4 derivative of a complex function. Derivative of a function

If g(x) And f(u) – differentiable functions of their arguments, respectively, at points x And u= g(x), then the complex function is also differentiable at the point x and is found by the formula

A typical mistake when solving derivative problems is mechanically transferring the rules for differentiating simple functions to complex functions. Let's learn to avoid this mistake.

Example 2. Find the derivative of a function

Wrong solution: calculate the natural logarithm of each term in parentheses and look for the sum of the derivatives:

Correct solution: again we determine where the “apple” is and where the “minced meat” is. Here the natural logarithm of the expression in parentheses is an “apple”, that is, a function over the intermediate argument u, and the expression in brackets is “minced meat”, that is, an intermediate argument u by independent variable x.

Then (using formula 14 from the derivatives table)

In many real-life problems, the expression with a logarithm can be somewhat more complicated, which is why there is a lesson

Example 3. Find the derivative of a function

Wrong solution:

Correct solution. Once again we determine where the “apple” is and where the “mincemeat” is. Here, the cosine of the expression in brackets (formula 7 in the table of derivatives) is an “apple”, it is prepared in mode 1, which affects only it, and the expression in brackets (the derivative of the degree is number 3 in the table of derivatives) is “minced meat”, it is prepared under mode 2, which affects only it. And as always, we connect two derivatives with the product sign. Result:

The derivative of a complex logarithmic function is a frequent task in tests, so we strongly recommend that you attend the lesson “Derivative of a logarithmic function.”

The first examples were on complex functions, in which the intermediate argument on the independent variable was a simple function. But in practical tasks it is often necessary to find the derivative of a complex function, where the intermediate argument is either itself a complex function or contains such a function. What to do in such cases? Find derivatives of such functions using tables and differentiation rules. When the derivative of the intermediate argument is found, it is simply substituted into the right place in the formula. Below are two examples of how this is done.

In addition, it is useful to know the following. If a complex function can be represented as a chain of three functions

then its derivative should be found as the product of the derivatives of each of these functions:

Many of your homework assignments may require you to open your guides in new windows. Actions with powers and roots And Operations with fractions .

Example 4. Find the derivative of a function

We apply the rule of differentiation of a complex function, not forgetting that in the resulting product of derivatives there is an intermediate argument with respect to the independent variable x does not change:

We prepare the second factor of the product and apply the rule for differentiating the sum:

The second term is the root, so

Thus, we found that the intermediate argument, which is a sum, contains a complex function as one of the terms: raising to a power is a complex function, and what is being raised to a power is an intermediate argument with respect to the independent variable x.

Therefore, we again apply the rule for differentiating a complex function:

We transform the degree of the first factor into a root, and when differentiating the second factor, do not forget that the derivative of the constant is equal to zero:

Now we can find the derivative of the intermediate argument needed to calculate the derivative of a complex function required in the problem statement y:

Example 5. Find the derivative of a function

First, we use the rule for differentiating the sum:

We obtained the sum of the derivatives of two complex functions. Let's find the first one:

Here, raising the sine to a power is a complex function, and the sine itself is an intermediate argument for the independent variable x. Therefore, we will use the rule of differentiation of a complex function, along the way taking the factor out of brackets :

Now we find the second term of the derivatives of the function y:

Here raising the cosine to a power is a complex function f, and the cosine itself is an intermediate argument in the independent variable x. Let us again use the rule for differentiating a complex function:

The result is the required derivative:

Table of derivatives of some complex functions

For complex functions, based on the rule of differentiation of a complex function, the formula for the derivative of a simple function takes a different form.

