It's very easy to remember.

Well, we will not go far, we will immediately consider the inverse function. What is the inverse of the exponential function? Logarithm:

In our case, the base is a number:

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

What is 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 exponent and the natural logarithm are functions that are uniquely simple in terms of the derivative. 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.

Differentiation rules

What rules? Another new term, again?!...

Differentiation is the process of finding the derivative.

Only and everything. What is another word for this process? Not proizvodnovanie... The differential of mathematics is called the very increment of the 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 sign of the derivative.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let, or easier.

Examples.

Find derivatives of functions:

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

Solutions:

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

Derivative of a product

Everything is similar here: we introduce a new function and find its increment:

Derivative:

Examples:

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

Solutions:

Derivative of exponential function

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

So where is some number.

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

To do this, we 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 the exponent: as it was, it remains, only a factor appeared, which is just a number, but not a variable.

Examples:
Find derivatives of functions:

Answers:

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

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

In this example, the product of two functions:

Derivative of a logarithmic function

Here it is similar: you already know the derivative of the natural logarithm:

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

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

Only now instead of we will write:

The denominator turned out to be just a constant (a constant number, without a variable). The derivative is very simple:

Derivatives of the exponential and logarithmic functions are almost never found in the exam, 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 arc tangent. These functions can be difficult to understand (although if the logarithm seems difficult to you, read the topic "Logarithms" and everything will work out), but in terms of mathematics, the word "complex" does not mean "difficult".

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

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then we will square the resulting number. So, they give us 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, in order to find its value, we do the first action directly with the variable, and then another second action with what happened as a result of the first.

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

For our example, .

We may well do the same actions in reverse order: first you square, and then I look for the cosine of the resulting number:. It is 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). .

The last action we do will be called "external" function, and the action performed first - respectively "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 is internal:

Answers: The separation of inner and outer functions is very similar to changing variables: for example, in the function

1. What action will we take first? First we calculate the sine, and only then we raise it to a cube. So it's an internal function, not 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 - 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. For 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 to be simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

(just don’t try to reduce by now! Nothing is taken out from under the cosine, remember?)

3) Internal: ;

External: ;

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

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

1. Radical expression. .

2. Root. .

3. Sinus. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN

Function derivative- the ratio of the increment of the function to the increment of the argument with an infinitesimal increment of the argument:

Basic derivatives:

Differentiation rules:

The constant is taken out of the sign of the derivative:

Derivative of sum:

Derivative 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, find its derivative.
2. We define the "external" function, find its derivative.
3. We multiply the results of the first and second points.

And the theorem on the derivative of a complex function, the formulation of which is as follows:

Let 1) the function $u=\varphi (x)$ has a derivative $u_(x)"=\varphi"(x_0)$ at some point $x_0$, 2) the function $y=f(u)$ has at the corresponding point $u_0=\varphi (x_0)$ the derivative $y_(u)"=f"(u)$. Then the complex function $y=f\left(\varphi (x) \right)$ at the mentioned point will also have a derivative equal to the product of the derivatives of the functions $f(u)$ and $\varphi (x)$:

$$ \left(f(\varphi (x))\right)"=f_(u)"\left(\varphi (x_0) \right)\cdot \varphi"(x_0) $$

or, in shorter notation: $y_(x)"=y_(u)"\cdot u_(x)"$.

In the examples of this section, all functions have the form $y=f(x)$ (ie, we consider only functions of one variable $x$). Accordingly, in all examples, the derivative $y"$ is taken with respect to the variable $x$. To emphasize that the derivative is taken with respect to the variable $x$, one often writes $y"_x$ instead of $y"$.

Examples #1, #2, and #3 provide a detailed process for finding the derivative of complex functions. Example No. 4 is intended for a more complete understanding of the derivatives table and it makes sense to familiarize yourself with it.

It is advisable, after studying the material in examples No. 1-3, to move on to independently solving examples No. 5, No. 6 and No. 7. Examples #5, #6 and #7 contain a short solution so that the reader can check the correctness of his result.

Example #1

Find the derivative of the function $y=e^(\cos x)$.

