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Complex differentiation. Complex function

Since you came here, you probably already saw this formula in the textbook

and make a face like this:

Friend, don't worry! In fact, everything is simply outrageous. You will definitely understand everything. Just one request - read the article slowly, try to understand every step. I wrote as simply and clearly as possible, but you still need to understand 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 packing things into large boxes. Suppose we need to collect some small items, for example, school writing materials. If you just throw them into 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 “complex” process is presented in the diagram below:

It would seem, what does mathematics have to do with it? Yes, despite the fact that a complex function is formed in EXACTLY THE SAME way! Only we “pack” not notebooks and pens, but \(x\), while the “packages” and “boxes” are different.

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”. Now let’s put it in a “box” - pack it, for example, into a cubic function.


What will happen in the end? Yes, that’s right, there will be a “bag of things in a box,” that is, “cosine of X cubed.”

The resulting design is a complex function. It differs from simple one in that SEVERAL “influences” (packages) are applied to one X in a row and it turns out as if “function from function” - “packaging within packaging”.

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

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

\(x → 7^x → tg⁡(7^x)\)

Now let’s “pack” x twice into trigonometric functions, first in and then in:

\(x → sin⁡x → cotg⁡ (sin⁡x)\)

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 logarithm to the base \(4\) , then to the power \(-2\).

Find the answers to this task at the end of the article.

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 luckier - theirs may be more complicated☺).

"Unpacking" a complex function

Look at the previous function again. Can you figure out the “packing” sequence? What X was stuffed into first, what then, and so on until the very end. That is, which function is nested within which? Take a piece of paper and write down what you think. You can do this with a chain with 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 a sine, it, in turn, was placed into the logarithm to the base \(2\), and in the end this whole construction was stuffed into a power fives.

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

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

\(x → \log_2⁡x → tg⁡(\log_2⁡x)\)

Another example: \(y=\cos⁡((x^3))\). Let's analyze - first we cubed X, and then took the cosine of the result. This means the sequence will be: \(x → x^3 → \cos⁡((x^3))\). Pay attention, the function seems to be similar to the very first one (where it has pictures). But this is a completely different function: here in the cube is x (that is, \(\cos⁡((x·x·x)))\), and there in the cube is 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 they first did arithmetic operations with x, then took the sine of the result: \(x → 2x+5 → \sin⁡((2x+5))\). And this 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 \(ctg x\). This means that all their combinations are simple functions:

\(x^7+ ctg x\) - simple,
\(x^7· cot x\) – simple,
\(\frac(x^7)(ctg x)\) – simple, etc.

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



Okay, go ahead 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 fact is that without such an analysis we will not be able to reliably find 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 remember our analogy at the very beginning, then the internal function is a “package”, and the external function is a “box”. Those. what X is “wrapped” in first is an internal function, and what the internal function is “wrapped” in is already external. Well, it’s clear why - she’s outside, that means external.

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

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

Complete the last practice of analyzing complex functions, and let's finally move on to what we were all started for - we will find derivatives of complex functions:

Fill in the blanks in the table:


Derivative of a complex function

Bravo to us, we finally 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 a constant internal function and the derivative of the internal function.

And immediately look at the “word by word” parsing diagram to understand what is what:

I hope the terms “derivative” and “product” do not cause any difficulties. “Complex function” - we have already sorted it out. 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 an external function, in which only the external function changes, and the internal one remains the same. Still not clear? Okay, let's use an example.

Let us have a function \(y=\sin⁡(x^3)\). It is clear that the internal function here is \(x^3\), and the external
. Let us now find the derivative of the exterior with respect to the constant interior.

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

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

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

Derivatives of elementary functions

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

So, derivatives of elementary functions:

Solving physical problems or examples in mathematics is completely impossible without knowledge of the derivative and methods for calculating it. The derivative is one of the most important concepts in mathematical analysis. We decided to devote today’s article to this fundamental topic. What is a derivative, what is its physical and geometric meaning, how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?

Geometric and physical meaning of derivative

Let there be a function f(x) , specified in a certain interval (a, b) . Points x and x0 belong to this interval. When x changes, the function itself changes. Changing the argument - the difference in its values x-x0 . This difference is written as delta x and is called argument increment. A change or increment of a function is the difference between the values ​​of a function at two points. Definition of derivative:

The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.