1. Derivative of a complex power function, where u x
2. Derivative of the root of the expression
3. Derivative of an exponential function
4. Special case of exponential function
5. Derivative of a logarithmic function with an arbitrary positive base A
6. Derivative of a complex logarithmic function, where u– differentiable function of the argument x
7. Derivative of sine
8. Derivative of cosine
9. Derivative of tangent
10. Derivative of cotangent
11. Derivative of arcsine
12. Derivative of arccosine
13. Derivative of arctangent
14. Derivative of arc cotangent

If you follow the definition, then the derivative of a function at a point is the limit of the ratio of the increment of the function Δ y to the argument increment Δ x:

Everything seems to be clear. But try using this formula to calculate, say, the derivative of the function f(x) = x 2 + (2x+ 3) · e x sin x. If you do everything by definition, then after a couple of pages of calculations you will simply fall asleep. Therefore, there are simpler and more effective ways.

To begin with, we note that from the entire variety of functions we can distinguish the so-called elementary functions. These are relatively simple expressions, the derivatives of which have long been calculated and tabulated. Such functions are quite easy to remember - along with their derivatives.

Derivatives of elementary functions

Elementary functions are all those listed below. The derivatives of these functions must be known by heart. Moreover, it is not at all difficult to memorize them - that’s why they are elementary.

So, derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, zero!)
Power with rational exponent	f(x) = x n	n · x n − 1
Sinus	f(x) = sin x	cos x
Cosine	f(x) = cos x	−sin x(minus sine)
Tangent	f(x) = tg x	1/cos 2 x
Cotangent	f(x) = ctg x	− 1/sin 2 x
Natural logarithm	f(x) = log x	1/x
Arbitrary logarithm	f(x) = log a x	1/(x ln a)
Exponential function	f(x) = e x	e x(nothing changed)

If an elementary function is multiplied by an arbitrary constant, then the derivative of the new function is also easily calculated:

(C · f)’ = C · f ’.

In general, constants can be taken out of the sign of the derivative. For example:

(2x 3)’ = 2 · ( x 3)’ = 2 3 x 2 = 6x 2 .

Obviously, elementary functions can be added to each other, multiplied, divided - and much more. This is how new functions will appear, no longer particularly elementary, but also differentiated according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions be given f(x) And g(x), the derivatives of which are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:

1. (f + g)’ = f ’ + g ’
2. (f − g)’ = f ’ − g ’

So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.

Strictly speaking, there is no concept of “subtraction” in algebra. There is a concept of “negative element”. Therefore the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.

f(x) = x 2 + sin x; g(x) = x 4 + 2x 2 − 3.

Function f(x) is the sum of two elementary functions, therefore:

f ’(x) = (x 2 + sin x)’ = (x 2)’ + (sin x)’ = 2x+ cos x;

We reason similarly for the function g(x). Only there are already three terms (from the point of view of algebra):

g ’(x) = (x 4 + 2x 2 − 3)’ = (x 4 + 2x 2 + (−3))’ = (x 4)’ + (2x 2)’ + (−3)’ = 4x 3 + 4x + 0 = 4x · ( x 2 + 1).

Answer:
f ’(x) = 2x+ cos x;
g ’(x) = 4x · ( x 2 + 1).

Derivative of the product

Mathematics is a logical science, so many people believe that if the derivative of a sum is equal to the sum of derivatives, then the derivative of the product strike">equal to the product of derivatives. But screw you! The derivative of a product is calculated using a completely different formula. Namely:

(f · g) ’ = f ’ · g + f · g ’

The formula is simple, but it is often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.

Task. Find derivatives of functions: f(x) = x 3 cos x; g(x) = (x 2 + 7x− 7) · e x .

Function f(x) is the product of two elementary functions, so everything is simple:

f ’(x) = (x 3 cos x)’ = (x 3)’ cos x + x 3 (cos x)’ = 3x 2 cos x + x 3 (− sin x) = x 2 (3cos x − x sin x)

Function g(x) the first multiplier is a little more complicated, but the general scheme does not change. Obviously, the first factor of the function g(x) is a polynomial and its derivative is the derivative of the sum. We have:

g ’(x) = ((x 2 + 7x− 7) · e x)’ = (x 2 + 7x− 7)’ · e x + (x 2 + 7x− 7) · ( e x)’ = (2x+ 7) · e x + (x 2 + 7x− 7) · e x = e x· (2 x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x+ 9) · e x .