We need to find the derivative of the complex function $y"$. Since $y=e^(\cos x)$, then $y"=\left(e^(\cos x)\right)"$. To find the derivative $ \left(e^(\cos x)\right)"$ use formula #6 from the table of derivatives. In order to use formula No. 6, you need to take into account that in our case $u=\cos x$. The further solution consists in a banal substitution of the expression $\cos x$ instead of $u$ into formula No. 6:

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)" \tag (1.1)$$

Now we need to find the value of the expression $(\cos x)"$. Again we turn to the table of derivatives, choosing formula No. 10 from it. Substituting $u=x$ into formula No. 10, we have: $(\cos x)"=-\ sin x\cdot x"$. Now we continue equality (1.1), supplementing it with the found result:

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x") \tag (1.2) $$

Since $x"=1$, we continue equality (1.2):

$$ y"=\left(e^(\cos x) \right)"=e^(\cos x)\cdot (\cos x)"= e^(\cos x)\cdot (-\sin x \cdot x")=e^(\cos x)\cdot (-\sin x\cdot 1)=-\sin x\cdot e^(\cos x) \tag (1.3) $$

So, from equality (1.3) we have: $y"=-\sin x\cdot e^(\cos x)$. Naturally, explanations and intermediate equalities are usually skipped, writing the derivative in one line, as in the equality ( 1.3) So, the derivative of the complex function has been found, it remains only to write down the answer.

Answer: $y"=-\sin x\cdot e^(\cos x)$.

Example #2

Find the derivative of the function $y=9\cdot \arctg^(12)(4\cdot \ln x)$.

We need to calculate the derivative $y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"$. To begin with, we note that the constant (i.e. the number 9) can be taken out of the sign of the derivative:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)" \tag (2.1) $$

Now let's turn to the expression $\left(\arctg^(12)(4\cdot \ln x) \right)"$. To make it easier to select the desired formula from the table of derivatives, I will present the expression in question in this form: $\left( \left(\arctg(4\cdot \ln x) \right)^(12)\right)"$. Now it is clear that it is necessary to use formula No. 2, i.e. $\left(u^\alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. Substitute $u=\arctg(4\cdot \ln x)$ and $\alpha=12$ into this formula:

Complementing equality (2.1) with the obtained result, we have:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"= 108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))" \tag (2.2) $$

In this situation, a mistake is often made when the solver at the first step chooses the formula $(\arctg \; u)"=\frac(1)(1+u^2)\cdot u"$ instead of the formula $\left(u^\ alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. The point is that the derivative of the external function must be found first. To understand which function will be external to the expression $\arctg^(12)(4\cdot 5^x)$, imagine that you are counting the value of the expression $\arctg^(12)(4\cdot 5^x)$ for some value of $x$. First you calculate the value of $5^x$, then multiply the result by 4 to get $4\cdot 5^x$. Now we take the arctangent from this result, getting $\arctg(4\cdot 5^x)$. Then we raise the resulting number to the twelfth power, getting $\arctg^(12)(4\cdot 5^x)$. The last action, i.e. raising to the power of 12, - and will be an external function. And it is from it that one should begin finding the derivative, which was done in equality (2.2).

Now we need to find $(\arctg(4\cdot \ln x))"$. We use formula No. 19 of the derivatives table, substituting $u=4\cdot \ln x$ into it:

$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)" $$

Let's slightly simplify the resulting expression, taking into account $(4\cdot \ln x)^2=4^2\cdot (\ln x)^2=16\cdot \ln^2 x$.

$$ (\arctg(4\cdot \ln x))"=\frac(1)(1+(4\cdot \ln x)^2)\cdot (4\cdot \ln x)"=\frac( 1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" $$

Equality (2.2) will now become:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" \tag (2.3) $$

It remains to find $(4\cdot \ln x)"$. We take the constant (i.e. 4) out of the sign of the derivative: $(4\cdot \ln x)"=4\cdot (\ln x)"$. For In order to find $(\ln x)"$, we use formula No. 8, substituting $u=x$ into it: $(\ln x)"=\frac(1)(x)\cdot x"$. Since $x"=1$, then $(\ln x)"=\frac(1)(x)\cdot x"=\frac(1)(x)\cdot 1=\frac(1)(x )$ Substituting the result obtained into formula (2.3), we obtain:

$$ y"=\left(9\cdot \arctg^(12)(4\cdot \ln x) \right)"=9\cdot\left(\arctg^(12)(4\cdot \ln x) \right)"=\\ =108\cdot\left(\arctg(4\cdot \ln x) \right)^(11)\cdot (\arctg(4\cdot \ln x))"=108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot (4\cdot \ln x)" =\\ =108\cdot \left(\arctg(4\cdot \ln x) \right)^(11)\cdot \frac(1)(1+16\cdot \ln^2 x)\cdot 4\ cdot \frac(1)(x)=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x)).$ $

Let me remind you that the derivative of a complex function is most often in one line, as written in the last equality. Therefore, when making standard calculations or tests, it is not at all necessary to paint the solution in the same detail.