Otherwise it can be written like this:

What's the point of finding such a limit? Here's what it is:

the derivative of a function at a point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at a given point.


Physical meaning of the derivative: the derivative of the path with respect to time is equal to the speed of rectilinear motion.

Indeed, since school days everyone knows that speed is a particular path x=f(t) and time t . Average speed over a certain period of time:

To find out the speed of movement at a moment in time t0 you need to calculate the limit:

Rule one: set a constant

The constant can be taken out of the derivative sign. Moreover, this must be done. When solving examples in mathematics, take it as a rule - If you can simplify an expression, be sure to simplify it .

Example. Let's calculate the derivative:

Rule two: derivative of the sum of functions

The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.

We will not give a proof of this theorem, but rather consider a practical example.

Find the derivative of the function:

Rule three: derivative of the product of functions

The derivative of the product of two differentiable functions is calculated by the formula:

Example: find the derivative of a function:

Solution:

It is important to talk about calculating derivatives of complex functions here. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable.

In the above example we come across the expression:

In this case, the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first calculate the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate argument itself with respect to the independent variable.

Rule four: derivative of the quotient of two functions

Formula for determining the derivative of the quotient of two functions:

We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it seems, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.

With any questions on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult test and understand the tasks, even if you have never done derivative calculations before.

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 the corresponding at the 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 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 the detailed process for finding the derivative of 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 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 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 the 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 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. $\left(u^\alpha \right)"=\alpha\cdot u^(\alpha-1)\cdot u"$. Let’s substitute $u=\arctg(4\cdot \ln x)$ and $\alpha=12$ into this formula:

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 $\arctg^(12)(4\cdot 5^x)$. The 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:

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

It remains to find $(4\cdot \ln x)"$. 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 obtained result 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:

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

It remains to find $(5\cdot 9^x)"$. First, 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 a special case of formula 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.

Complex derivatives. Logarithmic derivative.
Derivative of a power-exponential function

We continue to improve our differentiation technique. In this lesson, we will consolidate the material we have covered, look at more complex derivatives, and also get acquainted with new techniques and tricks for finding a 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? Examples of solutions, which will allow you to improve your skills almost from scratch. Next, you need to carefully study the page Derivative of a complex function, understand and resolve All the examples I gave. This lesson is logically the third, and after mastering it you will confidently differentiate fairly complex functions. It is undesirable to take the position of “Where else? That’s enough!”, since all examples and solutions are taken from real tests and are often encountered in practice.

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

According to the rule of differentiation of complex functions :

When studying other matan topics in the future, such a detailed recording is most often not required; it is assumed that the student knows how to find such 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’s?” This should be followed by an almost instant and polite response: .

The first example will be immediately intended for independent decision.

Example 1

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

, , ,
, , ,
, , ,

, , ,

, , ,

, , ,

, ,

Answers at the end of the lesson

Complex derivatives

After preliminary artillery preparation, examples with 3-4-5 nestings of functions will be less scary. The following two examples may seem complicated to some, but if you understand them (someone will suffer), 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 your investments. In cases where there are doubts, I remind you of a useful technique: we take the experimental value of “x”, for example, and try (mentally or in a draft) to substitute this value into the “terrible expression”.

1) First we need to calculate the expression, which means the sum is the deepest embedding.

2) Then you need to calculate the logarithm:

4) Then cube the cosine:

5) At the fifth step the difference:

6) And finally, the most external function is the square root:

Formula for differentiating a complex function are applied in reverse order, from the outermost function to the innermost. We decide:

There seem to be no errors...

(1) 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 zero. In the second term we take the derivative of the degree (cube).

(4) Take the derivative of the cosine.

(5) Take the derivative of the logarithm.

(6) And finally, we take the derivative of the deepest embedding .

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 beauty and simplicity of the analyzed derivative. I noticed that they like to give a similar thing in an exam to check whether a student understands how to find the derivative of a complex function or does not understand.

The following example is for you to solve on your own.

Example 3

Find the derivative of a function

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

Full solution and answer at the end of the lesson.

It's time to move on to something smaller and nicer.
It is not uncommon for an example to show the product of not two, but three functions. How to find the derivative of the product of three factors?

