Find the derivative of a complex function. Derivative of a complex function

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 - compound 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 meaning of the “y”, five operations are performed, that is, there is a composition of five functions: “external” (the last of them) - exponential - e  ;

then in reverse order, power. (♦) 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 obtain

5. For an arbitrary 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 of differentiation of quotients:

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 outermost 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:

It seems without errors:

1) Take the derivative of the square root.

2) Take the derivative of the difference using the rule

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

4) Take the derivative of the cosine.

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, let’s see if it’s possible to turn the product of three functions into the product of two functions? For example, if we had two polynomials in the product, then 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 really - 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 the draft to see if the answer can be simplified?

Let's reduce the expression of the numerator to a common denominator and get rid of the three-story structure of the 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

In this lesson we will learn how to find derivative of a complex function. The lesson is a logical continuation of the lesson How to find the derivative?, in which we examined the simplest derivatives, and also became acquainted with the rules of differentiation and some technical techniques for finding derivatives. Thus, if you are not very good with derivatives of functions or some points in this article are not entirely clear, then first read the above lesson. Please get in a serious mood - the material is not simple, 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 at the table at the rule (No. 5) for differentiating a complex function:

Let's figure it out. First of all, let's pay attention to the entry. Here we have two functions - and , and the function, figuratively speaking, is nested within the function . A function of this type (when one function is nested within another) is called a complex function.

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

! These definitions are not theoretical and should not appear in the final design of assignments. I use 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 an entire expression, so finding the derivative right away 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 the sine cannot be “torn into pieces”:

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

First step what you need to do 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 embedded under the sine. But what if everything is not obvious? How to accurately determine which function is external and which is internal? To do this, I suggest using the following technique, which can be done mentally or in a draft.

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

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

Secondly will need to be found, so sine – will be an external function:

After we SOLD OUT With internal and external functions, it's time to apply the rule of differentiation of complex functions.

Let's start deciding. From class How to find the derivative? we remember that the design of a solution to 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 table formulas are also applicable if “x” is replaced with a complex expression, in this case:

Please note that the inner function hasn't changed, we don't touch it.

Well, it's quite obvious that

The final result of applying the formula looks like this:

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

If there is any misunderstanding, write the solution down 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 down:

Let's figure out where we have an external function and where we have an internal one. To do this, we try (mentally or in a draft) to calculate the value of the expression at . What should you do first? First of all, you need to calculate what the base is equal to: therefore, the polynomial is the internal function:

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

According to the formula, you first need to find the derivative of the external function, in this case, the degree. We look for the required 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 for differentiating a complex function is as follows:

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

Now all that remains is to find a very simple derivative of the internal function and tweak the result a little:

Example 4

Find the derivative of a function

This is an example for you to solve on your own (answer at the end of the lesson).

To consolidate your 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 the external and where the internal function is, why the tasks are solved this way?

Example 5

a) Find the derivative of the 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 power. Thus, first we bring the function into the form appropriate for differentiation:

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

We again represent the degree 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 reduce the expression to a common denominator in brackets and write everything down as one fraction. It’s beautiful, of course, but when you get cumbersome long derivatives, 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 you to solve on your own (answer at the end of the lesson).

It is interesting to note that sometimes instead of the rule for differentiating a complex function, you can use the rule for differentiating a quotient , but such a solution will look like a funny perversion. 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 move the minus out of the derivative sign, and raise the cosine into the numerator:

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

We find the derivative of the internal function and reset the cosine back down:

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

Example 9

Find the derivative of a function

This is an example for you to solve on your own (answer at the end of the lesson).

So far we have looked at 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

Let's understand the attachments of this function. Let's try to calculate the expression using the experimental value. How would we count on a calculator?

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

This arcsine of one should then be squared:

And finally, we raise seven to a power:

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

Let's start deciding

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

Under the stroke we have a complex function again! But it’s already simpler. It is easy to verify that the inner function is the arcsine, the outer function is the degree. According to the rule for differentiating a complex function, you first need to take the derivative of the power.

Very easy to remember.

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

In our case, the base is the number:

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

What is it equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

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

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

Rules of differentiation

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

Differentiation is the process of finding the derivative.

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

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

There are 5 rules in total.

The constant is taken out of the derivative sign.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

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

Examples.

Find the derivatives of the functions:

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

Solutions:

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

Derivative of the product

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

Derivative:

Examples:

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:

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

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

Happened?

Here, check yourself:

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

Examples:
Find the derivatives of the functions:

Answers:

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

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

In this example, the product of two functions:

Derivative of a logarithmic function

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

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

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

Only now we will write instead:

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

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

Derivative of a complex function.

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

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

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

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

For our example, .

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

Second example: (same thing). .

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

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

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

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 (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.

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

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

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

Derivatives of elementary functions

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

So, derivatives of elementary functions:

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

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

(C · f)’ = C · f ’.

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

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

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

Derivative of sum and difference

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

(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 of “negative element”. Therefore the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.

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

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

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

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

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

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

Derivative of the product

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

(f · g) ’ = f ’ · g + f · g ’

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

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

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

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

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

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

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

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

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

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

Task. Find derivatives of functions:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(x n)’ = n · x n − 1

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