Answer: $y"=432\cdot \frac(\arctg^(11)(4\cdot \ln x))(x\cdot (1+16\cdot \ln^2 x))$.

Example #3

Find $y"$ of the function $y=\sqrt(\sin^3(5\cdot9^x))$.

First, let's slightly transform the $y$ function by expressing the radical (root) as a power: $y=\sqrt(\sin^3(5\cdot9^x))=\left(\sin(5\cdot 9^x) \right)^(\frac(3)(7))$. Now let's start finding the derivative. Since $y=\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))$, then:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)" \tag (3.1) $$

We use formula No. 2 from the table of derivatives, substituting $u=\sin(5\cdot 9^x)$ and $\alpha=\frac(3)(7)$ into it:

$$ \left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"= \frac(3)(7)\cdot \left( \sin(5\cdot 9^x)\right)^(\frac(3)(7)-1) (\sin(5\cdot 9^x))"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" $$

We continue equality (3.1) using the obtained result:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))" \tag (3.2) $$

Now we need to find $(\sin(5\cdot 9^x))"$. For this, we use formula No. 9 from the table of derivatives, substituting $u=5\cdot 9^x$ into it:

$$ (\sin(5\cdot 9^x))"=\cos(5\cdot 9^x)\cdot(5\cdot 9^x)" $$

Complementing equality (3.2) with the obtained result, we have:

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))"=\\ =\frac(3) (7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9^x)\cdot(5\cdot 9 ^x)" \tag (3.3) $$

It remains to find $(5\cdot 9^x)"$. First, we take the constant (the number $5$) out of the sign of the derivative, i.e. $(5\cdot 9^x)"=5\cdot (9^x) "$. To find the derivative $(9^x)"$, we apply formula No. 5 of the table of derivatives, substituting $a=9$ and $u=x$ into it: $(9^x)"=9^x\cdot \ ln9\cdot x"$. Since $x"=1$, then $(9^x)"=9^x\cdot \ln9\cdot x"=9^x\cdot \ln9$. Now we can continue equality (3.3):

$$ y"=\left(\left(\sin(5\cdot 9^x)\right)^(\frac(3)(7))\right)"=\frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) (\sin(5\cdot 9^x))"=\\ =\frac(3) (7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9^x)\cdot(5\cdot 9 ^x)"= \frac(3)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7)) \cos(5\cdot 9 ^x)\cdot 5\cdot 9^x\cdot \ln9=\\ =\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right) ^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x. $$

You can return from powers to radicals (i.e. roots) again by writing $\left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))$ as $\ frac(1)(\left(\sin(5\cdot 9^x)\right)^(\frac(4)(7)))=\frac(1)(\sqrt(\sin^4(5\ cdot 9^x)))$. Then the derivative will be written in the following form:

$$ y"=\frac(15\cdot \ln 9)(7)\cdot \left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))\cdot \cos(5\cdot 9^x)\cdot 9^x= \frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x) (\sqrt(\sin^4(5\cdot 9^x))). $$

Answer: $y"=\frac(15\cdot \ln 9)(7)\cdot \frac(\cos (5\cdot 9^x)\cdot 9^x)(\sqrt(\sin^4(5\ cdot 9^x)))$.

Example #4

Show that formulas No. 3 and No. 4 of the table of derivatives are a special case of formula No. 2 of this table.

In formula No. 2 of the table of derivatives, the derivative of the function $u^\alpha$ is written. Substituting $\alpha=-1$ into formula #2, we get:

$$(u^(-1))"=-1\cdot u^(-1-1)\cdot u"=-u^(-2)\cdot u"\tag (4.1)$$

Since $u^(-1)=\frac(1)(u)$ and $u^(-2)=\frac(1)(u^2)$, equality (4.1) can be rewritten as follows: $ \left(\frac(1)(u) \right)"=-\frac(1)(u^2)\cdot u"$. This is the formula number 3 of the derivatives table.

