Find the derivative of the root of x. Find the derivative: algorithm and examples of solutions

Instructions

Before finding the derivative of the root, pay attention to the other functions present in the example being solved. If the problem has many radical expressions, then use the following rule for finding the derivative of the square root:

(√x)" = 1 / 2√x.

And to find the derivative of the cube root, use the formula:

(³√x)" = 1 / 3(³√x)²,

where ³√x denotes the cube root of x.

If, intended for differentiation, there is a variable in fractional , then convert the root into a power function with the appropriate exponent. For a square root it will be a power of ½, and for a cube root it will be ⅓:

√x = x^½,
³√х = x ^ ⅓,

where ^ denotes exponentiation.

To find the derivative of a power function in general and x^1, x^⅓ in particular, use the following rule:

(x^n)" = n * x^(n-1).

For the derivative of a root, this relation implies:

(x^½)" = ½ x ^ (-½) and
(x^⅓)" = ⅓ x ^ (-⅔).

Having differentiated everything, look carefully at the rest of the example. If you have a very cumbersome expression in your answer, then you can probably simplify it. Most school examples are structured in such a way that the end result is a small number or a compact expression.

In many derivative problems, roots (square and cube) are found together with other functions. To find the derivative of the root in this case, use the following rules:
the derivative of a constant (constant number, C) equals zero: C" = 0;
the constant factor is taken out of the derivative sign: (k*f)" = k * (f)" (f is an arbitrary function);
the derivative of the sum of several functions is equal to the sum of the derivatives: (f + g)" = (f)" + (g)";
the derivative of the product of two functions is equal to... no, not the product of derivatives, but the following expression: (fg)" = (f)"g + f (g)";
the derivative of the quotient is also not equal to the quotient of the derivatives, but is found according to the following rule: (f/g)" = ((f)"g – f(g)") / g².

note

On this page you can calculate the derivative of a function online and obtain a detailed solution to the problem. The solution of the derivatives of a function is made using the rules of differentiation that students study in the course of mathematical analysis at the institute. In order to find the derivative of a function, you need to enter the function for differentiation in the "Function" field according to the data entry rules.

Helpful advice

The derivative of a function is the limit of the ratio of the increment of a function to the increment of the argument when the increment of the argument tends to zero: The mathematical meaning of this definition is not very easy to understand, since in a school algebra course the concept of the limit of a function is either not studied at all or is studied very superficially. But in order to learn how to find derivatives of various functions, this is not necessary.

Sources:

• derived root of x

1. General case of the formula for the derivative of a root of an arbitrary degree- a fraction in the numerator of which there is one, and in the denominator a number equal to the power of the root for which the derivative was calculated, multiplied by the root of the same power, the radical expression of which is a variable in the power of the root for which the derivative was calculated, reduced by one
2. Square root derivative- is a special case of the previous formula. Derivative of the square root of x is a fraction whose numerator is one and the denominator is two times the square root of x
3. Derivative of cube root, also a special case of the general formula. The derivative of a cube root is one divided by three cube roots of x squared.

Below are transformations that explain why the formulas for finding the derivatives of square and cubic roots are exactly the same as shown in the figure.

Of course, you don’t have to remember these formulas at all, if you take into account that extracting the root of a derivative power is the same as raising a fraction whose denominator is equal to the same power. Then finding the derivative of the root is reduced to applying the formula for finding the derivative of the power of the corresponding fraction.

Derivative of a variable under square root

Explanation:
(√x)" = (x 1/2)"

The square root is exactly the same operation as raising to the power of 1/2,This means that to find the derivative of a root, you can apply the formula from the rule for finding the derivative of a variable to an arbitrary degree:

(x 1/2)" = 1/2 x -1/2 = 1 / (2√x)

Derivative of cube root (derivative of third root)

Let's imagine the cube root as a power of 1/3 and find the derivative using the general rules of differentiation. The brief formula can be seen in the picture above, and below is an explanation of why this is so.

The power -2/3 is obtained by subtracting one from 1/3

Derivation of the formula for the derivative of a power function (x to the power of a). Derivatives from roots of x are considered. Formula for the derivative of a higher order power function. Examples of calculating derivatives.

