f"g + fg" $$ $$ ( Cf)"=Cf" $$ $$ \left(\frac(f)(g) \right) " = \frac(f"g-fg")(g^2) $$ $$ \left(\frac (C)(g) \right) " = -\frac(Cg")(g^2) $$ Compound function derivative:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$

Table of derivatives of some functions

$$ \left(\frac(1)(x) \right) " = -\frac(1)(x^2) $$ $$ (\sqrt(x)) " = \frac(1)(2\ sqrt(x)) $$ $$ \left(x^a \right) " = ax^(a-1) $$ $$ \left(a^x \right) " = a^x \cdot \ln a $$ $$ \left(e^x \right) " = e^x $$ $$ (\ln x)" = \frac(1)(x) $$ $$ (\log_a x)" = \frac (1)(x\ln a) $$ $$ (\sin x)" = \cos x $$ $$ (\cos x)" = -\sin x $$ $$ (\text(tg) x) " = \frac(1)(\cos^2 x) $$ $$ (\text(ctg) x)" = -\frac(1)(\sin^2 x) $$ $$ (\arcsin x) " = \frac(1)(\sqrt(1-x^2)) $$ $$ (\arccos x)" = \frac(-1)(\sqrt(1-x^2)) $$ $$ (\text(arctg) x)" = \frac(1)(1+x^2) $$ $$ (\text(arctg) x)" = \frac(-1)(1+x^2) $ $

Let the function be given implicitly as an equation
. Differentiating this equation with respect to X and solving the resulting equation with respect to the derivative , we find the derivative of the first order (the first derivative). Differentiating with respect to X the first derivative we get the second derivative of the implicit function. Substituting an already found value into the expression of the second derivative, we express across X and y. We proceed similarly to find the third-order derivative (and beyond).

Example.Find , if
.

Solution: Differentiate the equation with respect to X:
. From here we find
. Further .

Derivatives of higher orders from functions given parametrically.

Let the function
given by parametric equations
.

As you know, the first derivative is found according to the formula
. Let's find the second derivative
, i.e.
. Similarly
.

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 defined by the equality
. Attitude
at
, therefore different from the derivative
by the value of b.m., i.e. can be written
(
). Let's multiply everything by
, we get
. Function increment
consists of two terms. first term
- the main part of the increment, is the differential of the function.

Def. function differential
is called the product of the derivative and the increment of the argument. Denoted
.

The differential of an independent variable is the same as its increment
.

(). Thus, the formula for the differential can be written
. The differential of a function is equal to the product of the derivative and the differential of the independent variable. It follows from this relation that the derivative can be considered as the ratio of differentials
.

The differential is used in approximate calculations. Since in the expression
second term
an infinitesimal quantity use the approximate equality
or expanded

Example: calculate an approximate value
.

Function
has a derivative
.

According to the formula (*) : .

Example: find the differential of a function

The geometric meaning of the 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 ∆MAV
, hence
, but according to the geometric meaning of the tangent
. So
. Comparing this formula with the differential formula, we get that
, i.e. function differential
at the point X is equal to the increment of the ordinate of the tangent to the graph of the function at that point, when X gets an increment ∆х.

Differential calculation rules.

Since the function differential
differs from the derivative by a factor
, then all the rules for calculating the derivative are also 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 complex function
, where

(because
).

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.

Derivative applications.

Theorems on the mean value.

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 vanishes, i.e.
, a< c< b.

Geometrically, Rolle's theorem means that on the graph of the function
there is a point where 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 holds.

The formula is called the Lagrange formula or the finite increment formula: the increment of a differentiable function on the interval
is equal to the increment of the argument multiplied by the value of the derivative at some interior point of this segment.

The geometric meaning of Lagrange's theorem: on the graph of the function
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 functions
and
continuous on the segment
, are differentiable on the interval
, and
for
, then there is at least one point
such that the equality
.

Cauchy's theorem serves as 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
are continuous and differentiable in a neighborhood of a point X 0 and vanish at this point
. Let it go
in the vicinity of the point X 0 . if there is a limit
, then
.

Proof: Applicable to functions
and
Cauchy's theorem for the segment

Lying in the neighborhood of a point X 0 . Then
, where x 0 < c< x. Because
we get
. Let us pass to the limit at

. Because
, then
, That's why
.

So the limit of the ratio of two b.m. equals the limit of the ratio of their derivatives, if the latter exists
.

Theorem.(L'Hopital's rule for disclosure of uncertainties of the form
) Let the functions
and
are continuous and differentiable in a neighborhood of a point X 0 (except perhaps the dot X 0 ), in this neighborhood
,
. If there is a limit

, then
.

Uncertainties of the form (
) are reduced to two main ( ),
through identical transformations.

Example:

Let the function be given implicitly using the equation
(1) .
And let this equation, for some value , have only decision. Let the function be a differentiable function at the point , and
.
Then, for this value , there is a derivative , which is determined by the formula:
(2) .

Proof

For proof, consider the function as a complex function of the variable :
.
We apply the rule of differentiation of a complex function and find the derivative with respect to the variable of the left and right sides of the equation
(3) :
.
Since the derivative of the constant is equal to zero and , then
(4) ;
.

