Derivative of natural logarithm and logarithm to base a.

Proof and derivation of formulas for the derivative of the natural logarithm and the logarithm to base a. Examples of calculating derivatives of ln 2x, ln 3x and ln nx. Proof of the formula for the derivative of the nth order logarithm using the method of mathematical induction.

Derivation of formulas for the derivatives of the natural logarithm and the logarithm to base a

The derivative of the natural logarithm of x is equal to one divided by x:
(1) (ln x)′ =.

The derivative of the logarithm to base a is equal to one divided by the variable x multiplied by the natural logarithm of a:
(2) (log a x)′ =.

Proof

Let there be some positive number, not equal to one. Consider a function depending on a variable x, which is a logarithm to the base:
.
This function is defined at . Let's find its derivative with respect to the variable x. By definition, the derivative is the following limit:
(3) .

Let's transform this expression to reduce it to known mathematical properties and rules. To do this we need to know the following facts:
A) Properties of the logarithm. We will need the following formulas:
(4) ;
(5) ;
(6) ;
B) Continuity of the logarithm and the property of limits for a continuous function:
(7) .
Here is a function that has a limit and this limit is positive.
IN) The meaning of the second remarkable limit:
(8) .

Let's apply these facts to our limit. First we transform the algebraic expression
.
To do this, we apply properties (4) and (5).

.

Let us use property (7) and the second remarkable limit (8):
.

And finally, we apply property (6):
.
Logarithm to base e called natural logarithm. It is designated as follows:
.
Then ;
.

Thus, we obtained formula (2) for the derivative of the logarithm.

Derivative of the natural logarithm

Once again we write out the formula for the derivative of the logarithm to base a:
.
This formula has the simplest form for the natural logarithm, for which , . Then
(1) .

Because of this simplicity, the natural logarithm is very widely used in mathematical analysis and in other branches of mathematics related to differential calculus. Logarithmic functions with other bases can be expressed through the natural logarithm using property (6):
.

The derivative of the logarithm with respect to the base can be found from formula (1), if you take the constant out of the differentiation sign:
.

Other ways to prove the derivative of a logarithm

Here we assume that we know the formula for the derivative of the exponential:
(9) .
Then we can derive the formula for the derivative of the natural logarithm, given that the logarithm is the inverse function of the exponential.

Let us prove the formula for the derivative of the natural logarithm, applying the formula for the derivative of the inverse function:
.
In our case . Inverse function the exponential to the natural logarithm is:
.
Its derivative is determined by formula (9). Variables can be designated by any letter. In formula (9), replace the variable x with y:
.
Since then
.
Then
.
The formula is proven.

Now we prove the formula for the derivative of the natural logarithm using differentiation rules complex function . Since the functions and are inverse to each other, then
.
Let's differentiate this equation with respect to the variable x:
(10) .
The derivative of x is equal to one:
.
We apply the rule of differentiation of complex functions:
.
Here . Let's substitute in (10):
.
From here
.

Example

Find derivatives of ln 2x, ln 3x And lnnx.

Solution

The original functions have a similar form. Therefore we will find the derivative of the function y = log nx. Then we substitute n = 2 and n = 3. And, thus, we obtain formulas for the derivatives of ln 2x And ln 3x .

So, we are looking for the derivative of the function
y = log nx .
Let's imagine this function as a complex function consisting of two functions:
1) Functions depending on a variable: ;
2) Functions depending on a variable: .
Then the original function is composed of the functions and :
.

Let's find the derivative of the function with respect to the variable x:
.
Let's find the derivative of the function with respect to the variable:
.
We apply the formula for the derivative of a complex function.
.
Here we set it up.

So we found:
(11) .
We see that the derivative does not depend on n. This result is quite natural if we transform the original function using the formula for the logarithm of the product:
.
- this is a constant. Its derivative is zero. Then, according to the rule of differentiation of the sum, we have:
.

Answer

; ; .

Derivative of the logarithm of modulus x

Let's find the derivative of another very important function- natural logarithm of modulus x:
(12) .

Let's consider the case. Then the function looks like:
.
Its derivative is determined by formula (1):
.

Now let's consider the case. Then the function looks like:
,
Where .
But we also found the derivative of this function in the example above. It does not depend on n and is equal to
.
Then
.

We combine these two cases into one formula:
.

Accordingly, for the logarithm to base a, we have:
.

Derivatives of higher orders of the natural logarithm

Consider the function
.
We found its first-order derivative:
(13) .

Let's find the second-order derivative:
.
Let's find the third order derivative:
.
Let's find the fourth order derivative:
.

You can notice that the nth order derivative has the form:
(14) .
Let us prove this by mathematical induction.

Proof

Let us substitute the value n = 1 into formula (14):
.
Since , then when n = 1 , formula (14) is valid.

Let us assume that formula (14) is satisfied for n = k. Let us prove that this implies that the formula is valid for n = k + 1 .

Indeed, for n = k we have:
.
Differentiate with respect to the variable x:

.
So we got:
.
This formula coincides with formula (14) for n = k + 1 . Thus, from the assumption that formula (14) is valid for n = k, it follows that formula (14) is valid for n = k + 1 .

Therefore, formula (14), for the nth order derivative, is valid for any n.

Derivatives of higher orders of logarithm to base a

To find the nth order derivative of a logarithm to base a, you need to express it in terms of the natural logarithm:
.
Applying formula (14), we find the nth derivative:
.

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

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

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

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

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

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

Let's formulate it.

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

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

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

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

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

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

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

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

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

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

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

Rules of differentiation

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

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

Table of derivatives of some functions

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

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 appeared and exactly certain rules differentiation. 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 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 the algebraic sum of functions is equal to algebraic sum 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 error that occurs on initial stage studying derivatives, but as they solve several one- and two-part examples, the average student 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).

Other common mistake - mechanical solution derivative of a complex function as a 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:

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 others 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:

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 .