Let's turn again to formula No. 2 of the derivatives table. Substitute $\alpha=\frac(1)(2)$ into it:

$$\left(u^(\frac(1)(2))\right)"=\frac(1)(2)\cdot u^(\frac(1)(2)-1)\cdot u" =\frac(1)(2)u^(-\frac(1)(2))\cdot u"\tag (4.2) $$

Since $u^(\frac(1)(2))=\sqrt(u)$ and $u^(-\frac(1)(2))=\frac(1)(u^(\frac( 1)(2)))=\frac(1)(\sqrt(u))$, then equality (4.2) can be rewritten as follows:

$$ (\sqrt(u))"=\frac(1)(2)\cdot \frac(1)(\sqrt(u))\cdot u"=\frac(1)(2\sqrt(u) )\cdot u" $$

The resulting equality $(\sqrt(u))"=\frac(1)(2\sqrt(u))\cdot u"$ is formula No. 4 of the derivatives table. As you can see, formulas No. 3 and No. 4 of the derivatives table are obtained from formula No. 2 by substituting the corresponding value of $\alpha$.

complex derivatives. Logarithmic derivative.
Derivative of exponential function

We continue to improve our differentiation technique. In this lesson, we will consolidate the material covered, consider more complex derivatives, and also get acquainted with new tricks and tricks for finding the derivative, in particular, with the logarithmic derivative.

Those readers who have a low level of preparation should refer to the article How to find the derivative? Solution examples which will allow you to raise your skills almost from scratch. Next, you need to carefully study the page Derivative of a compound function, understand and resolve All the examples I have given. This lesson is logically the third in a row, and after mastering it, you will confidently differentiate fairly complex functions. It is undesirable to stick to the position “Where else? Yes, and that's enough! ”, Since all the examples and solutions are taken from real tests and are often found in practice.

Let's start with repetition. At the lesson Derivative of a compound function we have considered a number of examples with detailed comments. In the course of studying differential calculus and other sections of mathematical analysis, you will have to differentiate very often, and it is not always convenient (and not always necessary) to paint examples in great detail. Therefore, we will practice in the oral finding of derivatives. The most suitable "candidates" for this are derivatives of the simplest of complex functions, for example:

According to the rule of differentiation of a complex function :

When studying other matan topics in the future, such a detailed record is most often not required, it is assumed that the student is able to find similar derivatives on autopilot. Let's imagine that at 3 o'clock in the morning the phone rang, and a pleasant voice asked: "What is the derivative of the tangent of two x?". This should be followed by an almost instantaneous and polite response: .

The first example will be immediately intended for an independent solution.

Example 1

Find the following derivatives orally, in one step, for example: . To complete the task, you only need to use table of derivatives of elementary functions(if she hasn't already remembered). If you have any difficulties, I recommend re-reading the lesson Derivative of a compound function.

, , ,
, , ,
, , ,

, , ,

, , ,

, , ,

, ,

Answers at the end of the lesson

Complex derivatives

After preliminary artillery preparation, examples with 3-4-5 attachments of functions will be less scary. Perhaps the following two examples will seem complicated to some, but if they are understood (someone suffers), then almost everything else in differential calculus will seem like a child's joke.

Example 2

Find the derivative of a function

As already noted, when finding the derivative of a complex function, first of all, it is necessary Right UNDERSTAND INVESTMENTS. In cases where there are doubts, I remind you of a useful trick: we take the experimental value "x", for example, and try (mentally or on a draft) to substitute this value into the "terrible expression".

1) First we need to calculate the expression, so the sum is the deepest nesting.

2) Then you need to calculate the logarithm:

4) Then cube the cosine:

5) At the fifth step, the difference:

6) And finally, the outermost function is the square root:

Complex Function Differentiation Formula are applied in reverse order, from the outermost function to the innermost. We decide:

Seems to be no error...

(1) We take the derivative of the square root.

(2) We take the derivative of the difference using the rule

(3) The derivative of the triple is equal to zero. In the second term, we take the derivative of the degree (cube).

(4) We take the derivative of the cosine.

(5) We take the derivative of the logarithm.

(6) Finally, we take the derivative of the deepest nesting .

It may seem too difficult, but this is not the most brutal example. Take, for example, Kuznetsov's collection and you will appreciate all the charm and simplicity of the analyzed derivative. I noticed that they like to give a similar thing at the exam to check whether the student understands how to find the derivative of a complex function, or does not understand.

The following example is for a standalone solution.

Example 3

Find the derivative of a function

Hint: First we apply the rules of linearity and the rule of differentiation of the product

Full solution and answer at the end of the lesson.

It's time to move on to something more compact and prettier.
It is not uncommon for a situation where the product of not two, but three functions is given in an example. How to find the derivative of the product of three factors?

Example 4

Find the derivative of a function

First, we look, but is it possible to turn the product of three functions into a product of two functions? For example, if we had two polynomials in the product, then we could open the brackets. But in this example, all functions are different: degree, exponent and logarithm.