The formula has been proven.

Derivatives of higher orders

Let us rewrite equation (4) using other notation:
(4) .
Moreover, and are complex functions of the variable :
;
.
Dependence defines the equation (1):
(1) .

We find the derivative with respect to the variable from the left and right sides of equation (4).
According to the formula for the derivative of a complex function, we have:
;
.
According to the derivative product formula:

.
According to the derivative sum formula:


.

Since the derivative of the right side of equation (4) is equal to zero, then
(5) .
Substituting the derivative here, we obtain the value of the second-order derivative in implicit form.

Differentiating equation (5) in a similar way, we obtain an equation containing a third order derivative:
.
Substituting here the found values ​​of the derivatives of the first and second orders, we find the value of the third order derivative.

Continuing differentiation, one can find a derivative of any order.

Examples

Example 1

Find the first derivative of the function given implicitly by the equation:
(P1) .

Formula 2 Solution

We find the derivative by formula (2):
(2) .

Let's move all the variables to the left side so that the equation takes the form .
.
From here.

We find the derivative with respect to , assuming that it is constant.
;
;
;
.

We find the derivative with respect to the variable, assuming the variable is constant.
;
;
;
.

By formula (2) we find:
.

We can simplify the result if we note that according to the original equation (A.1), . Substitute :
.
Multiply the numerator and denominator by:
.

Solution in the second way

Let's solve this example in the second way. To do this, we find the derivative with respect to the variable of the left and right parts of the original equation (P1).

We apply:
.
We apply the formula for the derivative of a fraction:
;
.
We apply the formula for the derivative of a complex function:
.
We differentiate the original equation (P1).
(P1) ;
;
.
Multiply by and group the terms.
;
.

Substitute (from equation (P1)):
.
Let's multiply by:
.

Answer

Example 2

Find the second order derivative of the function given implicitly using the equation:
(P2.1) .

Solution

Differentiate the original equation with respect to the variable , assuming that it is a function of :
;
.
We apply the formula for the derivative of a complex function.
.

We differentiate the original equation (A2.1):
;
.
It follows from the original equation (A2.1) that . Substitute :
.
Expand the brackets and group the members:
;
(P2.2) .
We find the derivative of the first order:
(P2.3) .

To find the second order derivative, we differentiate equation (A2.2).
;
;
;
.
We substitute the expression for the first order derivative (A2.3):
.
Let's multiply by:

;
.
From here we find the derivative of the second order.

Answer

Example 3

Find the third order derivative for of the function given implicitly using the equation:
(P3.1) .

Solution

Differentiate the original equation with respect to the variable, assuming that is a function of .
;
;
;
;
;
;
(P3.2) ;

We differentiate equation (A3.2) with respect to the variable .
;
;
;
;
;
(P3.3) .

We differentiate equation (A3.3).
;
;
;
;
;
(P3.4) .

From equations (A3.2), (A3.3) and (A3.4) we find the values ​​of derivatives at .
;
;
.

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

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

Derivative of a function defined implicitly

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

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

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

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

Consider the function

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

Let's consider another function:

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

Allow me to introduce: - an example implicit function.

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

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

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

Example 1

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

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

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

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

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

The product is differentiable according to the usual rule :

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

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


If there are brackets, then open them:

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

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

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

The derivative has been found. Ready.

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

The second way to solve

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

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

We move all the terms to the left side:

And consider a function of two variables:

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

In this way:

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

Let's look at a few more examples.

Example 2

Find the derivative of a function given implicitly

We hang strokes on both parts:

We use the rules of linearity:

Finding derivatives:

Expanding all parentheses:

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

Final answer:

Example 3

Find the derivative of a function given implicitly

Complete Solution and sample design at the end of the lesson.

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

Example 4

Find the derivative of a function given implicitly

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

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

Expanding the brackets:

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

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

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

Final answer:

Example 5

Find the derivative of a function given implicitly

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

Derivative of a parametrically defined function

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

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

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

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

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

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

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

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

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

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

Example 6

We use the formula

In this case:

In this way:

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

Example 7

Find the derivative of a function given parametrically

This is a do-it-yourself example.

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

Example 8

Find the first and second derivatives of a function given parametrically

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

In this case:

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

Derivative of a function defined implicitly

Or, in short, the derivative of an implicit function. What is an implicit function? Since my lessons are practical, I try to avoid definitions, formulations of theorems, but here it would be appropriate to do so. What is a function anyway?

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

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

Roughly speaking, the letter "y" in this case is the function.

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

Consider the function

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

Let's consider another function:

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

Allow me to introduce: - an example implicit function.

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

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

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

Example 1

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

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

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

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

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

The product is differentiable according to the usual rule :

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

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


If there are brackets, then open them:

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

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

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

The derivative has been found. Ready.

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

The second way to solve

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

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

We move all the terms to the left side:

And consider a function of two variables:

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

In this way:

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

Let's look at a few more examples.

Example 2

Find the derivative of a function given implicitly

We hang strokes on both parts:

We use the rules of linearity:

Finding derivatives:

Expanding all parentheses:

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

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

Final answer:

Example 3

Find the derivative of a function given implicitly

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

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

Example 4

Find the derivative of a function given implicitly

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