See also: Power function and roots, formulas and graph
Power Function Graphs

Basic formulas

The derivative of x to the power of a is equal to a times x to the power of a minus one:
(1) .

The derivative of the nth root of x to the mth power is:
(2) .

Derivation of the formula for the derivative of a power function

Case x > 0

Consider a power function of the variable x with exponent a:
(3) .
Here a is an arbitrary real number. Let's first consider the case.

To find the derivative of function (3), we use the properties of a power function and transform it to the following form:
.

Now we find the derivative using:
;
.
Here .

Formula (1) has been proven.

Derivation of the formula for the derivative of a root of degree n of x to the degree of m

Now consider a function that is the root of the following form:
(4) .

To find the derivative, we transform the root to a power function:
.
Comparing with formula (3) we see that
.
Then
.

Using formula (1) we find the derivative:
(1) ;
;
(2) .

In practice, there is no need to memorize formula (2). It is much more convenient to first transform the roots to power functions, and then find their derivatives using formula (1) (see examples at the end of the page).

Case x = 0

If , then the power function is defined for the value of the variable x = 0 . Let's find the derivative of function (3) at x = 0 . To do this, we use the definition of a derivative:
.

Let's substitute x = 0 :
.
In this case, by derivative we mean the right-hand limit for which .

So we found:
.
From this it is clear that for , .
At , .
At , .
This result is also obtained from formula (1):
(1) .
Therefore, formula (1) is also valid for x = 0 .

Case x< 0

Consider function (3) again:
(3) .
For certain values ​​of the constant a, it is also defined for negative values ​​of the variable x. Namely, let a be a rational number. Then it can be represented as an irreducible fraction:
,
where m and n are integers that do not have a common divisor.

If n is odd, then the power function is also defined for negative values ​​of the variable x. For example, when n = 3 and m = 1 we have the cube root of x:
.
It is also defined for negative values ​​of the variable x.

Let us find the derivative of the power function (3) for and for rational values ​​of the constant a for which it is defined. To do this, let's represent x in the following form:
.
Then ,
.
We find the derivative by placing the constant outside the sign of the derivative and applying the rule for differentiating a complex function:

.
Here . But
.
Since then
.
Then
.
That is, formula (1) is also valid for:
(1) .

Higher order derivatives

Now let's find higher order derivatives of the power function
(3) .
We have already found the first order derivative:
.

Taking the constant a outside the sign of the derivative, we find the second-order derivative:
.
Similarly, we find derivatives of the third and fourth orders:
;

.

From this it is clear that derivative of arbitrary nth order has the following form:
.

notice, that if a is a natural number, then the nth derivative is constant:
.
Then all subsequent derivatives are equal to zero:
,
at .

Examples of calculating derivatives

Example

Find the derivative of the function:
.

Let's convert roots to powers:
;
.
Then the original function takes the form:
.

Finding derivatives of powers:
;
.
The derivative of the constant is zero:
.

The operation of finding the derivative is called differentiation.

As a result of solving problems of finding derivatives of the simplest (and not very simple) functions by defining the derivative as the limit of the ratio of the increment to the increment of the argument, a table of derivatives and precisely defined rules of differentiation appeared. The first to work in the field of finding derivatives were Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716).

Therefore, in our time, to find the derivative of any function, you do not need to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but you only need to use the table of derivatives and the rules of differentiation. The following algorithm is suitable for finding the derivative.

To find the derivative, you need an expression under the prime sign break down simple functions into components and determine what actions (product, sum, quotient) these functions are related. Next, we find the derivatives of elementary functions in the table of derivatives, and the formulas for the derivatives of the product, sum and quotient - in the rules of differentiation. The derivative table and differentiation rules are given after the first two examples.

Example 1. Find the derivative of a function

Solution. From the rules of differentiation we find out that the derivative of a sum of functions is the sum of derivatives of functions, i.e.

From the table of derivatives we find out that the derivative of "x" is equal to one, and the derivative of sine is equal to cosine. We substitute these values ​​into the sum of derivatives and find the derivative required by the condition of the problem:

Example 2. Find the derivative of a function

Solution. We differentiate as a derivative of a sum in which the second term has a constant factor; it can be taken out of the sign of the derivative:

If questions still arise about where something comes from, they are usually cleared up after familiarizing yourself with the table of derivatives and the simplest rules of differentiation. We are moving on to them right now.