Example 4

Find the derivative of a function

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

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

The trick is that by “y” we denote the product of two functions: , and by “ve” we denote the logarithm: . Why can this be done? Is it possible – this is not a 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 also get twisted and put something out of brackets, but in this case it’s better to leave the answer exactly in this form - it will be easier to check.

The considered 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 using the first method.

Let's look at similar examples with fractions.

Example 6

Find the derivative of a function

There are several ways you can go here:

Or like this:

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

In principle, the example is solved, and if it is left as is, it will not be an error. But if you have time, it is always advisable to check on a draft to see if the answer can be simplified? Let us reduce the expression of the numerator to common denominator And let's 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 the derivative, but during banal school transformations. On the other hand, teachers often reject the assignment and ask to “bring it to mind” the derivative.

A simpler example to solve on your own:

Example 7

Find the derivative of a function

We continue to master the methods of 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 the long way, using the rule for differentiating a complex function:

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

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



! If you have a practice notebook at hand, copy these formulas directly there. If you don't have a notebook, copy them onto a piece of paper, since the remaining examples of the lesson will revolve around these formulas.

The solution itself can be written something like this:

Let's transform the function:

Finding the derivative:

Pre-converting 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 you to solve on your own:

Example 9

Find the derivative of a function

Example 10

Find the derivative of a function

All transformations and answers are 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

We recently looked at similar examples. What to do? You can sequentially 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 end up with 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 a function can take negative values, then, generally speaking, you need to use modules: , which will disappear as a result of differentiation. However, the current design is also acceptable, where by default it is taken into account complex meanings. But if in all rigor, then in both cases a reservation should be made that.

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

Let's start with differentiation.
We conclude both parts under the prime:

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

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 game” - IS ITSELF A FUNCTION(if it is not very clear, refer to the article Derivative of a function specified implicitly). Therefore, the logarithm is an external function, and the “y” is an internal function. And we use the rule for differentiating a complex function :

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

And now let’s remember what kind of “player”-function we talked about during differentiation? Let's look at the condition:

Final answer:

Example 12

Find the derivative of a function

This is an example for you to solve on your own. A sample design of an example of this type is at the end of the lesson.

Using 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 a power-exponential function

We have not considered this function yet. A power-exponential function is a function for which both the degree and the base depend on the “x”. A classic example that will be given to you in any textbook or lecture:

How to find the derivative of a power-exponential function?

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

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

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

We find the derivative; to do this, we enclose both parts under strokes:

Further actions are simple:

Finally:

If any conversion is not entirely clear, please re-read the explanations of Example No. 11 carefully.

In practical tasks, the power-exponential function will always be more complex than the lecture example discussed.

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 logarithm x” (another logarithm is nested under the logarithm). When differentiating, as we remember, it is better to immediately move the constant out of the derivative sign so that it does not get in the way; and, of course, we apply the familiar rule :
Name Function Derivative
Constant f(x) = C, C ∈ R 0 (yes, zero!)
Power with rational exponent f(x) = x n n · x n − 1
Sinus f(x) = sin x cos x
Cosine f(x) = cos x −sin x(minus sine)
Tangent f(x) = tg x 1/cos 2 x
Cotangent f(x) = ctg x − 1/sin 2 x
Natural logarithm f(x) = log x 1/x
Arbitrary logarithm f(x) = log a x 1/(x ln a)
Exponential function f(x) = e x e x(nothing changed)

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

(C · f)’ = C · f ’.

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

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

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

Derivative of sum and difference

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

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

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

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

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

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

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

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

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

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

Derivative of the product

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

(f · g) ’ = f ’ · g + f · g ’

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

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

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

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

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

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

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

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

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

Not weak, huh? Where did the minus come from? Why g 2? And like this! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it 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 we get an elementary function f(x) = e x. Therefore, we make a replacement: let 2 x + 3 = t, f(x) = f(t) = e t. We look for the derivative of a complex function using the formula:

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

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

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

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

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

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

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

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

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

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

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

(x n)’ = n · x n − 1

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

Task. Find the derivative of the function:

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

f(x) = (x 2 + 8x − 7) 0,5 .

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

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

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

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

Finally, back to the roots:

 

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