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Derivative of the exponential to the x power. Derivative of e to the power of x and an exponential function Derivative of the function e to the power of x

05.06.2022

Many numbers acquired their magnitude and superstitious meaning in ancient times. Nowadays, new myths are being added to them. There are many legends about the number pi; the famous Fibonacci numbers are not much less famous than it. But perhaps the most surprising thing is the number e, which he cannot do without modern mathematics, physics and even economics.

The arithmetic value of e is approximately 2.718. Why not exactly, but approximately? Because this number is irrational and transcendental, it cannot be expressed as a fraction with natural integers or a polynomial with rational coefficients. For most calculations, the specified accuracy of 2.718 is sufficient, although the modern level of computing technology allows its value to be determined with an accuracy of more than a trillion decimal places.

The main feature of the number e is that the derivative of its exponential function f (x) = e x is equal to the value of the function e x itself. No other mathematical relationship has such an unusual property. Let's talk about this in a little more detail.

What is a limit

First, let's understand the concept of limit. Consider some mathematical expression, for example, i = 1/n. Can see, that as “n” increases", the value of "i" will decrease, and as "n" tends to infinity (which is indicated by the sign ∞), "i" will tend to the limit value (more often called just the limit) equal to zero. The expression for the limit (denoted as lim) for the case under consideration can be written as lim n →∞ (1/ n) = 0.

There are different limits for different expressions. One of these limits, included in Soviet and Russian textbooks as the second remarkable limit, is the expression lim n →∞ (1+1/ n) n. Already in the Middle Ages it was established that the limit of this expression is the number e.

The first remarkable limit includes the expression lim n →∞ (Sin n / n) = 1.

How to find the derivative of e x - in this video.

What is the derivative of a function

To explain the concept of a derivative, we should recall what a function is in mathematics. In order not to clutter the text with complex definitions, we will focus on the intuitive mathematical concept of a function, which consists in the fact that in it one or more quantities completely determine the value of another quantity if they are interrelated. For example, in the formula S = π ∙ r 2 the area of a circle, the value of the radius r completely and uniquely determines the area of the circle S.

Depending on the type, functions can be algebraic, trigonometric, logarithmic, etc. They can have two, three or more arguments interconnected. For example, the distance S traveled, which an object covered at a uniformly accelerated speed, is described by the function S = 0.5 ∙ a ∙ t 2 + V ∙ t, where “t” is the time of movement, the argument “a” is acceleration (can be either positive or and a negative value) and “V” is the initial speed of movement. Thus, the distance traveled depends on the values of three arguments, two of which (“a” and “V”) are constant.

Let us use this example to demonstrate the elementary concept of a derivative of a function. It characterizes the rate of change of the function at a given point. In our example, this will be the speed of movement of the object at a specific moment in time. With constant “a” and “V”, it depends only on time “t”, that is, in scientific language, you need to take the derivative of the function S with respect to time “t”.

This process is called differentiation and is performed by calculating the limit of the ratio of the growth of a function to the growth of its argument by a negligibly small amount. Solving such problems for individual functions is often difficult and is not discussed here. It's also worth noting that some functions at certain points have no such limits at all.

In our example, the derivative S over time “t” will take the form S" = ds/dt = a ∙ t + V, from which it can be seen that the speed S" changes according to a linear law depending on “t”.

Derivative of the exponent

An exponential function is called an exponential function, the base of which is the number e. It is usually displayed in the form F (x) = e x, where the exponent x is a variable quantity. This function has complete differentiability over the entire range of real numbers. As x grows, it constantly increases and is always greater than zero. Its inverse function is the logarithm.

The famous mathematician Taylor managed to expand this function into a series named after him e x = 1 + x/1! + x 2 /2! + x 3 /3! + … in the x range from - ∞ to + ∞.

Law based on this function, is called exponential. He describes:

increase in compound bank interest rates;
increase in animal populations and global population;
rigor mortis time and much more.

Let us repeat once again the remarkable property of this dependence - the value of its derivative at any point is always equal to the value of the function at this point, that is, (e x)" = e x.

