First level

Function derivative. Comprehensive Guide (2019)

Imagine a straight road passing through a hilly area. That is, it goes up and down, but does not turn right or left. If the axis is directed horizontally along the road, and vertically, then the road line will be very similar to the graph of some continuous function:

The axis is a certain level of zero height, in life we ​​use sea level as it.

Moving forward along such a road, we are also moving up or down. We can also say: when the argument changes (moving along the abscissa axis), the value of the function changes (moving along the ordinate axis). Now let's think about how to determine the "steepness" of our road? What could this value be? Very simple: how much will the height change when moving forward a certain distance. Indeed, on different sections of the road, moving forward (along the abscissa) one kilometer, we will rise or fall a different number of meters relative to sea level (along the ordinate).

We denote progress forward (read "delta x").

The Greek letter (delta) is commonly used in mathematics as a prefix meaning "change". That is - this is a change in magnitude, - a change; then what is it? That's right, a change in size.

Important: the expression is a single entity, one variable. You should never tear off the "delta" from the "x" or any other letter! That is, for example, .

So, we have moved forward, horizontally, on. If we compare the line of the road with the graph of a function, then how do we denote the rise? Certainly, . That is, when moving forward on we rise higher on.

It is easy to calculate the value: if at the beginning we were at a height, and after moving we were at a height, then. If the end point turned out to be lower than the start point, it will be negative - this means that we are not ascending, but descending.

Back to "steepness": this is a value that indicates how much (steeply) the height increases when moving forward per unit distance:

Suppose that on some section of the path, when advancing by km, the road rises up by km. Then the steepness in this place is equal. And if the road, when advancing by m, sank by km? Then the slope is equal.

Now consider the top of a hill. If you take the beginning of the section half a kilometer to the top, and the end - half a kilometer after it, you can see that the height is almost the same.

That is, according to our logic, it turns out that the slope here is almost equal to zero, which is clearly not true. A lot can change just a few miles away. Smaller areas need to be considered for a more adequate and accurate estimate of the steepness. For example, if you measure the change in height when moving one meter, the result will be much more accurate. But even this accuracy may not be enough for us - after all, if there is a pole in the middle of the road, we can simply slip through it. What distance should we choose then? Centimeter? Millimeter? Less is better!

In real life, measuring distance to the nearest millimeter is more than enough. But mathematicians always strive for perfection. Therefore, the concept was infinitesimal, that is, the modulo value is less than any number that we can name. For example, you say: one trillionth! How much less? And you divide this number by - and it will be even less. Etc. If we want to write that the value is infinitely small, we write like this: (we read “x tends to zero”). It is very important to understand that this number is not equal to zero! But very close to it. This means that it can be divided into.

The concept opposite to infinitely small is infinitely large (). You've probably already encountered it when you were working on inequalities: this number is greater in modulus than any number you can think of. If you come up with the largest possible number, just multiply it by two and you get even more. And infinity is even more than what happens. In fact, infinitely large and infinitely small are inverse to each other, that is, at, and vice versa: at.

Now back to our road. The ideally calculated slope is the slope calculated for an infinitely small segment of the path, that is:

I note that with an infinitely small displacement, the change in height will also be infinitely small. But let me remind you that infinitely small does not mean equal to zero. If you divide infinitesimal numbers by each other, you can get a completely ordinary number, for example,. That is, one small value can be exactly twice as large as another.

Why all this? The road, the steepness ... We are not going on a rally, but we are learning mathematics. And in mathematics everything is exactly the same, only called differently.

The concept of a derivative

The derivative of a function is the ratio of the increment of the function to the increment of the argument at an infinitesimal increment of the argument.

Increment in mathematics is called change. How much the argument () has changed when moving along the axis is called argument increment and denoted by How much the function (height) has changed when moving forward along the axis by a distance is called function increment and is marked.

So, the derivative of a function is the relation to when. We denote the derivative with the same letter as the function, only with a stroke from the top right: or simply. So, let's write the derivative formula using these notations:

As in the analogy with the road, here, when the function increases, the derivative is positive, and when it decreases, it is negative.