Answer:
f ’(x) = x 2 (3cos x − x sin x);
g ’(x) = x(x+ 9) · e x .

Please note that in the last step the derivative is factorized. Formally, this does not need to be done, but most derivatives are not calculated on their own, but to examine the function. This means that further the derivative will be equated to zero, its signs will be determined, and so on. For such a case, it is better to have an expression factorized.

If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set we are interested in, we can define a new function h(x) = f(x)/g(x). For such a function you can also find the derivative:

Not weak, huh? Where did the minus come from? Why g 2? And like this! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it with specific examples.

Task. Find derivatives of functions:

The numerator and denominator of each fraction contain elementary functions, so all we need is the formula for the derivative of the quotient:

According to tradition, let's factorize the numerator - this will greatly simplify the answer:

A complex function is not necessarily a half-kilometer-long formula. For example, it is enough to take the function f(x) = sin x and replace the variable x, say, on x 2 + ln x. It will work out f(x) = sin ( x 2 + ln x) - this is a complex function. It also has a derivative, but it will not be possible to find it using the rules discussed above.

What should I do? In such cases, replacing a variable and formula for the derivative of a complex function helps:

f ’(x) = f ’(t) · t', If x is replaced by t(x).

As a rule, the situation with understanding this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it using specific examples, with a detailed description of each step.

Task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) = sin ( x 2 + ln x)

Note that if in the function f(x) instead of expression 2 x+ 3 will be easy x, then we get an elementary function f(x) = e x. Therefore, we make a replacement: let 2 x + 3 = t, f(x) = f(t) = e t. We look for the derivative of a complex function using the formula:

f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’

And now - attention! We perform the reverse replacement: t = 2x+ 3. We get:

f ’(x) = e t · t ’ = e 2x+ 3 (2 x + 3)’ = e 2x+ 3 2 = 2 e 2x + 3

Now let's look at the function g(x). Obviously it needs to be replaced x 2 + ln x = t. We have:

g ’(x) = g ’(t) · t’ = (sin t)’ · t’ = cos t · t ’

Reverse replacement: t = x 2 + ln x. Then:

g ’(x) = cos ( x 2 + ln x) · ( x 2 + ln x)’ = cos ( x 2 + ln x) · (2 x + 1/x).

That's all! As can be seen from the last expression, the whole problem has been reduced to calculating the derivative sum.

Answer:
f ’(x) = 2 · e 2x + 3 ;
g ’(x) = (2x + 1/x) cos ( x 2 + ln x).

Very often in my lessons, instead of the term “derivative,” I use the word “prime.” For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.

Thus, calculating the derivative comes down to getting rid of these same strokes according to the rules discussed above. As a final example, let's return to the derivative power with a rational exponent:

(x n)’ = n · x n − 1

Few people know that in the role n may well be a fractional number. For example, the root is x 0.5. What if there is something fancy under the root? Again, the result will be a complex function - they like to give such constructions in tests and exams.

Task. Find the derivative of the function:

First, let's rewrite the root as a power with a rational exponent:

f(x) = (x 2 + 8x − 7) 0,5 .

Now we make a replacement: let x 2 + 8x − 7 = t. We find the derivative using the formula:

f ’(x) = f ’(t) · t ’ = (t 0.5)’ · t’ = 0.5 · t−0.5 · t ’.

Let's do the reverse replacement: t = x 2 + 8x− 7. We have:

f ’(x) = 0.5 · ( x 2 + 8x− 7) −0.5 · ( x 2 + 8x− 7)’ = 0.5 · (2 x+ 8) ( x 2 + 8x − 7) −0,5 .

Finally, back to the roots:

It is not entirely correct to call functions of a complex type the term “complex function”. For example, it looks very impressive, but this function is not complicated, unlike.

In this article, we will understand the concept of a complex function, learn how to identify it as part of elementary functions, give a formula for finding its derivative, and consider in detail the solution of typical examples.

When solving examples, we will constantly use the table of derivatives and differentiation rules, so keep them before your eyes.