Let us present the derivatives for the most general cases of the exponential:

(e ax)" = a ∙ e ax;
(e f (x))" = f"(x) ∙ e f (x) .

Using these dependencies, it is easy to find derivatives for other particular types of this function.

Some interesting facts about the number e

The names of such scientists as Napier, Oughtred, Huygens, Bernoulli, Leibniz, Newton, Euler, and others are associated with this number. The latter actually introduced the notation e for this number, and also found the first 18 signs, using the series e = 1 + 1/1 that he discovered for the calculation! + 2/2! + 3/3! ...

The number e appears in the most unexpected places. For example, it is included in the catenary equation, which describes the sag of a rope under its own weight when its ends are fixed to supports.

Video

The topic of the video lesson is the derivative of the exponential function.

Proof and derivation of the formulas for the derivative of the exponential (e to the x power) and the exponential function (a to the x power). Examples of calculating derivatives of e^2x, e^3x and e^nx. Formulas for derivatives of higher orders.

Content

See also: Exponential function - properties, formulas, graph
Exponent, e to the x power - properties, formulas, graph

Basic formulas

The derivative of an exponent is equal to the exponent itself (the derivative of e to the x power is equal to e to the x power):
(1) (e x )′ = e x.

The derivative of an exponential function with a base a is equal to the function itself multiplied by the natural logarithm of a:
(2) .

An exponential is an exponential function whose base is equal to the number e, which is the following limit:
.
Here it can be either a natural number or a real number. Next, we derive formula (1) for the derivative of the exponential.

Derivation of the exponential derivative formula

Consider the exponential, e to the x power:
y = e x .
This function is defined for everyone.
(3) .

Let's find its derivative with respect to the variable x.
By definition, the derivative is the following limit: Let's transform this expression to reduce it to known mathematical properties and rules. To do this we need the following facts:
(4) ;
A) Exponent property:
(5) ;
B) Property of logarithm:
(6) .
IN)
Continuity of the logarithm and the property of limits for a continuous function: The meaning of the second remarkable limit:
(7) .

Let's apply these facts to our limit (3). We use property (4):
;
.

Let's make a substitution.
Then ; .
.
Due to the continuity of the exponential,
.

Therefore, when , .
.

As a result we get:
Let's make a substitution.
.

Then . At , . And we have:
.
Let's apply the logarithm property (5):
.

Then

Let's apply property (6). Since there is a positive limit and the logarithm is continuous, then:
(8)
Here we also used the second remarkable limit (7). Then

Thus, we obtained formula (1) for the derivative of the exponential.
;
.
Derivation of the formula for the derivative of an exponential function
.

Now we derive formula (2) for the derivative of the exponential function with a base of degree a.

We believe that and .
(14) .
(1) .

Then the exponential function
;
.

Defined for everyone.
.

Let's transform formula (8). To do this, we will use the properties of the exponential function and logarithm.

So, we transformed formula (8) to the following form:
.
Higher order derivatives of e to the x power
(15) .

Now let's find derivatives of higher orders. Let's look at the exponent first:
;
.

We see that the derivative of function (14) is equal to function (14) itself. Differentiating (1), we obtain derivatives of the second and third order:
.

This shows that the nth order derivative is also equal to the original function:

Derivatives of higher orders of the exponential function

Now consider an exponential function with a base of degree a:

We found its first-order derivative:
Differentiating (15), we obtain derivatives of the second and third order:
We see that each differentiation leads to the multiplication of the original function by .
Therefore, the nth order derivative has the following form:

Derivation of the derivative of an exponential to the power $x$

We have $e^x$, where $x: -\infty

$y"=\lim\limits_(\Delta x\to 0) \frac(e^(x+\Delta x)-e^x)(\Delta x)$.

By the property of the exponent $e^(a+bx)=e^a*e^b$ we can transform the numerator of the limit:

$e^(x+\Delta x)-e^x = e^x*e^(\Delta x)-e^x = e^x(e^(\Delta x)-1)$.