In such cases, it is necessary successively apply the product differentiation rule twice

The trick is that for "y" we denote the product of two functions: , and for "ve" - ​​the logarithm:. Why can this be done? Is it - this is not the product of two factors and the rule does not work?! There is nothing complicated:

Now it remains to apply the rule a second time to bracket:

You can still pervert and take something out of the brackets, but in this case it is better to leave the answer in this form - it will be easier to check.

The above example can be solved in the second way:

Both solutions are absolutely equivalent.

Example 5

Find the derivative of a function

This is an example for an independent solution, in the sample it is solved in the first way.

Consider similar examples with fractions.

Example 6

Find the derivative of a function

Here you can go in several ways:

Or like this:

But the solution can be written more compactly if, first of all, we use the rule of differentiation of the quotient , taking for the whole numerator:

In principle, the example is solved, and if it is left in this form, it will not be a mistake. But if you have time, it is always advisable to check on a draft, but is it possible to simplify the answer? We bring the expression of the numerator to a common denominator and get rid of the three-story fraction:

The disadvantage of additional simplifications is that there is a risk of making a mistake not when finding a derivative, but when banal school transformations. On the other hand, teachers often reject the task and ask to “bring it to mind” the derivative.

A simpler example for a do-it-yourself solution:

Example 7

Find the derivative of a function

We continue to master the techniques for finding the derivative, and now we will consider a typical case when a “terrible” logarithm is proposed for differentiation

Example 8

Find the derivative of a function

Here you can go a long way, using the rule of differentiation of a complex function:

But the very first step immediately plunges you into despondency - you have to take an unpleasant derivative of a fractional degree, and then also from a fraction.

That's why before how to take the derivative of the “fancy” logarithm, it is previously simplified using well-known school properties:

! If you have a practice notebook handy, copy these formulas right there. If you don't have a notebook, draw them on a piece of paper, as the rest of the lesson's examples will revolve around these formulas.

The solution itself can be formulated like this:

Let's transform the function:

We find the derivative:

The preliminary transformation of the function itself greatly simplified the solution. Thus, when a similar logarithm is proposed for differentiation, it is always advisable to “break it down”.

And now a couple of simple examples for an independent solution:

Example 9

Find the derivative of a function

Example 10

Find the derivative of a function

All transformations and answers at the end of the lesson.

logarithmic derivative

If the derivative of logarithms is such sweet music, then the question arises, is it possible in some cases to organize the logarithm artificially? Can! And even necessary.

Example 11

Find the derivative of a function

Similar examples we have recently considered. What to do? One can successively apply the rule of differentiation of the quotient, and then the rule of differentiation of the product. The disadvantage of this method is that you get a huge three-story fraction, which you don’t want to deal with at all.

But in theory and practice there is such a wonderful thing as the logarithmic derivative. Logarithms can be organized artificially by "hanging" them on both sides:

Note : because function can take negative values, then, generally speaking, you need to use modules: , which disappear as a result of differentiation. However, the current design is also acceptable, where by default the complex values. But if with all rigor, then in both cases it is necessary to make a reservation that.

Now you need to “break down” the logarithm of the right side as much as possible (formulas in front of your eyes?). I will describe this process in great detail:

Let's start with the differentiation.
We conclude both parts with a stroke:

The derivative of the right side is quite simple, I will not comment on it, because if you are reading this text, you should be able to handle it with confidence.

What about the left side?

On the left side we have complex function. I foresee the question: “Why, is there one letter “y” under the logarithm?”.

The fact is that this “one letter y” - IS A FUNCTION IN ITSELF(if it is not very clear, refer to the article Derivative of a function implicitly specified). Therefore, the logarithm is an external function, and "y" is an internal function. And we use the compound function differentiation rule :

On the left side, as if by magic, we have a derivative. Further, according to the rule of proportion, we throw the “y” from the denominator of the left side to the top of the right side:

And now we remember what kind of "game"-function we talked about when differentiating? Let's look at the condition:

Final answer:

Example 12

Find the derivative of a function

This is a do-it-yourself example. Sample design of an example of this type at the end of the lesson.

With the help of the logarithmic derivative, it was possible to solve any of examples No. 4-7, another thing is that the functions there are simpler, and, perhaps, the use of the logarithmic derivative is not very justified.

Derivative of exponential function

We have not considered this function yet. An exponential function is a function that has and the degree and base depend on "x". A classic example that will be given to you in any textbook or at any lecture:

How to find the derivative of an exponential function?