But is the derivative equal to zero? Certainly. For example, if we are driving on a flat horizontal road, the steepness is zero. Indeed, the height does not change at all. So with the derivative: the derivative of a constant function (constant) is equal to zero:

since the increment of such a function is zero for any.

Let's take the hilltop example. It turned out that it was possible to arrange the ends of the segment on opposite sides of the vertex in such a way that the height at the ends turns out to be the same, that is, the segment is parallel to the axis:

But large segments are a sign of inaccurate measurement. We will raise our segment up parallel to itself, then its length will decrease.

In the end, when we are infinitely close to the top, the length of the segment will become infinitely small. But at the same time, it remained parallel to the axis, that is, the height difference at its ends is equal to zero (does not tend, but is equal to). So the derivative

This can be understood as follows: when we are standing at the very top, a small shift to the left or right changes our height negligibly.

There is also a purely algebraic explanation: to the left of the top, the function increases, and to the right, it decreases. As we have already found out earlier, when the function increases, the derivative is positive, and when it decreases, it is negative. But it changes smoothly, without jumps (because the road does not change its slope sharply anywhere). Therefore, there must be between negative and positive values. It will be where the function neither increases nor decreases - at the vertex point.

The same is true for the valley (the area where the function decreases on the left and increases on the right):

A little more about increments.

So we change the argument to a value. We change from what value? What has he (argument) now become? We can choose any point, and now we will dance from it.

Consider a point with a coordinate. The value of the function in it is equal. Then we do the same increment: increase the coordinate by. What is the argument now? Very easy: . What is the value of the function now? Where the argument goes, the function goes there: . What about function increment? Nothing new: this is still the amount by which the function has changed:

Practice finding increments:

1. Find the increment of the function at a point with an increment of the argument equal to.
2. The same for a function at a point.

Solutions:

At different points, with the same increment of the argument, the increment of the function will be different. This means that the derivative at each point has its own (we discussed this at the very beginning - the steepness of the road at different points is different). Therefore, when we write a derivative, we must indicate at what point:

Power function.

A power function is called a function where the argument is to some extent (logical, right?).

And - to any extent: .

The simplest case is when the exponent is:

Let's find its derivative at a point. Remember the definition of a derivative:

So the argument changes from to. What is the function increment?

Increment is. But the function at any point is equal to its argument. So:

The derivative is:

The derivative of is:

b) Now consider the quadratic function (): .

Now let's remember that. This means that the value of the increment can be neglected, since it is infinitely small, and therefore insignificant against the background of another term:

So, we have another rule:

c) We continue the logical series: .

This expression can be simplified in different ways: open the first bracket using the formula for abbreviated multiplication of the cube of the sum, or decompose the entire expression into factors using the formula for the difference of cubes. Try to do it yourself in any of the suggested ways.

So, I got the following:

And let's remember that again. This means that we can neglect all terms containing:

We get: .

d) Similar rules can be obtained for large powers:

e) It turns out that this rule can be generalized for a power function with an arbitrary exponent, not even an integer:

(2)

You can formulate the rule with the words: “the degree is brought forward as a coefficient, and then decreases by”.

We will prove this rule later (almost at the very end). Now let's look at a few examples. Find the derivative of functions:

1. (in two ways: by the formula and using the definition of the derivative - by counting the increment of the function);

1. . Believe it or not, this is a power function. If you have questions like “How is it? And where is the degree? ”, Remember the topic“ ”!
   Yes, yes, the root is also a degree, only a fractional one:.
   So our square root is just a power with an exponent:
   .
   We are looking for the derivative using the recently learned formula:

   If at this point it became unclear again, repeat the topic "" !!! (about a degree with a negative indicator)

2. . Now the exponent:

   And now through the definition (have you forgotten yet?):
   ;
   .
   Now, as usual, we neglect the term containing:
   .

3. . Combination of previous cases: .

trigonometric functions.