Complex function is a function whose argument is also a function.

From our point of view, this definition is the most understandable. Conventionally, it can be denoted as f(g(x)) . That is, g(x) is like an argument of the function f(g(x)) .

For example, let f be the arctangent function and g(x) = lnx be the natural logarithm function, then the complex function f(g(x)) is arctan(lnx) . Another example: f is the function of raising to the fourth power, and is an entire rational function (see ), then .

In turn, g(x) can also be a complex function. For example, . Conventionally, such an expression can be denoted as . Here f is the sine function, is the square root function, - fractional rational function. It is logical to assume that the degree of nesting of functions can be any finite natural number.

You can often hear a complex function called composition of functions.

Formula for finding the derivative of a complex function.

Example.

Find the derivative of a complex function.

Solution.

In this example, f is the squaring function and g(x) = 2x+1 is the linear function.

Here is the detailed solution using the complex function derivative formula:

Let's find this derivative by first simplifying the form of the original function.

Hence,

As you can see, the results are the same.

Try not to confuse which function is f and which is g(x) .

Let's illustrate this with an example to show your attention.

Example.

Find derivatives of complex functions and .

Solution.

In the first case, f is the squaring function and g(x) is the sine function, so
.

In the second case, f is a sine function, and is a power function. Therefore, by the formula for the product of a complex function we have

The derivative formula for a function has the form

Example.

Differentiate function .

Solution.

In this example, the complex function can be conventionally written as , where is the sine function, the third power function, the base e logarithm function, the arctangent function and the linear function, respectively.

According to the formula for the derivative of a complex function

Now we find

Let's put together the obtained intermediate results:

There is nothing scary, analyze complex functions like nesting dolls.

This could be the end of the article, if not for one thing...

It is advisable to clearly understand when to apply the rules of differentiation and the table of derivatives, and when to apply the formula for the derivative of a complex function.

BE EXTREMELY CAREFUL NOW. We will talk about the difference between complex functions and complex functions. Your success in finding derivatives will depend on how much you see this difference.

Let's start with simple examples. Function can be considered as complex: g(x) = tanx , . Therefore, you can immediately apply the formula for the derivative of a complex function

And here is the function It can no longer be called complex.

This function is the sum of three functions, 3tgx and 1. Although - is a complex function: - a power function (quadratic parabola), and f is a tangent function. Therefore, first we apply the sum differentiation formula:

It remains to find the derivative of the complex function:

That's why .

We hope you get the gist.

If we look more broadly, it can be argued that functions of a complex type can be part of complex functions, and complex functions can be components of functions of a complex type.

As an example, let us analyze the function into its component parts .

Firstly, this is a complex function that can be represented as , where f is the base 3 logarithm function, and g(x) is the sum of two functions And . That is, .

Secondly, let's deal with the function h(x) . It represents a relationship to .

This is the sum of two functions and , Where - a complex function with a numerical coefficient of 3. - cube function, - cosine function, - linear function.

This is the sum of two functions and , where - complex function, - exponential function, - power function.

Thus, .

Third, go to , which is the product of a complex function and the whole rational function

The squaring function is the logarithm function to base e.

Hence, .

Let's summarize:

Now the structure of the function is clear and it has become clear which formulas and in what sequence to apply when differentiating it.

In the section on differentiating a function (finding the derivative) you can familiarize yourself with the solution to similar problems.

Very easy to remember.

Well, let’s not go far, let’s immediately consider the inverse function. Which function is the inverse of the exponential function? Logarithm:

In our case, the base is the number:

Such a logarithm (that is, a logarithm with a base) is called “natural”, and we use a special notation for it: we write instead.

What is it equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

1. Find the derivative of the function.
2. What is the derivative of the function?

Answers: The exponential and natural logarithm are uniquely simple functions from a derivative perspective. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.

Rules of differentiation

Rules of what? Again a new term, again?!...

Differentiation is the process of finding the derivative.

That's all. What else can you call this process in one word? Not derivative... Mathematicians call the differential the same increment of a function at. This term comes from the Latin differentia - difference. Here.

