Decide physical tasks or examples in mathematics is completely impossible without knowledge about the derivative and methods for calculating it. Derivative is one of the most important concepts mathematical analysis. We decided to devote today’s article to this fundamental topic. What is a derivative, what is its physical and geometric meaning how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?
Geometric and physical meaning of derivative
Let there be a function f(x) , specified in a certain interval (a, b) . Points x and x0 belong to this interval. When x changes, the function itself changes. Changing the argument - the difference in its values x-x0 . This difference is written as delta x and is called argument increment. A change or increment of a function is the difference between the values of a function at two points. Definition of derivative:
The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.
Otherwise it can be written like this:
What's the point of finding such a limit? And here's what it is:
the derivative of a function at a point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at a given point.
Physical meaning derivative: the derivative of the path with respect to time is equal to the speed of rectilinear motion.
Indeed, since school days everyone knows that speed is a particular path x=f(t) and time t . Average speed for a certain period of time:
To find out the speed of movement at a moment in time t0 you need to calculate the limit:
Rule one: set a constant
The constant can be taken out of the derivative sign. Moreover, this must be done. When solving examples in mathematics, take it as a rule - If you can simplify an expression, be sure to simplify it .
Example. Let's calculate the derivative:
Rule two: derivative of the sum of functions
The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.
We will not give a proof of this theorem, but rather consider a practical example.
Find the derivative of the function:
Rule three: derivative of the product of functions
The derivative of the product of two differentiable functions is calculated by the formula:
Example: find the derivative of a function:
Solution: 
It is important to talk about calculating derivatives of complex functions here. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable.
In the above example we come across the expression:
IN in this case the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first calculate the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate argument itself with respect to the independent variable.
Rule four: derivative of the quotient of two functions
Formula for determining the derivative of the quotient of two functions:
We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it seems, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.
With any questions on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult test and understand the tasks, even if you have never done derivative calculations before.
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:
For this we will use simple rule: . Then:
Well, it worked. Now try to find the derivative, and don't forget that this function is complex.
Did it work?
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. Important Feature 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.
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 = ax (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 1In 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 the 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 us 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 + 0 = p ! (p - 1) !
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
On at this stage received 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 – negative number.
If the indicator p There is even number, then the power function is also 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 4Let's 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 5Let 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 6Let'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 function cos x 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
Derivative Section 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 7We 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 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.
From the table of derivatives we see that 
And 
.
Let’s make sure that the formulas for finding the derivatives of the inverse function lead us to the same results: 
As you can see, we got the same results as in the derivatives table.
Now we have the knowledge to prove formulas for the derivatives of inverse trigonometric functions.
Let's start with the derivative of the arcsine.
. Then, using the formula for the derivative of the inverse function, we get
All that remains is to carry out the transformations.
Since the arcsine range is the interval 
, That 
(see the section on basic elementary functions, their properties and graphs). Therefore, we are not considering it.
Hence, 
. The domain of definition of the arcsine derivative is the interval (-1;
1)
.
For the arc cosine, everything is done in exactly the same way: 
Let's find the derivative of the arctangent.
For the inverse function is 
.
Let's express the arctangent in terms of arccosine to simplify the resulting expression.
Let arctgx = z, Then 
Hence, 
The derivative of the arc cotangent is found in a similar way: 
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