Here we will use one fact from higher mathematics:

When expression.

You will learn the proof in the first year of the institute (and to get there, you need to pass the exam well). Now I'll just show it graphically:

We see that when the function does not exist - the point on the graph is punctured. But the closer to the value, the closer the function is to. This is the very “strives”.

Additionally, you can check this rule with a calculator. Yes, yes, do not be shy, take a calculator, we are not at the exam yet.

So let's try: ;

Don't forget to switch the calculator to Radians mode!

etc. We see that the smaller, the closer the value of the ratio to.

a) Consider a function. As usual, we find its increment:

Let's turn the difference of sines into a product. To do this, we use the formula (remember the topic ""):.

Now the derivative:

Let's make a substitution: . Then, for infinitely small, it is also infinitely small: . The expression for takes the form:

And now we remember that with the expression. And also, what if an infinitely small value can be neglected in the sum (that is, at).

So we get the following rule: the derivative of the sine is equal to the cosine:

These are basic (“table”) derivatives. Here they are in one list:

Later we will add a few more to them, but these are the most important, as they are used most often.

Practice:

1. Find the derivative of a function at a point;
2. Find the derivative of the function.

Solutions:

1. First, we find the derivative in a general form, and then we substitute its value instead:
   ;
   .
2. Here we have something similar to a power function. Let's try to bring her to
   normal view:
   .
   Ok, now you can use the formula:
   .
   .
3. . Eeeeeee….. What is it????

Okay, you're right, we still don't know how to find such derivatives. Here we have a combination of several types of functions. To work with them, you need to learn a few more rules:

Exponent and natural logarithm.

There is such a function in mathematics, the derivative of which for any is equal to the value of the function itself for the same. It is called "exponent", and is an exponential function

The base of this function - a constant - is an infinite decimal fraction, that is, an irrational number (such as). It is called the "Euler number", which is why it is denoted by a letter.

So the rule is:

It's very easy to remember.

Well, we will not go far, we will immediately consider the inverse function. What is the inverse of the exponential function? Logarithm:

In our case, the base is a number:

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

What is equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

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

Answers: The exponent and the natural logarithm are functions that are uniquely simple in terms of the derivative. 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.

Differentiation rules

What rules? Another new term, again?!...

Differentiation is the process of finding the derivative.

Only and everything. What is another word for this process? Not proizvodnovanie... The differential of mathematics is called the very increment of the 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 sign of the derivative.

If - some constant number (constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let, or easier.

Examples.

Find derivatives of functions:

1. at the point;
2. at the point;
3. at the point;
4. at the point.

Solutions:

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

Derivative of a product

Everything is similar here: we introduce a new function and find its increment:

Derivative:

Examples:

1. Find derivatives of functions and;
2. Find the derivative of a function at a point.

Solutions:

Derivative of exponential function

Now your knowledge is enough to learn how to find the derivative of any exponential function, and not just the exponent (have you forgotten what it is yet?).

So where is some number.

We already know the derivative of the function, so let's try to bring our function to a new base:

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

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

Happened?

Here, check yourself:

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

Examples:
Find derivatives of functions:

Answers:

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

Derivative of a logarithmic function

Here it is similar: you already know the derivative of the natural logarithm:

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

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

Only now instead of we will write:

The denominator turned out to be just a constant (a constant number, without a variable). The derivative is very simple:

Derivatives of the exponential and logarithmic functions are almost never found in the exam, 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 arc tangent. These functions can be difficult to understand (although if the logarithm seems difficult to you, read the topic "Logarithms" and everything will work out), but in terms of mathematics, the word "complex" does not mean "difficult".

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

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then we will square the resulting number. So, they give us 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, in order to find its value, we do the first action directly with the variable, and then another second action with what happened as a result of the first.

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

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

For the first example, .

Second example: (same). .