When deriving all these rules, we will use two functions, for example, and. We will also need formulas for their increments:

There are 5 rules in total.

The constant is taken out of the derivative sign.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let it be, or simpler.

Examples.

Find the derivatives of the functions:

1. at a point;
2. at a point;
3. at a point;
4. at the point.

Solutions:

1. (the derivative is the same at all points, since it is a linear function, remember?);

Derivative of the product

Everything is similar here: let’s introduce a new function and find its increment:

Derivative:

Examples:

1. Find the derivatives of the functions and;
2. Find the derivative of the function at a point.

Solutions:

Derivative of an exponential function

Now your knowledge is enough to learn how to find the derivative of any exponential function, and not just exponents (have you forgotten what that is yet?).

So, where is some number.

We already know the derivative of the function, so let's try to reduce our function to a new base:

To do this, we will use a simple rule: . Then:

Well, it worked. Now try to find the derivative, and don't forget that this function is complex.

Happened?

Here, check yourself:

The formula turned out to be very similar to the derivative of an exponent: as it was, it remains the same, only a factor appeared, which is just a number, but not a variable.

Examples:
Find the derivatives of the functions:

Answers:

This is just a number that cannot be calculated without a calculator, that is, it cannot be written down in a simpler form. Therefore, we leave it in this form in the answer.

Note that here is the quotient of two functions, so we apply the corresponding differentiation rule:

In this example, the product of two functions:

Derivative of a logarithmic function

It’s similar here: you already know the derivative of the natural logarithm:

Therefore, to find an arbitrary logarithm with a different base, for example:

We need to reduce this logarithm to the base. How do you change the base of a logarithm? I hope you remember this formula:

Only now we will write instead:

The denominator is simply a constant (a constant number, without a variable). The derivative is obtained very simply:

Derivatives of exponential and logarithmic functions are almost never found in the Unified State Examination, but it will not be superfluous to know them.

Derivative of a complex function.

What is a "complex function"? No, this is not a logarithm, and not an arctangent. These functions can be difficult to understand (although if you find the logarithm difficult, read the topic “Logarithms” and you will be fine), but from a mathematical point of view, the word “complex” does not mean “difficult”.

Imagine a small conveyor belt: two people are sitting and doing some actions with some objects. For example, the first one wraps a chocolate bar in a wrapper, and the second one ties it with a ribbon. The result is a composite object: a chocolate bar wrapped and tied with a ribbon. To eat a chocolate bar, you need to do the reverse steps in reverse order.

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then square the resulting number. So, we are given a number (chocolate), I find its cosine (wrapper), and then you square what I got (tie it with a ribbon). What happened? Function. This is an example of a complex function: when, to find its value, we perform the first action directly with the variable, and then a second action with what resulted from the first.

In other words, a complex function is a function whose argument is another function: .

For our example, .

We can easily do the same steps in reverse order: first you square it, and I then look for the cosine of the resulting number: . It’s easy to guess that the result will almost always be different. An important feature of complex functions: when the order of actions changes, the function changes.

Second example: (same thing). .

The action we do last will be called "external" function, and the action performed first - accordingly "internal" function(these are informal names, I use them only to explain the material in simple language).

Try to determine for yourself which function is external and which internal:

Answers: Separating inner and outer functions is very similar to changing variables: for example, in a function

1. What action will we perform first? First, let's calculate the sine, and only then cube it. This means that it is an internal function, but an external one.
   And the original function is their composition: .
2. Internal: ; external: .
   Examination: .
3. Internal: ; external: .
   Examination: .
4. Internal: ; external: .
   Examination: .
5. Internal: ; external: .
   Examination: .

We change variables and get a function.

Well, now we will extract our chocolate bar and look for the derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. In relation to the original example, it looks like this:

Another example:

So, let's finally formulate the official rule:

Algorithm for finding the derivative of a complex function:

It seems simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

(Just don’t try to cut it by now! Nothing comes out from under the cosine, remember?)