That is, $y"=\lim\limits_(\Delta x\to 0) \frac(e^(x+\Delta x)-e^x)(\Delta x)=\lim\limits_(\Delta x\to 0) \frac(e^x(e^(\Delta x)-1))(\Delta x)$.

Let us denote $t=e^(\Delta x)-1$. We get $e^(\Delta x)=t+1$, and by the property of the logarithm it turns out that $\Delta x = ln(t+1)$.

Since the exponential is continuous, we have $\lim\limits_(\Delta x\to 0) e^(\Delta x)=e^0=1.$ Therefore, if $\Delta x\to 0$, then $t \ to 0$.

As a result, we show the transformation:

$y"=\lim\limits_(\Delta x\to 0) \frac(e^(\Delta x)-1)(\Delta x)=e^x\lim\limits_(t\to 0)\frac (t)(ln(t+1))$.

Let us denote $n=\frac (1)(t)$, then $t=\frac(1)(n)$. It turns out that if $t\to 0$, then $n\to\infty$.

Let's transform our limit:

$y"=e^x\lim\limits_(t\to 0)\frac(t)(ln(t+1))=e^x\lim\limits_(n\to\infty)\frac(1) (n\cdot ln(\frac(1)(n)+1)^n)$.

By the property of the logarithm $b\cdot ln c=ln c^b$ we have

$n\cdot ln (\frac(1)(n)+1)=ln(\frac(1)(n)+1)^n=ln(1+\frac(1)(n))^n$ .

The limit is converted as follows:

$y"=e^x\lim\limits_(n\to\infty)\frac(1)(n\cdot ln(\frac(1)(n)+1)) = e^x\lim\limits_( n\to\infty)\frac(1)(ln(\frac(1)(n)+1)^n)= e^x\frac(1)(\lim\limits_(n\to\infty) ln (\frac(1)(n)+1)^n)$.

According to the property of continuity of the logarithm and the property of limits for a continuous function: $\lim\limits_(x\to x_0)ln(f(x))=ln(\lim\limits_f(x))$, where $f(x)$ has positive limit $\lim\limits_(x\to x_0)f(x)$. So, due to the fact that the logarithm is continuous and there is a positive limit $\lim\limits_(n\to\infty)(\frac(1)(n)+1)^n$, we can deduce:

$\lim\limits_(n\to\infty)ln(1+\frac(1)(n))^n=ln\lim\limits_(n\to\infty)ln(1+\frac(1)( n))^n=ln e=1$.

Let's use the value of the second remarkable limit $\lim\limits_(n\to\infty)(1+\frac(1)(n))^n=e$. We get:

$y"= e^x\frac(1)(\lim\limits_(n\to\infty) ln(\frac(1)(n)+1)^n) = e^x\cdot\frac(1 )(ln e) = e^x\cdot\frac(1)(1)=e^x$.

Thus, we have derived the formula for the derivative of an exponential and can claim that the derivative of an exponential to the power of $x$ is equivalent to the derivative of an exponential to the power of $x$:

There are also other ways to derive this formula using other formulas and rules.

Example 1

Let's look at an example of finding the derivative of a function.

Condition: Find the derivative of the function $y=2^x + 3^x + 10^x + e^x$.

Solution: To the terms $2^x, 3^x$ and $10^x$ we apply the formula $(a^x)"=a^x\cdot ln a$. According to the derived formula $(e^x)"=e^x$ the fourth term $e^x$ does not change.

Answer: $y" = 2^x\cdot ln 2 + 3^x\cdot ln 3 + 10^x\cdot ln 10 + e^x$.

Thus, we have derived the formula $(e^x)"=e^x$, while giving definitions to the basic concepts, and analyzed an example of finding the derivative of a function with an exponent as one of the terms.

We present a summary table for convenience and clarity when studying the topic.

