Derivative of the root of a complex function. Find the derivative: algorithm and examples of solutions

On which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some techniques for finding derivatives. Thus, if you are not very good with derivatives of functions or some points of this article are not entirely clear, then first read the above lesson. Please tune in to a serious mood - the material is not easy, but I will still try to present it simply and clearly.

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

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

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

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

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

To clarify the situation, consider:

Example 1

Find the derivative of a function

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

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

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

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

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

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

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

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

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

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

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

Well, it is quite obvious that

The result of applying the formula clean looks like this:

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

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

Example 2

Find the derivative of a function

Example 3

Find the derivative of a function

As always, we write:

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

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

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

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

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

Example 4

Find the derivative of a function

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

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

Example 5

a) Find the derivative of a function

b) Find the derivative of the function

Example 6

Find the derivative of a function

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

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

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

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

Example 7

Find the derivative of a function

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

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

Example 8

Find the derivative of a function

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

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

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

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

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

Example 9

Find the derivative of a function

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

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

Example 10

Find the derivative of a function

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

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

This arcsine of unity should then be squared:

And finally, we raise the seven to the power:

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

We start to decide

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

Examples of calculating derivatives using the formula for the derivative of a complex function are given.

Content

See also: Proof of the formula for the derivative of a complex function

Basic Formulas

Here we give examples of calculating derivatives of the following functions:
; ; ; ; .

If a function can be represented as a complex function in the following form:
,
then its derivative is determined by the formula:
.
In the examples below, we will write this formula in the following form:
.
where .
Here, the subscripts or , located under the sign of the derivative, denote the variable with respect to which differentiation is performed.

Usually, in tables of derivatives, the derivatives of functions from the variable x are given. However, x is a formal parameter. The variable x can be replaced by any other variable. Therefore, when differentiating a function from a variable , we simply change, in the table of derivatives, the variable x to the variable u .

Simple examples

Example 1

Find the derivative of a complex function
.

We write the given function in an equivalent form:
.
In the table of derivatives we find:
;
.

According to the formula for the derivative of a complex function, we have:
.
Here .

Example 2

Find derivative
.

We take out the constant 5 beyond the sign of the derivative and from the table of derivatives we find:
.

.
Here .

Example 3

Find the derivative
.

We take out the constant -1 for the sign of the derivative and from the table of derivatives we find:
;
From the table of derivatives we find:
.

We apply the formula for the derivative of a complex function:
.
Here .

More complex examples

In more complex examples, we apply the compound function differentiation rule several times. In doing so, we calculate the derivative from the end. That is, we break the function into its component parts and find the derivatives of the simplest parts using derivative table. We also apply sum differentiation rules, products and fractions . Then we make substitutions and apply the formula for the derivative of a complex function.

Example 4

Find the derivative
.

We select the simplest part of the formula and find its derivative. .

.
Here we have used the notation
.

We find the derivative of the next part of the original function, applying the results obtained. We apply the rule of differentiation of the sum:
.

Once again, we apply the rule of differentiation of a complex function.

.
Here .

Example 5

Find the derivative of a function
.

We select the simplest part of the formula and find its derivative from the table of derivatives. .

We apply the rule of differentiation of a complex function.
.
Here
.

We differentiate the next part, applying the results obtained.
.
Here
.

Let's differentiate the next part.

.
Here
.

Now we find the derivative of the desired function.

.
Here
.

See also:

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

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

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

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

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

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

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

Example #1

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

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

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

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

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

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

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

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

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

Example #2

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

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

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

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

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

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

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

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

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

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

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

Equality (2.2) will now become:

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

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

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

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

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

Example #3

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

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

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

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

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

We continue equality (3.1) using the obtained result:

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

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

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

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

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

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

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

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

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

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

Example #4

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

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

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

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

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

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

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

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

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

Functions of a complex form are not entirely correct to call the term "complex function". For example, it looks very impressive, but this function is not complicated, unlike.

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

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

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

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

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

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

You can often hear that a complex function is called function composition.

The formula for finding the derivative of a complex function.

Example.

Find the derivative of a complex function.

Solution.

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

Here is a detailed solution using the formula for the derivative of a complex function:

Let's find this derivative, after simplifying the form of the original function.

Consequently,

As you can see, the results match.

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

Let us explain this with an example for attention.

Example.

Find derivatives of complex functions and .

Solution.

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

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

The derivative formula for a function has the form

Example.

Differentiate Function .

Solution.

In this example, the complex function can be conditionally written as , where is the sine function, the function of raising to the third power, the logarithm function to the base e, the function of taking the arc tangent and the linear function, respectively.

According to the formula for the derivative of a complex function

Now we find

Putting together the obtained intermediate results:

There is nothing terrible, disassemble complex functions like nesting dolls.

This could have ended the article, if not for one but ...

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

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

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

And here is the function can no longer be called difficult.

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

It remains to find the derivative of a complex function:

That's why .

We hope you get the gist.

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

As an example, let's analyze the component parts of the function .

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

Secondly, let's deal with the function h(x) . It is related to .

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

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

In this way, .

Thirdly, go to , which is the product of a complex function and an entire rational function

The squaring function is the logarithm function to base e.

Consequently, .

To summarize:

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

In the section differentiation of a function (finding a derivative) you can find the solution of such problems.