3) Internal: ;

External: ;

It is immediately clear that this is a three-level complex function: after all, this is already a complex function in itself, and we also extract the root from it, that is, we perform the third action (put the chocolate in a wrapper and with a ribbon in the briefcase). But there is no reason to be afraid: we will still “unpack” this function in the same order as usual: from the end.

That is, first we differentiate the root, then the cosine, and only then the expression in brackets. And then we multiply it all.

In such cases, it is convenient to number the actions. That is, let's imagine what we know. In what order will we perform actions to calculate the value of this expression? Let's look at an example:

The later the action is performed, the more “external” the corresponding function will be. The sequence of actions is the same as before:

Here the nesting is generally 4-level. Let's determine the course of action.

1. Radical expression. .

2. Root. .

3. Sine. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN THINGS

Derivative of a function- the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument:

Basic derivatives:

Rules of differentiation:

The constant is taken out of the derivative sign:

Derivative of the sum:

Derivative of the product:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

1. We define the “internal” function and find its derivative.
2. We define the “external” function and find its derivative.
3. We multiply the results of the first and second points.

Functions of a complex type do not always fit the definition of a complex function. If there is a function of the form y = sin x - (2 - 3) · a r c t g x x 5 7 x 10 - 17 x 3 + x - 11, then it cannot be considered complex, unlike y = sin 2 x.

This article will show the concept of a complex function and its identification. Let's work with formulas for finding the derivative with examples of solutions in the conclusion. The use of the derivative table and differentiation rules significantly reduces the time for finding the derivative.

Basic definitions

Definition 1

A complex function is one whose argument is also a function.

It is denoted this way: f (g (x)). We have that the function g (x) is considered an argument f (g (x)).

Definition 2

If there is a function f and it is a cotangent function, then g(x) = ln x is the natural logarithm function. We find that the complex function f (g (x)) will be written as arctg(lnx). Or a function f, which is a function raised to the 4th power, where g (x) = x 2 + 2 x - 3 is considered an entire rational function, we obtain that f (g (x)) = (x 2 + 2 x - 3) 4 .

Obviously g(x) can be complex. From the example y = sin 2 x + 1 x 3 - 5 it is clear that the value of g has the cube root of the fraction. This expression can be denoted as y = f (f 1 (f 2 (x))). From where we have that f is a sine function, and f 1 is a function located under the square root, f 2 (x) = 2 x + 1 x 3 - 5 is a fractional rational function.

Definition 3

The degree of nesting is determined by any natural number and is written as y = f (f 1 (f 2 (f 3 (. . . (f n (x)))))) .

Definition 4

The concept of function composition refers to the number of nested functions according to the conditions of the problem. To solve, use the formula for finding the derivative of a complex function of the form

(f (g (x))) " = f " (g (x)) g " (x)

Examples

Example 1

Find the derivative of a complex function of the form y = (2 x + 1) 2.

Solution

The condition shows that f is a squaring function, and g(x) = 2 x + 1 is considered a linear function.

Let's apply the derivative formula for a complex function and write:

f " (g (x)) = ((g (x)) 2) " = 2 (g (x)) 2 - 1 = 2 g (x) = 2 (2 x + 1) ; g " (x) = (2 x + 1) " = (2 x) " + 1 " = 2 x " + 0 = 2 1 x 1 - 1 = 2 ⇒ (f (g (x))) " = f " (g (x)) g " (x) = 2 (2 x + 1) 2 = 8 x + 4

It is necessary to find the derivative with a simplified original form of the function. We get:

y = (2 x + 1) 2 = 4 x 2 + 4 x + 1

From here we have that

y " = (4 x 2 + 4 x + 1) " = (4 x 2) " + (4 x) " + 1 " = 4 (x 2) " + 4 (x) " + 0 = = 4 · 2 · x 2 - 1 + 4 · 1 · x 1 - 1 = 8 x + 4

The results were the same.

When solving problems of this type, it is important to understand where the function of the form f and g (x) will be located.