Constanty = C

Power function y = x p

(x p) " = p x p - 1

Exponential functiony = a x

(a x) " = a x ln a

In particular, whena = ewe have y = e x

(e x) " = e x

Logarithmic function

(log a x) " = 1 x ln a

In particular, whena = ewe have y = logx

(ln x) " = 1 x

Trigonometric functions

(sin x) " = cos x (cos x) " = - sin x (t g x) " = 1 cos 2 x (c t g x) " = - 1 sin 2 x

Inverse trigonometric functions

(a r c sin x) " = 1 1 - x 2 (a r c cos x) " = - 1 1 - x 2 (a r c t g x) " = 1 1 + x 2 (a r c c t g x) " = - 1 1 + x 2

Hyperbolic functions

(s h x) " = c h x (c h x) " = s h x (t h x) " = 1 c h 2 x (c t h x) " = - 1 s h 2 x

Let us analyze how the formulas of the specified table were obtained or, in other words, we will prove the derivation of derivative formulas for each type of function.

Derivative of a constant

Evidence 1

In order to derive this formula, we take as a basis the definition of the derivative of a function at a point. We use x 0 = x, where x takes the value of any real number, or, in other words, x is any number from the domain of the function f (x) = C. Let's write down the limit of the ratio of the increment of a function to the increment of the argument as ∆ x → 0:

lim ∆ x → 0 ∆ f (x) ∆ x = lim ∆ x → 0 C - C ∆ x = lim ∆ x → 0 0 ∆ x = 0

Please note that the expression 0 ∆ x falls under the limit sign. It is not the uncertainty “zero divided by zero,” since the numerator does not contain an infinitesimal value, but precisely zero. In other words, the increment of a constant function is always zero.

So, the derivative of the constant function f (x) = C is equal to zero throughout the entire domain of definition.

Example 1

The constant functions are given:

f 1 (x) = 3, f 2 (x) = a, a ∈ R, f 3 (x) = 4. 13 7 22 , f 4 (x) = 0 , f 5 (x) = - 8 7

Solution

Let us describe the given conditions. In the first function we see the derivative of the natural number 3. In the following example, you need to take the derivative of A, Where A- any real number. The third example gives us the derivative of the irrational number 4. 13 7 22, the fourth is the derivative of zero (zero is an integer). Finally, in the fifth case we have the derivative of the rational fraction - 8 7.

Answer: derivatives of given functions are zero for any real x(over the entire definition area)

f 1 " (x) = (3) " = 0 , f 2 " (x) = (a) " = 0 , a ∈ R , f 3 " (x) = 4 . 13 7 22 " = 0 , f 4 " (x) = 0 " = 0 , f 5 " (x) = - 8 7 " = 0

Derivative of a power function

Let's move on to the power function and the formula for its derivative, which has the form: (x p) " = p x p - 1, where the exponent p is any real number.

Evidence 2

Here is a proof of the formula when the exponent is a natural number: p = 1, 2, 3, …

We again rely on the definition of a derivative. Let's write down the limit of the ratio of the increment of a power function to the increment of the argument:

(x p) " = lim ∆ x → 0 = ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x

To simplify the expression in the numerator, we use Newton’s binomial formula:

(x + ∆ x) p - x p = C p 0 + x p + C p 1 · x p - 1 · ∆ x + C p 2 · x p - 2 · (∆ x) 2 + . . . + + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p - x p = = C p 1 x p - 1 ∆ x + C p 2 x p - 2 (∆ x) 2 + . . . + C p p - 1 x (∆ x) p - 1 + C p p (∆ x) p

Thus:

(x p) " = lim ∆ x → 0 ∆ (x p) ∆ x = lim ∆ x → 0 (x + ∆ x) p - x p ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 ∆ x + C p 2 · x p - 2 · (∆ x) 2 + . + C p p - 1 · x · (∆ x) p - 1 + C p p · (∆ x) p) ∆ x = = lim ∆ x → 0 (C p 1 x p - 1 + C p 2 x p - 2 ∆ x + . . + C p p - 1 x (∆ x) p - 2 + C p p (∆ x) p - 1) = = C p 1 · x p - 1 + 0 + .

Thus, we have proven the formula for the derivative of a power function when the exponent is a natural number.