It is necessary to use the technique just considered - the logarithmic derivative. We hang logarithms on both sides:

As a rule, the degree is taken out from under the logarithm on the right side:

As a result, on the right side we have a product of two functions, which will be differentiated according to the standard formula .

We find the derivative, for this we enclose both parts under strokes:

The next steps are easy:

Finally:

If some transformation is not entirely clear, please re-read the explanations of Example 11 carefully.

In practical tasks, the exponential function will always be more complicated than the considered lecture example.

Example 13

Find the derivative of a function

We use the logarithmic derivative.

On the right side we have a constant and the product of two factors - "x" and "logarithm of the logarithm of x" (another logarithm is nested under the logarithm). When differentiating a constant, as we remember, it is better to immediately take it out of the sign of the derivative so that it does not get in the way; and, of course, apply the familiar rule :

In this lesson, we will learn how to find derivative of a complex function. The lesson is a logical continuation of the lesson How to find the derivative?, on which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some technical methods for finding derivatives. Thus, if you are not very good with derivatives of functions or some points of this article are not entirely clear, then first read the above lesson. Please tune in to a serious mood - the material is not easy, but I will still try to present it simply and clearly.

In practice, you have to deal with the derivative of a complex function very often, I would even say almost always, when you are given tasks to find derivatives.

We look in the table at the rule (No. 5) for differentiating a complex function:

We understand. First of all, let's take a look at the notation. Here we have two functions - and , and the function, figuratively speaking, is nested in the function . A function of this kind (when one function is nested within another) is called a complex function.

I will call the function external function, and the function – inner (or nested) function.

! These definitions are not theoretical and should not appear in the final design of assignments. I use the informal expressions "external function", "internal" function only to make it easier for you to understand the material.

To clarify the situation, consider:

Example 1

Find the derivative of a function

Under the sine, we have not just the letter "x", but the whole expression, so finding the derivative immediately from the table will not work. We also notice that it is impossible to apply the first four rules here, there seems to be a difference, but the fact is that it is impossible to “tear apart” the sine:

In this example, already from my explanations, it is intuitively clear that the function is a complex function, and the polynomial is an internal function (embedding), and an external function.

First step, which must be performed when finding the derivative of a complex function is to understand which function is internal and which is external.

In the case of simple examples, it seems clear that a polynomial is nested under the sine. But what if it's not obvious? How to determine exactly which function is external and which is internal? To do this, I propose to use the following technique, which can be carried out mentally or on a draft.

Let's imagine that we need to calculate the value of the expression with a calculator (instead of one, there can be any number).

What do we calculate first? First of all you will need to perform the following action: , so the polynomial will be an internal function:

Secondly you will need to find, so the sine - will be an external function:

After we UNDERSTAND With inner and outer functions, it's time to apply the compound function differentiation rule.

We start to decide. From the lesson How to find the derivative? we remember that the design of the solution of any derivative always begins like this - we enclose the expression in brackets and put a stroke at the top right:

At first we find the derivative of the external function (sine), look at the table of derivatives of elementary functions and notice that . All tabular formulas are applicable even if "x" is replaced by a complex expression, in this case:

Note that the inner function has not changed, we do not touch it.

Well, it is quite obvious that

The final result of applying the formula looks like this:

The constant factor is usually placed at the beginning of the expression:

If there is any misunderstanding, write down the decision on paper and read the explanations again.

Example 2

Find the derivative of a function

Example 3

Find the derivative of a function

As always, we write:

We figure out where we have an external function, and where is an internal one. To do this, we try (mentally or on a draft) to calculate the value of the expression for . What needs to be done first? First of all, you need to calculate what the base is equal to:, which means that the polynomial is the internal function:

And, only then exponentiation is performed, therefore, the power function is an external function:

According to the formula, first you need to find the derivative of the external function, in this case, the degree. We are looking for the desired formula in the table:. We repeat again: any tabular formula is valid not only for "x", but also for a complex expression. Thus, the result of applying the rule of differentiation of a complex function is the following:

I emphasize again that when we take the derivative of the outer function, the inner function does not change:

Now it remains to find a very simple derivative of the inner function and “comb” the result a little:

Example 4

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

To consolidate the understanding of the derivative of a complex function, I will give an example without comments, try to figure it out on your own, reason, where is the external and where is the internal function, why are the tasks solved that way?

Example 5

a) Find the derivative of a function

b) Find the derivative of the function

Example 6

Find the derivative of a function

Here we have a root, and in order to differentiate the root, it must be represented as a degree. Thus, we first bring the function into the proper form for differentiation:

Analyzing the function, we come to the conclusion that the sum of three terms is an internal function, and exponentiation is an external function. We apply the rule of differentiation of a complex function:

The degree is again represented as a radical (root), and for the derivative of the internal function, we apply a simple rule for differentiating the sum:

Ready. You can also bring the expression to a common denominator in brackets and write everything as one fraction. It’s beautiful, of course, but when cumbersome long derivatives are obtained, it’s better not to do this (it’s easy to get confused, make an unnecessary mistake, and it will be inconvenient for the teacher to check).

Example 7

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

It is interesting to note that sometimes, instead of the rule for differentiating a complex function, one can use the rule for differentiating a quotient , but such a solution would look like a perversion funny. Here is a typical example:

Example 8

Find the derivative of a function

Here you can use the rule of differentiation of the quotient , but it is much more profitable to find the derivative through the rule of differentiation of a complex function:

We prepare the function for differentiation - we take out the minus sign of the derivative, and raise the cosine to the numerator:

Cosine is an internal function, exponentiation is an external function.
Let's use our rule:

We find the derivative of the inner function, reset the cosine back down:

Ready. In the considered example, it is important not to get confused in the signs. By the way, try to solve it with the rule , the answers must match.

Example 9

Find the derivative of a function

This is an example for self-solving (answer at the end of the lesson).

So far, we have considered cases where we had only one nesting in a complex function. In practical tasks, you can often find derivatives, where, like nesting dolls, one inside the other, 3 or even 4-5 functions are nested at once.

Example 10

Find the derivative of a function

We understand the attachments of this function. We try to evaluate the expression using the experimental value . How would we count on a calculator?

First you need to find, which means that the arcsine is the deepest nesting:

This arcsine of unity should then be squared:

And finally, we raise the seven to the power:

That is, in this example we have three different functions and two nestings, while the innermost function is the arcsine, and the outermost function is the exponential function.

We start to decide

According to the rule, you first need to take the derivative of the external function. We look at the table of derivatives and find the derivative of the exponential function: The only difference is that instead of "x" we have a complex expression, which does not negate the validity of this formula. So, the result of applying the rule of differentiation of a complex function is the following:

Under the dash, we have a tricky function again! But it's already easier. It is easy to see that the inner function is the arcsine and the outer function is the degree. According to the rule of differentiation of a complex function, you first need to take the derivative of the degree.

Since you came here, you probably already managed to see this formula in the textbook

and make a face like this:

Friend, don't worry! In fact, everything is simple to disgrace. You will definitely understand everything. Only one request - read the article slowly try to understand every step. I wrote as simply and clearly as possible, but you still need to delve into the idea. And be sure to solve the tasks from the article.

What is a complex function?

Imagine that you are moving to another apartment and therefore you are packing things in big boxes. Let it be necessary to collect some small items, for example, school stationery. If you just throw them in a huge box, they will get lost among other things. To avoid this, you first put them, for example, in a bag, which you then put in a large box, after which you seal it. This "hardest" process is shown in the diagram below:

It would seem, where does the mathematics? And besides, a complex function is formed in EXACTLY THE SAME way! Only we “pack” not notebooks and pens, but \ (x \), while different “packages” and “boxes” serve.

For example, let's take x and "pack" it into a function:

As a result, we get, of course, \(\cos⁡x\). This is our "bag of things". And now we put it in a "box" - we pack it, for example, into a cubic function.

What will happen in the end? Yes, that's right, there will be a "package with things in a box", that is, "cosine of x cubed."

The resulting construction is a complex function. It differs from the simple one in that SEVERAL “impacts” (packages) are applied to one X in a row and it turns out, as it were, “a function from a function” - “a package in a package”.

In the school course, there are very few types of these same “packages”, only four:

Let's now "pack" x first into an exponential function with base 7, and then into a trigonometric function. We get:

\(x → 7^x → tg⁡(7^x)\)

And now let's “pack” x twice into trigonometric functions, first into and then into:

\(x → sin⁡x → ctg⁡ (sin⁡x)\)

Simple, right?

Now write the functions yourself, where x:
- first it is “packed” into a cosine, and then into an exponential function with base \(3\);
- first to the fifth power, and then to the tangent;
- first to the base logarithm \(4\) , then to the power \(-2\).

See the answers to this question at the end of the article.

But can we "pack" x not two, but three times? No problem! And four, and five, and twenty-five times. Here, for example, is a function in which x is "packed" \(4\) times:

\(y=5^(\log_2⁡(\sin⁡(x^4)))\)

But such formulas will not be found in school practice (students are more fortunate - they can be more difficult☺).

"Unpacking" a complex function

Look at the previous function again. Can you figure out the sequence of "packing"? What X was stuffed into first, what then, and so on until the very end. That is, which function is nested in which? Take a piece of paper and write down what you think. You can do this with a chain of arrows, as we wrote above, or in any other way.

Now the correct answer is: first x was “packed” into the \(4\)th power, then the result was packed into the sine, it, in turn, was placed in the logarithm base \(2\), and in the end the whole construction was shoved into the power fives.

That is, it is necessary to unwind the sequence IN THE REVERSE ORDER. And here is a hint how to do it easier: just look at the X - you have to dance from it. Let's look at a few examples.

For example, here is a function: \(y=tg⁡(\log_2⁡x)\). We look at X - what happens to him first? Taken from him. And then? The tangent of the result is taken. And the sequence will be the same:

\(x → \log_2⁡x → tg⁡(\log_2⁡x)\)

Another example: \(y=\cos⁡((x^3))\). We analyze - first x was cubed, and then the cosine was taken from the result. So the sequence will be: \(x → x^3 → \cos⁡((x^3))\). Pay attention, the function seems to be similar to the very first one (where with pictures). But this is a completely different function: here in the cube x (that is, \(\cos⁡((x x x)))\), and there in the cube the cosine \(x\) (that is, \(\cos⁡ x·\cos⁡x·\cos⁡x\)). This difference arises from different "packing" sequences.

The last example (with important information in it): \(y=\sin⁡((2x+5))\). It is clear that here we first performed arithmetic operations with x, then the sine was taken from the result: \(x → 2x+5 → \sin⁡((2x+5))\). And this is an important point: despite the fact that arithmetic operations are not functions in themselves, here they also act as a way of “packing”. Let's delve a little deeper into this subtlety.

As I said above, in simple functions x is "packed" once, and in complex functions - two or more. Moreover, any combination of simple functions (that is, their sum, difference, multiplication or division) is also a simple function. For example, \(x^7\) is a simple function, and so is \(ctg x\). Hence, all their combinations are simple functions:

\(x^7+ ctg x\) - simple,
\(x^7 ctg x\) is simple,
\(\frac(x^7)(ctg x)\) is simple, and so on.

However, if one more function is applied to such a combination, it will already be a complex function, since there will be two “packages”. See diagram:

Okay, let's get on with it now. Write the sequence of "wrapping" functions:
\(y=cos(⁡(sin⁡x))\)
\(y=5^(x^7)\)
\(y=arctg⁡(11^x)\)
\(y=log_2⁡(1+x)\)
The answers are again at the end of the article.

Internal and external functions

Why do we need to understand function nesting? What does this give us? The point is that without such an analysis we will not be able to reliably find the derivatives of the functions discussed above.

And in order to move on, we will need two more concepts: internal and external functions. This is a very simple thing, moreover, in fact, we have already analyzed them above: if we recall our analogy at the very beginning, then the inner function is the “package”, and the outer one is the “box”. Those. what X is “wrapped” in first is an internal function, and what the internal is “wrapped” in is already external. Well, it’s understandable why - it’s outside, it means external.

Here in this example: \(y=tg⁡(log_2⁡x)\), the function \(\log_2⁡x\) is internal, and
- external.

And in this one: \(y=\cos⁡((x^3+2x+1))\), \(x^3+2x+1\) is internal, and
- external.

Perform the last practice of analyzing complex functions, and finally, let's move on to the point for which everything was started - we will find derivatives of complex functions:

Fill in the gaps in the table:

Derivative of a compound function

Bravo to us, we still got to the "boss" of this topic - in fact, the derivative of a complex function, and specifically, to that very terrible formula from the beginning of the article.☺

\((f(g(x)))"=f"(g(x))\cdot g"(x)\)

This formula reads like this:

The derivative of a complex function is equal to the product of the derivative of the external function with respect to the constant internal function and the derivative of the internal function.

And immediately look at the parsing scheme "by words" to understand what to relate to:

I hope the terms "derivative" and "product" do not cause difficulties. "Complex function" - we have already dismantled. The catch is in the "derivative of an external function with respect to a constant internal function." What it is?

Answer: this is the usual derivative of the outer function, in which only the outer function changes, while the inner one remains the same. Still unclear? Okay, let's take an example.

Let's say we have a function \(y=\sin⁡(x^3)\). It is clear that the inner function here is \(x^3\), and the outer
. Let us now find the derivative of the outer with respect to the constant inner.