Example 2

You should find the derivatives of complex functions of the form y = sin 2 x and y = sin x 2.

Solution

The first function notation says that f is the squaring function and g(x) is the sine function. Then we get that

y " = (sin 2 x) " = 2 sin 2 - 1 x (sin x) " = 2 sin x cos x

The second entry shows that f is a sine function, and g(x) = x 2 denotes a power function. It follows that we write the product of a complex function as

y " = (sin x 2) " = cos (x 2) (x 2) " = cos (x 2) 2 x 2 - 1 = 2 x cos (x 2)

The formula for the derivative y = f (f 1 (f 2 (f 3 (. . . (f n (x))))) will be written as y " = f " (f 1 (f 2 (f 3 (. . . ( f n (x))))) · f 1 " (f 2 (f 3 (. . . (f n (x)))) · · f 2 " (f 3 (. . . (f n (x))) )) · . . . fn "(x)

Example 3

Find the derivative of the function y = sin (ln 3 a r c t g (2 x)).

Solution

This example shows the difficulty of writing and determining the location of functions. Then y = f (f 1 (f 2 (f 3 (f 4 (x))))) denote where f , f 1 , f 2 , f 3 , f 4 (x) is the sine function, the function of raising to 3 degree, function with logarithm and base e, arctangent and linear function.

From the formula for defining a complex function we have that

y " = f " (f 1 (f 2 (f 3 (f 4 (x)))) f 1 " (f 2 (f 3 (f 4 (x)))) f 2 " (f 3 (f 4 (x)) f 3 " (f 4 (x)) f 4 " (x)

We get what we need to find

1. f " (f 1 (f 2 (f 3 (f 4 (x))))) as the derivative of the sine according to the table of derivatives, then f " (f 1 (f 2 (f 3 (f 4 (x)))) ) = cos (ln 3 a r c t g (2 x)) .
2. f 1 " (f 2 (f 3 (f 4 (x)))) as the derivative of a power function, then f 1 " (f 2 (f 3 (f 4 (x)))) = 3 ln 3 - 1 a r c t g (2 x) = 3 ln 2 a r c t g (2 x) .
3. f 2 " (f 3 (f 4 (x))) as a logarithmic derivative, then f 2 " (f 3 (f 4 (x))) = 1 a r c t g (2 x) .
4. f 3 " (f 4 (x)) as the derivative of the arctangent, then f 3 " (f 4 (x)) = 1 1 + (2 x) 2 = 1 1 + 4 x 2.
5. When finding the derivative f 4 (x) = 2 x, remove 2 from the sign of the derivative using the formula for the derivative of a power function with an exponent equal to 1, then f 4 " (x) = (2 x) " = 2 x " = 2 · 1 · x 1 - 1 = 2 .

We combine the intermediate results and get that

y " = f " (f 1 (f 2 (f 3 (f 4 (x)))) f 1 " (f 2 (f 3 (f 4 (x)))) f 2 " (f 3 (f 4 (x)) f 3 " (f 4 (x)) f 4 " (x) = = cos (ln 3 a r c t g (2 x)) 3 ln 2 a r c t g (2 x) 1 a r c t g (2 x) 1 1 + 4 x 2 2 = = 6 cos (ln 3 a r c t g (2 x)) ln 2 a r c t g (2 x) a r c t g (2 x) (1 + 4 x 2)

Analysis of such functions is reminiscent of nesting dolls. Differentiation rules cannot always be applied explicitly using a derivative table. Often you need to use a formula for finding derivatives of complex functions.

There are some differences between complex appearance and complex functions. With a clear ability to distinguish this, finding derivatives will be especially easy.