The last action we do will be called "external" function, and the action performed first - respectively "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 is internal:

Answers: The separation of inner and outer functions is very similar to changing variables: for example, in the function

1. What action will we take first? First we calculate the sine, and only then we raise it to a cube. So it's an internal function, not an external one.
   And the original function is their composition: .
2. Internal: ; external: .
   Examination: .
3. Internal: ; external: .
   Examination: .
4. Internal: ; external: .
   Examination: .
5. Internal: ; external: .
   Examination: .

we change variables and get a function.

Well, now we will extract our chocolate - 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. For 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:

Everything seems to be simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

(just don’t try to reduce by now! Nothing is taken out from under the cosine, remember?)

3) Internal: ;

External: ;

It is immediately clear that there is a three-level complex function here: after all, this is already a complex function in itself, and we still extract the root from it, that is, we perform the third action (put chocolate in a wrapper and with a ribbon in a briefcase). But there is no reason to be afraid: anyway, we will “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 - as before:

Here the nesting is generally 4-level. Let's determine the course of action.

1. Radical expression. .

2. Root. .

3. Sinus. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN

Function derivative- the ratio of the increment of the function to the increment of the argument with an infinitesimal increment of the argument:

Basic derivatives:

Differentiation rules:

The constant is taken out of the sign of the derivative:

Derivative of sum:

Derivative product:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

1. We define the "internal" function, find its derivative.
2. We define the "external" function, find its derivative.
3. We multiply the results of the first and second points.

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

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

To begin with, we note that the so-called elementary functions can be distinguished from the whole variety of functions. These are relatively simple expressions, the derivatives of which have long been calculated and entered in the table. Such functions are easy enough to remember, along with their derivatives.

Derivatives of elementary functions

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

So, the derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, yes, zero!)
Degree with rational exponent	f(x) = x n	n · x n − 1
Sinus	f(x) = sin x	cos x
Cosine	f(x) = cos x	− sin x(minus sine)
Tangent	f(x) = tg x	1/cos 2 x
Cotangent	f(x) = ctg x	− 1/sin2 x
natural logarithm	f(x) = log x	1/x
Arbitrary logarithm	f(x) = log a x	1/(x ln a)
Exponential function	f(x) = e x	e x(nothing changed)

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

(C · f)’ = C · f ’.

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

(2x 3)' = 2 ( x 3)' = 2 3 x 2 = 6x 2 .

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

Derivative of sum and difference

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

1. (f + g)’ = f ’ + g ’
2. (f − g)’ = f ’ − g ’

So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.

Strictly speaking, there is no concept of "subtraction" in algebra. There is a concept of "negative element". Therefore, the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.

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

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

f ’(x) = (x 2+ sin x)’ = (x 2)' + (sin x)’ = 2x+ cosx;

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

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

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

Derivative of a product

Mathematics is a logical science, so many people believe that if the derivative of the sum is equal to the sum of the derivatives, then the derivative of the product strike"\u003e equal to the product of derivatives. But figs to you! The derivative of the product is calculated using a completely different formula. Namely:

(f · g) ’ = f ’ · g + f · g ’

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

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

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

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

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

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

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

Note that in the last step, the derivative is factorized. Formally, this is not necessary, but most derivatives are not calculated on their own, but to explore the function. This means that further the derivative will be equated to zero, its signs will be found out, and so on. For such a case, it is better to have an expression decomposed into factors.

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

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

Task. Find derivatives of functions:

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

By tradition, we factor the numerator into factors - this will greatly simplify the answer:

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

How to be? In such cases, the replacement of a variable and the formula for the derivative of a complex function help:

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

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

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

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

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

And now - attention! Performing a reverse substitution: t = 2x+ 3. We get:

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

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

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

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

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

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

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

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

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

(x n)’ = n · x n − 1

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

Task. Find the derivative of a function:

First, let's rewrite the root as a power with a rational exponent:

f(x) = (x 2 + 8x − 7) 0,5 .

Now we make a substitution: let x 2 + 8x − 7 = t. We find the derivative by the formula:

f ’(x) = f ’(t) · t ’ = (t 0.5)' t' = 0.5 t−0.5 t ’.