Evidence 3

To provide proof for the case when p- any real number other than zero, we use the logarithmic derivative (here we should understand the difference from the derivative of a logarithmic function). To have a more complete understanding, it is advisable to study the derivative of a logarithmic function and further understand the derivative of an implicit function and the derivative of a complex function.

Let's consider two cases: when x positive and when x negative.

So x > 0. Then: x p > 0 . Let us logarithm the equality y = x p to base e and apply the property of the logarithm:

y = x p ln y = ln x p ln y = p · ln x

At this stage, we have obtained an implicitly specified function. Let's define its derivative:

(ln y) " = (p · ln x) 1 y · y " = p · 1 x ⇒ y " = p · y x = p · x p x = p · x p - 1

Now we consider the case when x – a negative number.

If the indicator p is an even number, then the power function is defined for x< 0 , причем является четной: y (x) = - y ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p · x p - 1

Then x p< 0 и возможно составить доказательство, используя логарифмическую производную.

If p is an odd number, then the power function is defined for x< 0 , причем является нечетной: y (x) = - y (- x) = - (- x) p . Тогда x p < 0 , а значит логарифмическую производную задействовать нельзя. В такой ситуации возможно взять за основу доказательства правила дифференцирования и правило нахождения производной сложной функции:

y " (x) = (- (- x) p) " = - ((- x) p) " = - p · (- x) p - 1 · (- x) " = = p · (- x) p - 1 = p x p - 1

The last transition is possible due to the fact that if p is an odd number, then p - 1 either an even number or zero (for p = 1), therefore, for negative x the equality (- x) p - 1 = x p - 1 is true.

So, we have proven the formula for the derivative of a power function for any real p.

Example 2

Functions given:

f 1 (x) = 1 x 2 3 , f 2 (x) = x 2 - 1 4 , f 3 (x) = 1 x log 7 12

Determine their derivatives.

Solution

We transform some of the given functions into tabular form y = x p , based on the properties of the degree, and then use the formula:

f 1 (x) = 1 x 2 3 = x - 2 3 ⇒ f 1 " (x) = - 2 3 x - 2 3 - 1 = - 2 3 x - 5 3 f 2 " (x) = x 2 - 1 4 = 2 - 1 4 x 2 - 1 4 - 1 = 2 - 1 4 x 2 - 5 4 f 3 (x) = 1 x log 7 12 = x - log 7 12 ⇒ f 3" ( x) = - log 7 12 x - log 7 12 - 1 = - log 7 12 x - log 7 12 - log 7 7 = - log 7 12 x - log 7 84

Derivative of an exponential function

Proof 4

Let us derive the derivative formula using the definition as a basis:

(a x) " = lim ∆ x → 0 a x + ∆ x - a x ∆ x = lim ∆ x → 0 a x (a ∆ x - 1) ∆ x = a x lim ∆ x → 0 a ∆ x - 1 ∆ x = 0 0

We got uncertainty. To expand it, let's write a new variable z = a ∆ x - 1 (z → 0 as ∆ x → 0). In this case, a ∆ x = z + 1 ⇒ ∆ x = log a (z + 1) = ln (z + 1) ln a . For the last transition, the formula for transition to a new logarithm base was used.

Let us substitute into the original limit:

(a x) " = a x · lim ∆ x → 0 a ∆ x - 1 ∆ x = a x · ln a · lim ∆ x → 0 1 1 z · ln (z + 1) = = a x · ln a · lim ∆ x → 0 1 ln (z + 1) 1 z = a x · ln a · 1 ln lim ∆ x → 0 (z + 1) 1 z

Let us remember the second remarkable limit and then we obtain the formula for the derivative of the exponential function:

(a x) " = a x · ln a · 1 ln lim z → 0 (z + 1) 1 z = a x · ln a · 1 ln e = a x · ln a

Example 3

The exponential functions are given:

f 1 (x) = 2 3 x , f 2 (x) = 5 3 x , f 3 (x) = 1 (e) x

It is necessary to find their derivatives.