Example 4

It is necessary to consider giving such an example. If there is a function of the form y = t g 2 x + 3 t g x + 1, then it can be considered as a complex function of the form g (x) = t g x, f (g) = g 2 + 3 g + 1. Obviously, it is necessary to use the formula for a complex derivative:

f " (g (x)) = (g 2 (x) + 3 g (x) + 1) " = (g 2 (x)) " + (3 g (x)) " + 1 " = = 2 · g 2 - 1 (x) + 3 g " (x) + 0 = 2 g (x) + 3 1 g 1 - 1 (x) = = 2 g (x) + 3 = 2 t g x + 3 ; g " (x) = (t g x) " = 1 cos 2 x ⇒ y " = (f (g (x))) " = f " (g (x)) g " (x) = (2 t g x + 3 ) · 1 cos 2 x = 2 t g x + 3 cos 2 x

A function of the form y = t g x 2 + 3 t g x + 1 is not considered complex, since it has the sum of t g x 2, 3 t g x and 1. However, t g x 2 is considered a complex function, then we obtain a power function of the form g (x) = x 2 and f, which is a tangent function. To do this, differentiate by amount. We get that

y " = (t g x 2 + 3 t g x + 1) " = (t g x 2) " + (3 t g x) " + 1 " = = (t g x 2) " + 3 (t g x) " + 0 = (t g x 2) " + 3 cos 2 x

Let's move on to finding the derivative of a complex function (t g x 2) ":

f " (g (x)) = (t g (g (x))) " = 1 cos 2 g (x) = 1 cos 2 (x 2) g " (x) = (x 2) " = 2 x 2 - 1 = 2 x ⇒ (t g x 2) " = f " (g (x)) g " (x) = 2 x cos 2 (x 2)

We get that y " = (t g x 2 + 3 t g x + 1) " = (t g x 2) " + 3 cos 2 x = 2 x cos 2 (x 2) + 3 cos 2 x

Functions of a complex type can be included in complex functions, and complex functions themselves can be components of functions of a complex type.

Example 5

For example, consider a complex function of the form y = log 3 x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 + ln 2 x (x 2 + 1)

This function can be represented as y = f (g (x)), where the value of f is a function of the base 3 logarithm, and g (x) is considered the sum of two functions of the form h (x) = x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 and k (x) = ln 2 x · (x 2 + 1) . Obviously, y = f (h (x) + k (x)).

Consider the function h(x). This is the ratio l (x) = x 2 + 3 cos 3 (2 x + 1) + 7 to m (x) = e x 2 + 3 3

We have that l (x) = x 2 + 3 cos 2 (2 x + 1) + 7 = n (x) + p (x) is the sum of two functions n (x) = x 2 + 7 and p (x) = 3 cos 3 (2 x + 1) , where p (x) = 3 p 1 (p 2 (p 3 (x))) is a complex function with numerical coefficient 3, and p 1 is a cube function, p 2 by a cosine function, p 3 (x) = 2 x + 1 by a linear function.

We found that m (x) = e x 2 + 3 3 = q (x) + r (x) is the sum of two functions q (x) = e x 2 and r (x) = 3 3, where q (x) = q 1 (q 2 (x)) is a complex function, q 1 is a function with an exponential, q 2 (x) = x 2 is a power function.

This shows that h (x) = l (x) m (x) = n (x) + p (x) q (x) + r (x) = n (x) + 3 p 1 (p 2 ( p 3 (x))) q 1 (q 2 (x)) + r (x)

When moving to an expression of the form k (x) = ln 2 x · (x 2 + 1) = s (x) · t (x), it is clear that the function is presented in the form of a complex s (x) = ln 2 x = s 1 ( s 2 (x)) with a rational integer t (x) = x 2 + 1, where s 1 is a squaring function, and s 2 (x) = ln x is logarithmic with base e.

It follows that the expression will take the form k (x) = s (x) · t (x) = s 1 (s 2 (x)) · t (x).

Then we get that

y = log 3 x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 + ln 2 x (x 2 + 1) = = f n (x) + 3 p 1 (p 2 (p 3 (x))) q 1 (q 2 (x)) = r (x) + s 1 (s 2 (x)) t (x)

Based on the structures of the function, it became clear how and what formulas need to be used to simplify the expression when differentiating it. To become familiar with such problems and for the concept of their solution, it is necessary to turn to the point of differentiating a function, that is, finding its derivative.

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