We make a reverse substitution: t = x 2 + 8x− 7. We have:

f ’(x) = 0.5 ( x 2 + 8x− 7) −0.5 ( x 2 + 8x− 7)' = 0.5 (2 x+ 8) ( x 2 + 8x − 7) −0,5 .

Finally, back to the roots:

The operation of finding a 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. Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716) were the first to work in the field of finding derivatives.

Therefore, in our time, in order to find the derivative of any function, it is not necessary to calculate the above-mentioned limit of the ratio of the increment of the function to the increment of the argument, but 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 stroke sign break down simple functions and determine what actions (product, sum, quotient) these functions are related. Further, 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 table of derivatives and differentiation rules are given after the first two examples.

Example 1 Find the derivative of a function

Decision. From the rules of differentiation we find out that the derivative of the 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 the sine is cosine. We substitute these values ​​in the sum of derivatives and find the derivative required by the condition of the problem:

Example 2 Find the derivative of a function

Decision. Differentiate as a derivative of the sum, in which the second term with a constant factor, it can be taken out of the sign of the derivative:

If there are still questions about where something comes from, they, as a rule, become clear after reading the table of derivatives and the simplest rules of differentiation. We are going 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 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
3. Derivative of degree. When solving problems, you need to convert non-square roots to a power.
4. Derivative of a variable to the power of -1
5. Derivative of the square root
6. Sine derivative
7. Cosine derivative
8. Tangent derivative
9. Derivative of cotangent
10. Derivative of the arcsine
11. Derivative of arc cosine
12. Derivative of arc tangent
13. Derivative of the inverse tangent
14. Derivative of natural logarithm
15. Derivative of a logarithmic function
16. Derivative of the exponent
17. Derivative of exponential function

Differentiation rules

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

Rule 1If functions

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

and

those. the derivative of the 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, then their derivatives are, i.e.

Rule 2If functions

are differentiable at some point , then their product is also 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.

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

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

For example, for three multipliers:

Rule 3If functions

differentiable at some point and , then at this point their quotient is also differentiable.u/v , and

those. the derivative of a quotient of two functions is equal to a fraction whose numerator 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 on other pages

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

Comment. You should not confuse a constant (that is, a number) as a term in the 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-two-component examples, this mistake no longer makes.

And if, when differentiating a product or a quotient, you have a term u"v, wherein 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 (such a case is analyzed in example 10).

Another common mistake is the mechanical solution of the derivative of a complex function as the derivative of a simple function. So derivative of a complex function devoted to a separate article. But first we will learn to find derivatives of simple functions.

Along the way, you can not do without transformations of expressions. To do this, you may need to open in new windows manuals Actions with powers and roots and Actions with fractions .

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

If you have a task like , then you are in the lesson "Derivatives of simple trigonometric functions".

Step by step examples - how to find the derivative

Example 3 Find the derivative of a function

Decision. We determine the parts of the function expression: the entire expression represents the 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 and 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 with 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 - 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 get the following values ​​of derivatives:

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

Decision. We are required to find the derivative of the quotient. We apply the formula for differentiating a quotient: the derivative of a quotient of two functions is equal to a fraction whose numerator 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's also not forget that the product, which is the second factor in the numerator, is taken with a minus sign in the current example:

If you are looking for solutions to such problems in which you need to find the derivative of a function, where there is a continuous pile of roots and degrees, such as, for example, then welcome to class "The derivative of the sum 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 you have a lesson "Derivatives of simple trigonometric functions" .

Example 5 Find the derivative of a function

Decision. In this function, we see a product, one of the factors of which is the square root of the independent variable, with the derivative of which we familiarized ourselves in the table of derivatives. According to the product differentiation rule and the tabular value of the derivative of the square root, we get:

Example 6 Find the derivative of a function

Decision. In this function, we see the quotient, the dividend of which is the square root of the independent variable. According to the rule of differentiation of the quotient, which we repeated and applied in example 4, and the tabular value of the derivative of the square root, we get:

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

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