Solution

We use the formula for the derivative of the exponential function and the properties of the logarithm:

f 1 " (x) = 2 3 x " = 2 3 x ln 2 3 = 2 3 x (ln 2 - ln 3) f 2 " (x) = 5 3 x " = 5 3 x ln 5 1 3 = 1 3 5 3 x ln 5 f 3 " (x) = 1 (e) x " = 1 e x " = 1 e x ln 1 e = 1 e x ln e - 1 = - 1 e x

Derivative of a logarithmic function

Evidence 5

Let us provide a proof of the formula for the derivative of a logarithmic function for any x in the domain of definition and any permissible values of the base a of the logarithm. Based on the definition of derivative, we get:

(log a x) " = lim ∆ x → 0 log a (x + ∆ x) - log a x ∆ x = lim ∆ x → 0 log a x + ∆ x x ∆ x = = lim ∆ x → 0 1 ∆ x log a 1 + ∆ x x = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x = = lim ∆ x → 0 log a 1 + ∆ x x 1 ∆ x · x x = lim ∆ x → 0 1 x · log a 1 + ∆ x x x ∆ x = = 1 x · log a lim ∆ x → 0 1 + ∆ x x x ∆ x = 1 x · log a e = 1 x · ln e ln a = 1 x · ln a

From the indicated chain of equalities it is clear that the transformations were based on the property of the logarithm. The equality lim ∆ x → 0 1 + ∆ x x x ∆ x = e is true in accordance with the second remarkable limit.

Example 4

Logarithmic functions are given:

f 1 (x) = log ln 3 x , f 2 (x) = ln x

It is necessary to calculate their derivatives.

Solution

Let's apply the derived formula:

f 1 " (x) = (log ln 3 x) " = 1 x · ln (ln 3) ; f 2 " (x) = (ln x) " = 1 x ln e = 1 x

So, the derivative of the natural logarithm is one divided by x.

Derivatives of trigonometric functions

Proof 6

Let's use some trigonometric formulas and the first wonderful limit to derive the formula for the derivative of a trigonometric function.

According to the definition of the derivative of the sine function, we get:

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x

The formula for the difference of sines will allow us to perform the following actions:

(sin x) " = lim ∆ x → 0 sin (x + ∆ x) - sin x ∆ x = = lim ∆ x → 0 2 sin x + ∆ x - x 2 cos x + ∆ x + x 2 ∆ x = = lim ∆ x → 0 sin ∆ x 2 · cos x + ∆ x 2 ∆ x 2 = = cos x + 0 2 · lim ∆ x → 0 sin ∆ x 2 ∆ x 2

Finally, we use the first wonderful limit:

sin " x = cos x + 0 2 · lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = cos x

So, the derivative of the function sin x will cos x.

We will also prove the formula for the derivative of the cosine:

cos " x = lim ∆ x → 0 cos (x + ∆ x) - cos x ∆ x = = lim ∆ x → 0 - 2 sin x + ∆ x - x 2 sin x + ∆ x + x 2 ∆ x = = - lim ∆ x → 0 sin ∆ x 2 sin x + ∆ x 2 ∆ x 2 = = - sin x + 0 2 lim ∆ x → 0 sin ∆ x 2 ∆ x 2 = - sin x

Those. the derivative of the cos x function will be – sin x.

We derive the formulas for the derivatives of tangent and cotangent based on the rules of differentiation:

t g " x = sin x cos x " = sin " x · cos x - sin x · cos " x cos 2 x = = cos x · cos x - sin x · (- sin x) cos 2 x = sin 2 x + cos 2 x cos 2 x = 1 cos 2 x c t g " x = cos x sin x " = cos " x · sin x - cos x · sin " x sin 2 x = = - sin x · sin x - cos x · cos x sin 2 x = - sin 2 x + cos 2 x sin 2 x = - 1 sin 2 x

Derivatives of inverse trigonometric functions

The section on the derivative of inverse functions provides comprehensive information on the proof of the formulas for the derivatives of arcsine, arccosine, arctangent and arccotangent, so we will not duplicate the material here.