Table of derivatives of simple functions

1. Derivative of a constant (number). Any number (1, 2, 5, 200...) that is in the function expression. Always equal to zero. This is very important to remember, as it is required very often
2. Derivative of the independent variable. Most often "X". Always equal to one. This is also important to remember for a long time
3. Derivative of degree. When solving problems, you need to convert non-square roots into powers.
4. Derivative of a variable to the power -1
5. Derivative of square root
6. Derivative of sine
7. Derivative of cosine
8. Derivative of tangent
9. Derivative of cotangent
10. Derivative of arcsine
11. Derivative of arc cosine
12. Derivative of arctangent
13. Derivative of arc cotangent
14. Derivative of the natural logarithm
15. Derivative of a logarithmic function
16. Derivative of the exponent
17. Derivative of an exponential function

Rules of differentiation

1. Derivative of a sum or difference
2. Derivative of the product
2a. Derivative of an expression multiplied by a constant factor
3. Derivative of the quotient
4. Derivative of a complex function

Rule 1.If the functions

are differentiable at some point, then the functions are differentiable at the same point

and

those. the derivative of an algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions.

Consequence. If two differentiable functions differ by a constant term, then their derivatives are equal, i.e.

Rule 2.If the functions

are differentiable at some point, then their product is differentiable at the same point

and

those. The derivative of the product of two functions is equal to the sum of the products of each of these functions and the derivative of the other.

Corollary 1. The constant factor can be taken out of the sign of the derivative:

Corollary 2. The derivative of the product of several differentiable functions is equal to the sum of the products of the derivative of each factor and all the others.

For example, for three multipliers:

Rule 3.If the functions

differentiable at some point And , then at this point their quotient is also differentiableu/v , and

those. the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator.

Where to look for things on other pages

When finding the derivative of a product and a quotient in real problems, it is always necessary to apply several differentiation rules at once, so there are more examples on these derivatives in the article"Derivative of the product and quotient of functions".

Comment. You should not confuse a constant (that is, a number) as a term in a sum and as a constant factor! In the case of a term, its derivative is equal to zero, and in the case of a constant factor, it is taken out of the sign of the derivatives. This is a typical mistake that occurs at the initial stage of studying derivatives, but as the average student solves several one- and two-part examples, he no longer makes this mistake.

And if, when differentiating a product or quotient, you have a term u"v, in which u- a number, for example, 2 or 5, that is, a constant, then the derivative of this number will be equal to zero and, therefore, the entire term will be equal to zero (this case is discussed in example 10).

Another common mistake is mechanically solving the derivative of a complex function as the derivative of a simple function. That's why derivative of a complex function a separate article is devoted. But first we will learn to find derivatives of simple functions.

Along the way, you can’t do without transforming expressions. To do this, you may need to open the manual in new windows. Actions with powers and roots And Operations with fractions .

If you are looking for solutions to derivatives of fractions with powers and roots, that is, when the function looks like , then follow the lesson “Derivative of sums of fractions with powers and roots.”

If you have a task like , then you will take the lesson “Derivatives of simple trigonometric functions”.

Step-by-step examples - how to find the derivative

Example 3. Find the derivative of a function

Solution. We define the parts of the function expression: the entire expression represents a product, and its factors are sums, in the second of which one of the terms contains a constant factor. We apply the product differentiation rule: the derivative of the product of two functions is equal to the sum of the products of each of these functions by the derivative of the other:

Next, we apply the rule of differentiation of the sum: the derivative of the algebraic sum of functions is equal to the algebraic sum of the derivatives of these functions. In our case, in each sum the second term has a minus sign. In each sum we see both an independent variable, the derivative of which is equal to one, and a constant (number), the derivative of which is equal to zero. So, “X” turns into one, and minus 5 turns into zero. In the second expression, "x" is multiplied by 2, so we multiply two by the same unit as the derivative of "x". We obtain the following derivative values:

We substitute the found derivatives into the sum of products and obtain the derivative of the entire function required by the condition of the problem:

And you can check the solution to the derivative problem on.

Example 4. Find the derivative of a function

Solution. We are required to find the derivative of the quotient. We apply the formula for differentiating the quotient: the derivative of the quotient of two functions is equal to a fraction, the numerator of which is the difference between the products of the denominator and the derivative of the numerator and the numerator and the derivative of the denominator, and the denominator is the square of the former numerator. We get:

We have already found the derivative of the factors in the numerator in example 2. Let us also not forget that the product, which is the second factor in the numerator in the current example, is taken with a minus sign:

If you are looking for solutions to problems in which you need to find the derivative of a function, where there is a continuous pile of roots and powers, such as, for example, , then welcome to class "Derivative of sums of fractions with powers and roots" .

If you need to learn more about the derivatives of sines, cosines, tangents and other trigonometric functions, that is, when the function looks like , then a lesson for you "Derivatives of simple trigonometric functions" .

Example 5. Find the derivative of a function

Solution. In this function we see a product, one of the factors of which is the square root of the independent variable, the derivative of which we familiarized ourselves with in the table of derivatives. Using the rule for differentiating the product and the tabular value of the derivative of the square root, we obtain:

You can check the solution to the derivative problem at online derivatives calculator .

Example 6. Find the derivative of a function

Solution. In this function we see a quotient whose dividend is the square root of the independent variable. Using the rule of differentiation of quotients, which we repeated and applied in example 4, and the tabulated value of the derivative of the square root, we obtain:

To get rid of a fraction in the numerator, multiply the numerator and denominator by .

Functions of a complex type do not always fit the definition of a complex function. If there is a function of the form y = sin x - (2 - 3) · a r c t g x x 5 7 x 10 - 17 x 3 + x - 11, then it cannot be considered complex, unlike y = sin 2 x.

This article will show the concept of a complex function and its identification. Let's work with formulas for finding the derivative with examples of solutions in the conclusion. The use of the derivative table and differentiation rules significantly reduces the time for finding the derivative.

Basic definitions

A complex function is one whose argument is also a function.

It is denoted this way: f (g (x)). We have that the function g (x) is considered an argument f (g (x)).

Definition 2

If there is a function f and it is a cotangent function, then g(x) = ln x is the natural logarithm function. We find that the complex function f (g (x)) will be written as arctg(lnx). Or a function f, which is a function raised to the 4th power, where g (x) = x 2 + 2 x - 3 is considered an entire rational function, we obtain that f (g (x)) = (x 2 + 2 x - 3) 4 .

Obviously g(x) can be complex. From the example y = sin 2 x + 1 x 3 - 5 it is clear that the value of g has the cube root of the fraction. This expression can be denoted as y = f (f 1 (f 2 (x))). From where we have that f is a sine function, and f 1 is a function located under the square root, f 2 (x) = 2 x + 1 x 3 - 5 is a fractional rational function.

Definition 3

The degree of nesting is determined by any natural number and is written as y = f (f 1 (f 2 (f 3 (. . . (f n (x)))))) .

Definition 4

The concept of function composition refers to the number of nested functions according to the conditions of the problem. To solve, use the formula for finding the derivative of a complex function of the form

(f (g (x))) " = f " (g (x)) g " (x)

Examples

Find the derivative of a complex function of the form y = (2 x + 1) 2.

Solution

The condition shows that f is a squaring function, and g(x) = 2 x + 1 is considered a linear function.

Let's apply the derivative formula for a complex function and write:

f " (g (x)) = ((g (x)) 2) " = 2 (g (x)) 2 - 1 = 2 g (x) = 2 (2 x + 1) ; g " (x) = (2 x + 1) " = (2 x) " + 1 " = 2 x " + 0 = 2 1 x 1 - 1 = 2 ⇒ (f (g (x))) " = f " (g (x)) g " (x) = 2 (2 x + 1) 2 = 8 x + 4

It is necessary to find the derivative with a simplified original form of the function. We get:

y = (2 x + 1) 2 = 4 x 2 + 4 x + 1

From here we have that

y " = (4 x 2 + 4 x + 1) " = (4 x 2) " + (4 x) " + 1 " = 4 (x 2) " + 4 (x) " + 0 = = 4 · 2 · x 2 - 1 + 4 · 1 · x 1 - 1 = 8 x + 4

The results were the same.

When solving problems of this type, it is important to understand where the function of the form f and g (x) will be located.

Example 2

You should find the derivatives of complex functions of the form y = sin 2 x and y = sin x 2.

Solution

The first function notation says that f is the squaring function and g(x) is the sine function. Then we get that

y " = (sin 2 x) " = 2 sin 2 - 1 x (sin x) " = 2 sin x cos x

The second entry shows that f is a sine function, and g(x) = x 2 denotes a power function. It follows that we write the product of a complex function as

y " = (sin x 2) " = cos (x 2) (x 2) " = cos (x 2) 2 x 2 - 1 = 2 x cos (x 2)

The formula for the derivative y = f (f 1 (f 2 (f 3 (. . . (f n (x))))) will be written as y " = f " (f 1 (f 2 (f 3 (. . . ( f n (x))))) · f 1 " (f 2 (f 3 (. . . (f n (x)))) · · f 2 " (f 3 (. . . (f n (x))) )) · . . . fn "(x)

Example 3

Find the derivative of the function y = sin (ln 3 a r c t g (2 x)).

Solution

This example shows the difficulty of writing and determining the location of functions. Then y = f (f 1 (f 2 (f 3 (f 4 (x))))) denote where f , f 1 , f 2 , f 3 , f 4 (x) is the sine function, the function of raising to 3 degree, function with logarithm and base e, arctangent and linear function.

From the formula for defining a complex function we have that

y " = f " (f 1 (f 2 (f 3 (f 4 (x)))) f 1 " (f 2 (f 3 (f 4 (x)))) f 2 " (f 3 (f 4 (x)) f 3 " (f 4 (x)) f 4 " (x)

We get what we need to find

1. f " (f 1 (f 2 (f 3 (f 4 (x))))) as the derivative of the sine according to the table of derivatives, then f " (f 1 (f 2 (f 3 (f 4 (x)))) ) = cos (ln 3 a r c t g (2 x)) .
2. f 1 " (f 2 (f 3 (f 4 (x)))) as the derivative of a power function, then f 1 " (f 2 (f 3 (f 4 (x)))) = 3 ln 3 - 1 a r c t g (2 x) = 3 ln 2 a r c t g (2 x) .
3. f 2 " (f 3 (f 4 (x))) as a logarithmic derivative, then f 2 " (f 3 (f 4 (x))) = 1 a r c t g (2 x) .
4. f 3 " (f 4 (x)) as the derivative of the arctangent, then f 3 " (f 4 (x)) = 1 1 + (2 x) 2 = 1 1 + 4 x 2.
5. When finding the derivative f 4 (x) = 2 x, remove 2 from the sign of the derivative using the formula for the derivative of a power function with an exponent equal to 1, then f 4 " (x) = (2 x) " = 2 x " = 2 · 1 · x 1 - 1 = 2 .

We combine the intermediate results and get that

y " = f " (f 1 (f 2 (f 3 (f 4 (x)))) f 1 " (f 2 (f 3 (f 4 (x)))) f 2 " (f 3 (f 4 (x)) f 3 " (f 4 (x)) f 4 " (x) = = cos (ln 3 a r c t g (2 x)) 3 ln 2 a r c t g (2 x) 1 a r c t g (2 x) 1 1 + 4 x 2 2 = = 6 cos (ln 3 a r c t g (2 x)) ln 2 a r c t g (2 x) a r c t g (2 x) (1 + 4 x 2)

Analysis of such functions is reminiscent of nesting dolls. Differentiation rules cannot always be applied explicitly using a derivative table. Often you need to use a formula for finding derivatives of complex functions.

There are some differences between complex appearance and complex functions. With a clear ability to distinguish this, finding derivatives will be especially easy.

Example 4

It is necessary to consider giving such an example. If there is a function of the form y = t g 2 x + 3 t g x + 1, then it can be considered as a complex function of the form g (x) = t g x, f (g) = g 2 + 3 g + 1. Obviously, it is necessary to use the formula for a complex derivative:

f " (g (x)) = (g 2 (x) + 3 g (x) + 1) " = (g 2 (x)) " + (3 g (x)) " + 1 " = = 2 · g 2 - 1 (x) + 3 g " (x) + 0 = 2 g (x) + 3 1 g 1 - 1 (x) = = 2 g (x) + 3 = 2 t g x + 3 ; g " (x) = (t g x) " = 1 cos 2 x ⇒ y " = (f (g (x))) " = f " (g (x)) g " (x) = (2 t g x + 3 ) · 1 cos 2 x = 2 t g x + 3 cos 2 x

A function of the form y = t g x 2 + 3 t g x + 1 is not considered complex, since it has the sum of t g x 2, 3 t g x and 1. However, t g x 2 is considered a complex function, then we obtain a power function of the form g (x) = x 2 and f, which is a tangent function. To do this, differentiate by amount. We get that

y " = (t g x 2 + 3 t g x + 1) " = (t g x 2) " + (3 t g x) " + 1 " = = (t g x 2) " + 3 (t g x) " + 0 = (t g x 2) " + 3 cos 2 x

Let's move on to finding the derivative of a complex function (t g x 2) ":

f " (g (x)) = (t g (g (x))) " = 1 cos 2 g (x) = 1 cos 2 (x 2) g " (x) = (x 2) " = 2 x 2 - 1 = 2 x ⇒ (t g x 2) " = f " (g (x)) g " (x) = 2 x cos 2 (x 2)

We get that y " = (t g x 2 + 3 t g x + 1) " = (t g x 2) " + 3 cos 2 x = 2 x cos 2 (x 2) + 3 cos 2 x

Functions of a complex type can be included in complex functions, and complex functions themselves can be components of functions of a complex type.

Example 5

For example, consider a complex function of the form y = log 3 x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 + ln 2 x (x 2 + 1)

This function can be represented as y = f (g (x)), where the value of f is a function of the base 3 logarithm, and g (x) is considered the sum of two functions of the form h (x) = x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 and k (x) = ln 2 x · (x 2 + 1) . Obviously, y = f (h (x) + k (x)).

Consider the function h(x). This is the ratio l (x) = x 2 + 3 cos 3 (2 x + 1) + 7 to m (x) = e x 2 + 3 3

We have that l (x) = x 2 + 3 cos 2 (2 x + 1) + 7 = n (x) + p (x) is the sum of two functions n (x) = x 2 + 7 and p (x) = 3 cos 3 (2 x + 1) , where p (x) = 3 p 1 (p 2 (p 3 (x))) is a complex function with numerical coefficient 3, and p 1 is a cube function, p 2 by a cosine function, p 3 (x) = 2 x + 1 by a linear function.

We found that m (x) = e x 2 + 3 3 = q (x) + r (x) is the sum of two functions q (x) = e x 2 and r (x) = 3 3, where q (x) = q 1 (q 2 (x)) is a complex function, q 1 is a function with an exponential, q 2 (x) = x 2 is a power function.

This shows that h (x) = l (x) m (x) = n (x) + p (x) q (x) + r (x) = n (x) + 3 p 1 (p 2 ( p 3 (x))) q 1 (q 2 (x)) + r (x)

When moving to an expression of the form k (x) = ln 2 x · (x 2 + 1) = s (x) · t (x), it is clear that the function is presented in the form of a complex s (x) = ln 2 x = s 1 ( s 2 (x)) with a rational integer t (x) = x 2 + 1, where s 1 is a squaring function, and s 2 (x) = ln x is logarithmic with base e.

It follows that the expression will take the form k (x) = s (x) · t (x) = s 1 (s 2 (x)) · t (x).

Then we get that

y = log 3 x 2 + 3 cos 3 (2 x + 1) + 7 e x 2 + 3 3 + ln 2 x (x 2 + 1) = = f n (x) + 3 p 1 (p 2 (p 3 (x))) q 1 (q 2 (x)) = r (x) + s 1 (s 2 (x)) t (x)

Based on the structures of the function, it became clear how and what formulas need to be used to simplify the expression when differentiating it. To become familiar with such problems and for the concept of their solution, it is necessary to turn to the point of differentiating a function, that is, finding its derivative.

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