Derivative of a function. Detailed theory with examples

What is a derivative function - this is a basic mathematical concept that is on the same level as integrals in analysis. This function at a certain point gives a characteristic of the rate of change of the function at this point.
Concepts such as differentiation and integration, the first is deciphered as the action of searching for a derivative, the second, on the contrary, restores a function starting from a given derivative.
Derivative calculations play an important part in differential calculations.
For a clear example, let's depict the derivative on the coordinate plane.

in the function y=f(x) we fix points M at which (x0; f(X0)) and N f (x0+?x) to each abscissa there is an increment in the form?x. