Finding the derivative of a complex function, examples with solutions. Examples of using the formula for the derivative of a complex function

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

Let 1) the function $u=\varphi (x)$ have at some point $x_0$ the derivative $u_(x)"=\varphi"(x_0)$, 2) the function $y=f(u)$ have at corresponding point$u_0=\varphi (x_0)$ 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 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 in this section, all functions have the form $y=f(x)$ (i.e., 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 $x$, $y"_x$ is often written instead of $y"$.

Examples No. 1, No. 2 and No. 3 outline detailed process finding the derivative complex functions. Example No. 4 is intended for a more complete understanding of the derivative 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 independent decision examples No. 5, No. 6 and No. 7. Examples No. 5, No. 6 and No. 7 contain short solution so that the reader can check the correctness of his result.

Example No. 1

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

We need to find the derivative of a complex function $y"$. Since $y=e^(\cos x)$, then $y"=\left(e^(\cos x)\right)"$. To find the derivative $ \left(e^(\cos x)\right)"$ we use formula No. 6 from the table of derivatives. In order to use formula No. 6, we need to take into account that in our case $u=\cos x$. The further solution consists in simply substituting 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 $(\cos x)"$. We turn again 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 let's continue equality (1.1), supplementing it with the result found:

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

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 down the finding of the derivative in one line, as in the equality ( 1.3) So, the derivative of a complex function has been found, all that remains is to write down the answer.

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

Example No. 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 derivative sign:

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

Now let's turn to the expression $\left(\arctg^(12)(4\cdot \ln x) \right)"$. To select the required formula from the table of derivatives it was easier, 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. $\left(u^\alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. In this formula we substitute $u=\arctg(4 \cdot \ln x)$ and $\alpha=12$:

Supplementing equality (2.1) with the result obtained, 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 come first. To understand which function will be external to the expression $\arctg^(12)(4\cdot 5^x)$, imagine that you are calculating the value of the expression $\arctg^(12)(4\cdot 5^x)$ at some value $x$. First you will calculate the value of $5^x$, then multiply the result by 4, getting $4\cdot 5^x$. Now we take the arctangent from this result, obtaining $\arctg(4\cdot 5^x)$. Then we raise the resulting number to the twelfth power, getting $\arctg^(12)(4\cdot 5^x)$. Last action, - i.e. raising to the power of 12 will be an external function. And it is from this that we must begin to find 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 simplify the resulting expression a little, 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:

It remains to find $(4\cdot \ln x)"$. Let's take the constant (i.e. 4) out of the derivative sign: $(4\cdot \ln x)"=4\cdot (\ln x)"$. 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 found in one line, as written in the last equality. Therefore, when preparing standard calculations or control work, it is not at all necessary to describe the solution in such detail.

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

Example No. 3

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

First, let's slightly transform the function $y$, 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:

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

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

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

Let us continue equality (3.1) using the result obtained:

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

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:

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

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

It remains to find $(5\cdot 9^x)"$. First, let's take the constant (the number $5$) outside the derivative sign, i.e. $(5\cdot 9^x)"=5\cdot (9^x) "$. To find the derivative $(9^x)"$, 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):

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

We can again return from powers to radicals (i.e., roots), writing $\left(\sin(5\cdot 9^x)\right)^(-\frac(4)(7))$ in the form $\ frac(1)(\left(\sin(5\cdot 9^x)\right)^(\frac(4)(7)))=\frac(1)(\sqrt(\sin^4(5\ cdot 9^x)))$. Then the derivative will be written in this 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 No. 4

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

Formula No. 2 of the table of derivatives contains the derivative of the function $u^\alpha$. Substituting $\alpha=-1$ into formula No. 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 $u^(-2)=\frac(1)(u^2)$, then equality (4.1) can be rewritten as follows: $ \left(\frac(1)(u) \right)"=-\frac(1)(u^2)\cdot u"$. This is formula No. 3 of the table of derivatives.

Let us turn again to formula No. 2 of the table of derivatives. Let's 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 table of derivatives. As you can see, formulas No. 3 and No. 4 of the derivative table are obtained from formula No. 2 by substituting the corresponding $\alpha$ value.

In “old” textbooks it is also called the “chain” rule. So if y = f (u), and u = φ (x), that is

y = f (φ (x))

complex - composite function (composition of functions) then

Where , after calculation is considered at u = φ (x).

Note that here we took “different” compositions from the same functions, and the result of differentiation naturally turned out to depend on the order of “mixing”.

The chain rule naturally extends to compositions of three or more functions. In this case, there will be three or more “links” in the “chain” that makes up the derivative. Here is an analogy with multiplication: “we have” a table of derivatives; “there” - multiplication table; “with us” is the chain rule and “there” is the “column” multiplication rule. When calculating such “complex” derivatives, no auxiliary arguments (u¸v, etc.), of course, are introduced, but, having noted for themselves the number and sequence of functions involved in the composition, the corresponding links are “strung” in the indicated order.

. Here, with the “x” to obtain the value of the “y”, five operations are performed, that is, there is a composition of five functions: “external” (the last of them) - exponential - e  ; further in reverse order sedate. (♦) 2 ; trigonometric sin(); sedate. () 3 and finally logarithmic ln.(). That's why

With the following examples we will “kill a couple of birds with one stone”: we will practice differentiating complex functions and add to the table of derivatives of elementary functions. So:

4. For a power function - y = x α - rewriting it using the well-known “basic logarithmic identity" - b=e ln b - in the form x α = x α ln x we get

5. For free exponential function using the same technique we will have

6. For an arbitrary logarithmic function, using the well-known formula for transition to a new base, we consistently obtain

7. To differentiate the tangent (cotangent), we use the rule for differentiating quotients:

To obtain the derivatives of inverse trigonometric functions, we use the relation that is satisfied by the derivatives of two mutually inverse functions, that is, the functions φ (x) and f (x) related by the relations:

This is the ratio

It is from this formula for mutually inverse functions

And
,

Finally, let us summarize these and some other derivatives that are also easily obtained in the following table.

First level

Derivative of a function. Comprehensive Guide (2019)

Let's imagine a straight road passing through a hilly area. That is, it goes up and down, but does not turn right or left. If the axis is directed horizontally along the road and vertically, then the road line will be very similar to the graph of some continuous function:

The axis is a certain level of zero altitude; in life we use sea level as it.

As we move forward along such a road, we also move up or down. We can also say: when the argument changes (movement along the abscissa axis), the value of the function changes (movement along the ordinate axis). Now let's think about how to determine the “steepness” of our road? What kind of value could this be? It’s very simple: how much the height will change when moving forward a certain distance. After all, on different areas roads, moving forward (along the x-axis) by one kilometer, we will rise or fall by different quantities meters relative to sea level (along the ordinate axis).

Let’s denote progress (read “delta x”).

The Greek letter (delta) is commonly used in mathematics as a prefix meaning "change". That is - this is a change in quantity, - a change; then what is it? That's right, a change in magnitude.

Important: an expression is a single whole, one variable. Never separate the “delta” from the “x” or any other letter! That is, for example, .

So, we have moved forward, horizontally, by. If we compare the line of the road with the graph of the function, then how do we denote the rise? Certainly, . That is, as we move forward, we rise higher.

The value is easy to calculate: if at the beginning we were at a height, and after moving we found ourselves at a height, then. If end point turned out to be lower than the initial one, it will be negative - this means that we are not ascending, but descending.

Let's return to "steepness": this is a value that shows how much (steeply) the height increases when moving forward one unit of distance:

Let us assume that on some section of the road, when moving forward by a kilometer, the road rises up by a kilometer. Then the slope at this place is equal. And if the road, while moving forward by m, dropped by km? Then the slope is equal.

Now let's look at the top of a hill. If you take the beginning of the section half a kilometer before the summit, and the end half a kilometer after it, you can see that the height is almost the same.

That is, according to our logic, it turns out that the slope here is almost equal to zero, which is clearly not true. Just over a distance of kilometers a lot can change. Smaller areas need to be considered for more adequate and accurate assessment steepness. For example, if you measure the change in height as you move one meter, the result will be much more accurate. But even this accuracy may not be enough for us - after all, if there is a pole in the middle of the road, we can simply pass it. What distance should we choose then? Centimeter? Millimeter? Less is better!

IN real life Measuring distances to the nearest millimeter is more than enough. But mathematicians always strive for perfection. Therefore, the concept was invented infinitesimal, that is, the absolute value is less than any number that we can name. For example, you say: one trillionth! How much less? And you divide this number by - and it will be even less. And so on. If we want to write that a quantity is infinitesimal, we write like this: (we read “x tends to zero”). It is very important to understand that this number is not zero! But very close to it. This means that you can divide by it.

The concept opposite to infinitesimal is infinitely large (). You've probably already come across it when you were working on inequalities: this number is modulo greater than any number you can think of. If you come up with the biggest number possible, just multiply it by two and you'll get an even bigger number. And infinity still Furthermore what will happen. In fact, the infinitely large and the infinitely small are the inverse of each other, that is, at, and vice versa: at.

Now let's get back to our road. The ideally calculated slope is the slope calculated for an infinitesimal segment of the path, that is:

I note that with an infinitesimal displacement, the change in height will also be infinitesimal. But let me remind you that infinitesimal does not mean equal to zero. If you divide infinitesimal numbers by each other, you can get quite regular number, For example, . That is, one small value can be exactly times larger than another.

What is all this for? The road, the steepness... We’re not going on a car rally, but we’re teaching mathematics. And in mathematics everything is exactly the same, only called differently.

Concept of derivative

The derivative of a function is the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument.

Incrementally in mathematics they call change. The extent to which the argument () changes as it moves along the axis is called argument increment and is designated. How much the function (height) has changed when moving forward along the axis by a distance is called function increment and is designated.

So, the derivative of a function is the ratio to when. We denote the derivative with the same letter as the function, only with a prime on the top right: or simply. So, let's write the derivative formula using these notations:

As in the analogy with the road, here when the function increases, the derivative is positive, and when it decreases, it is negative.

Can the derivative be equal to zero? Certainly. For example, if we are driving on a flat horizontal road, the steepness is zero. And it’s true, the height doesn’t change at all. So it is with the derivative: the derivative of a constant function (constant) is equal to zero:

since the increment of such a function is equal to zero for any.

Let's remember the hilltop example. It turned out that it was possible to arrange the ends of the segment along different sides from the top, so that the height at the ends is the same, that is, the segment is parallel to the axis:

But large segments are a sign of inaccurate measurement. We will raise our segment up parallel to itself, then its length will decrease.

Eventually, when we are infinitely close to the top, the length of the segment will become infinitesimal. But at the same time, it remained parallel to the axis, that is, the difference in heights at its ends is equal to zero (it does not tend to, but is equal to). So the derivative

This can be understood this way: when we stand at the very top, a small shift to the left or right changes our height negligibly.

There is also a purely algebraic explanation: to the left of the vertex the function increases, and to the right it decreases. As we found out earlier, when a function increases, the derivative is positive, and when it decreases, it is negative. But it changes smoothly, without jumps (since the road does not change its slope sharply anywhere). Therefore, between negative and positive values there must definitely be. It will be where the function neither increases nor decreases - at the vertex point.

The same is true for the trough (the area where the function on the left decreases and on the right increases):

A little more about increments.

So we change the argument to magnitude. We change from what value? What has it (the argument) become now? We can choose any point, and now we will dance from it.

Consider a point with a coordinate. The value of the function in it is equal. Then we do the same increment: we increase the coordinate by. What is the argument now? Very easy: . What is the value of the function now? Where the argument goes, so does the function: . What about function increment? Nothing new: this is still the amount by which the function has changed:

Practice finding increments:

Find the increment of the function at a point when the increment of the argument is equal to.
The same goes for the function at a point.

Solutions:

IN different points with the same argument increment, the function increment will be different. This means that the derivative at each point is different (we discussed this at the very beginning - the steepness of the road is different at different points). Therefore, when we write a derivative, we must indicate at what point:

Power function.

A power function is a function where the argument is to some degree (logical, right?).

Moreover - to any extent: .

The simplest case- this is when the exponent:

Let's find its derivative at a point. Let's recall the definition of a derivative:

So the argument changes from to. What is the increment of the function?

Increment is this. But a function at any point is equal to its argument. That's why:

The derivative is equal to:

The derivative of is equal to:

b) Now consider quadratic function (): .

Now let's remember that. This means that the value of the increment can be neglected, since it is infinitesimal, and therefore insignificant against the background of the other term:

So, we came up with another rule:

c) We continue the logical series: .