Derivatives of hyperbolic functions

Evidence 7

We can derive the formulas for the derivatives of the hyperbolic sine, cosine, tangent and cotangent using the differentiation rule and the formula for the derivative of the exponential function:

s h " x = e x - e - x 2 " = 1 2 e x " - e - x " = = 1 2 e x - - e - x = e x + e - x 2 = c h x c h " x = e x + e - x 2 " = 1 2 e x " + e - x " = = 1 2 e x + - e - x = e x - e - x 2 = s h x t h " x = s h x c h x " = s h " x · c h x - s h x · c h " x c h 2 x = c h 2 x - s h 2 x c h 2 x = 1 c h 2 x c t h " x = c h x s h x " = c h " x · s h x - c h x · s h " x s h 2 x = s h 2 x - c h 2 x s h 2 x = - 1 s h 2 x

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When deriving the very first formula of the table, we will proceed from the definition of the derivative function at a point. Let's take where x– any real number, that is, x– any number from the domain of definition of the function. Let us write down the limit of the ratio of the increment of the function to the increment of the argument at :

It should be noted that under the limit sign the expression is obtained, which is not the uncertainty of zero divided by zero, since the numerator does not contain an infinitesimal value, but precisely zero. In other words, the increment of a constant function is always zero.

Thus, derivative of a constant functionis equal to zero throughout the entire domain of definition.

Derivative of a power function.

The formula for the derivative of a power function has the form , where the exponent p– any real number.

Let us first prove the formula for the natural exponent, that is, for p = 1, 2, 3, …

We will use the definition of a derivative. Let us write down the limit of the ratio of the increment of a power function to the increment of the argument:

To simplify the expression in the numerator, we turn to the Newton binomial formula:

Hence,

This proves the formula for the derivative of a power function for a natural exponent.

Derivative of an exponential function.

We present the derivation of the derivative formula based on the definition:

We have arrived at uncertainty. To expand it, we introduce a new variable, and at . Then . In the last transition, we used the formula for transitioning to a new logarithmic base.

Let's substitute into the original limit:

If we recall the second remarkable limit, we arrive at the formula for the derivative of the exponential function:

Derivative of a logarithmic function.

Let us prove the formula for the derivative of a logarithmic function for all x from the domain of definition and all valid values of the base a logarithm By definition of derivative we have:

As you noticed, during the proof the transformations were carried out using the properties of the logarithm. Equality is true due to the second remarkable limit.

Derivatives of trigonometric functions.

To derive formulas for derivatives of trigonometric functions, we will have to recall some trigonometry formulas, as well as the first remarkable limit.

By definition of the derivative for the sine function we have .

Let's use the difference of sines formula:

It remains to turn to the first remarkable limit:

Thus, the derivative of the function sin x There is cos x.

The formula for the derivative of the cosine is proved in exactly the same way.

Therefore, the derivative of the function cos x There is –sin x.

We will derive formulas for the table of derivatives for tangent and cotangent using proven rules of differentiation (derivative of a fraction).

Derivatives of hyperbolic functions.

The rules of differentiation and the formula for the derivative of the exponential function from the table of derivatives allow us to derive formulas for the derivatives of the hyperbolic sine, cosine, tangent and cotangent.

Derivative of the inverse function.

To avoid confusion during presentation, let's denote in subscript the argument of the function by which differentiation is performed, that is, it is the derivative of the function f(x) By x.

Now let's formulate rule for finding the derivative of an inverse function.

Let the functions y = f(x) And x = g(y) mutually inverse, defined on the intervals and respectively. If at a point there is a finite non-zero derivative of the function f(x), then at the point there is a finite derivative of the inverse function g(y), and . In another post .

This rule can be reformulated for any x from the interval , then we get .

Let's check the validity of these formulas.

Let's find the inverse function for the natural logarithm (Here y is a function, and x- argument). Having resolved this equation for x, we get (here x is a function, and y– her argument). That is, and mutually inverse functions.