This expression can be simplified in different ways: open the first bracket using the formula for abbreviated multiplication of the cube of the sum, or factorize the entire expression using the difference of cubes formula. Try to do it yourself using any of the suggested methods.

So, I got the following:

And again let's remember that. This means that we can neglect all terms containing:

We get: .

d) Similar rules can be obtained for large powers:

e) It turns out that this rule can be generalized for a power function with an arbitrary exponent, not even an integer:

(2)

The rule can be formulated in the words: “the degree is brought forward as a coefficient, and then reduced by .”

We will prove this rule later (almost at the very end). Now let's look at a few examples. Find the derivative of the functions:

(in two ways: by formula and using the definition of derivative - by calculating the increment of the function);

. Believe it or not, this is a power function. If you have questions like “How is this? Where is the degree?”, remember the topic “”!
Yes, yes, the root is also a degree, only fractional: .
So ours Square root- this is just a degree with an indicator:
.
We look for the derivative using the recently learned formula:
If at this point it becomes unclear again, repeat the topic “”!!! (about degree with negative indicator)
. Now the exponent:
And now through the definition (have you forgotten yet?):
;
.
Now, as usual, we neglect the term containing:
.
. Combination of previous cases: .

Trigonometric functions.

Here we will use one fact from higher mathematics:

With expression.

You will learn the proof in the first year of institute (and to get there, you need to pass the Unified State Exam well). Now I’ll just show it graphically:

We see that when the function does not exist - the point on the graph is cut out. But the closer to the value, the closer the function is to. This is what “aims.”

Additionally, you can check this rule using a calculator. Yes, yes, don’t be shy, take a calculator, we’re not at the Unified State Exam yet.

So, let's try: ;

Don't forget to switch your calculator to Radians mode!

etc. We see that the less, the closer value relationship to

a) Consider the function. As usual, let's find its increment:

Let's turn the difference of sines into a product. To do this, we use the formula (remember the topic “”): .

Now the derivative:

Let's make a replacement: . Then for infinitesimal it is also infinitesimal: . The expression for takes the form:

And now we remember that with the expression. And also, what if an infinitesimal quantity can be neglected in the sum (that is, at).

So, we get the following rule: the derivative of the sine is equal to the cosine:

These are basic (“tabular”) derivatives. Here they are in one list:

Later we will add a few more to them, but these are the most important, since they are used most often.

Practice:

Find the derivative of the function at a point;
Find the derivative of the function.

Solutions:

First, let's find the derivative in general view, and then substitute its value:
;
.
Here we have something similar to power function. Let's try to bring her to
normal view:
.
Great, now you can use the formula:
.
.
. Eeeeeee….. What is this????

Okay, you're right, we don't yet know how to find such derivatives. Here we have a combination of several types of functions. To work with them, you need to learn a few more rules:

Exponent and natural logarithm.

There is a function in mathematics whose derivative for any value is equal to the value of the function itself at the same time. It is called “exponent”, and is an exponential function

The basis of this function is a constant - it is infinite decimal, that is, an irrational number (such as). It is called the “Euler number”, which is why it is denoted by a letter.

So, the rule:

Very easy to remember.

Well, let’s not go far, let’s look at it right away 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:

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

Answers: Exhibitor and natural logarithm- functions are uniquely simple in terms of derivatives. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after let's go through the rules differentiation.

Rules of differentiation

Rules of what? Again 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:

at a point;
at a point;
at a point;
at the point.

Solutions:

(the derivative is the same at all points, since this linear function, remember?);

Derivative of the product

Everything is similar here: let’s enter new feature and find its increment:

Derivative:

Examples:

Find the derivatives of the functions and;
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:

For this we will use 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 any more in simple form. Therefore, we leave it in this form in the answer.

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 almost never appear in the Unified State Examination, but it wouldn’t hurt 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 chocolate, you need to do reverse actions 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.

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. Important Feature complex functions: when the order of actions changes, the function changes.

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

For the first example, .

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

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: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
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 (we 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:

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

Definition. Let the function $y = f(x)$ be defined in a certain interval containing the point $x_0$. Let's give the argument an increment $\Delta x $ such that it does not leave this interval. Let's find the corresponding increment of the function $\Delta y $ (when moving from the point $x_0 $ to the point $x_0 + \Delta x $) and compose the relation $\frac(\Delta y)(\Delta x) $. If there is a limit to this ratio at $\Delta x \rightarrow 0$, then the specified limit is called derivative of a function$y=f(x) $ at the point $x_0 $ and denote $f"(x_0) $.

$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x_0) $$

The symbol y is often used to denote the derivative. Note that y" = f(x) is a new function, but naturally related to the function y = f(x), defined at all points x at which the above limit exists . This function is called like this: derivative of the function y = f(x).

Geometric meaning of derivative is as follows. If it is possible to draw a tangent to the graph of the function y = f(x) at the point with abscissa x=a, which is not parallel to the y-axis, then f(a) expresses the slope of the tangent:
$k = f"(a)$

Since $k = tg(a) $, then the equality $f"(a) = tan(a) $ is true.

Now let’s interpret the definition of derivative from the point of view of approximate equalities. Let the function $y = f(x)$ have a derivative at a specific point $x$:
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x) $$
This means that near the point x the approximate equality $\frac(\Delta y)(\Delta x) \approx f"(x)$, i.e. $\Delta y \approx f"(x) \cdot\Delta x$. The meaningful meaning of the resulting approximate equality is as follows: the increment of the function is “almost proportional” to the increment of the argument, and the coefficient of proportionality is the value of the derivative in given point X. For example, for the function $y = x^2$ the approximate equality $\Delta y \approx 2x \cdot \Delta x $ is valid. If we carefully analyze the definition of a derivative, we will find that it contains an algorithm for finding it.

Let's formulate it.

How to find the derivative of the function y = f(x)?

1. Fix the value of $x$, find $f(x)$
2. Give the argument $x$ an increment $\Delta x$, go to new point$x+ \Delta x $, find $f(x+ \Delta x) $
3. Find the increment of the function: $\Delta y = f(x + \Delta x) - f(x) $
4. Create the relation $\frac(\Delta y)(\Delta x) $
5. Calculate $$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) $$
This limit is the derivative of the function at point x.

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

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

Let the function y = f(x) be differentiable at the point x. Then a tangent can be drawn to the graph of the function at point M(x; f(x)), and, recall, the angular coefficient of the tangent is equal to f "(x). Such a graph cannot “break” at point M, i.e. the function must be continuous at point x.

These were “hands-on” arguments. Let us give a more rigorous reasoning. If the function y = f(x) is differentiable at the point x, then the approximate equality $\Delta y \approx f"(x) \cdot \Delta x $ holds. If in this equality $\Delta x $ tends to zero, then $\Delta y $ will tend to zero, and this is the condition for the continuity of the function at a point.

So, if a function is differentiable at a point x, then it is continuous at that point.

The reverse statement is not true. For example: function y = |x| is continuous everywhere, in particular at the point x = 0, but the tangent to the graph of the function at the “junction point” (0; 0) does not exist. If at some point a tangent cannot be drawn to the graph of a function, then the derivative does not exist at that point.

One more example. The function $y=\sqrt(x)$ is continuous on the entire number line, including at the point x = 0. And the tangent to the graph of the function exists at any point, including at the point x = 0. But at this point the tangent coincides with the y-axis, i.e., it is perpendicular to the abscissa axis, its equation has the form x = 0. Slope coefficient such a line does not have, which means that $f"(0) $ does not exist either

So, we got acquainted with a new property of a function - differentiability. How can one conclude from the graph of a function that it is differentiable?

The answer is actually given above. If at some point it is possible to draw a tangent to the graph of a function that is not perpendicular to the abscissa axis, then at this point the function is differentiable. If at some point the tangent to the graph of a function does not exist or it is perpendicular to the abscissa axis, then at this point the function is not differentiable.

Rules of differentiation

The operation of finding the derivative is called differentiation. When performing this operation, you often have to work with quotients, sums, products of functions, as well as “functions of functions,” that is, complex functions. Based on the definition of derivative, we can derive differentiation rules that make this work easier. If C is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:

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

Table of derivatives of some functions

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

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. It's relative simple expressions, the derivatives of which have long been calculated and listed in the table. 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 differentiable with respect 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:

(f + g)’ = f ’ + g ’
(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 " 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 factor is a little more complicated, but general scheme this doesn't 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 at 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 with specific examples, with detailed description every 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 it will work out 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 last expression, the whole problem was 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, a prime from the amount equal to the sum 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 last 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 act 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 to 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: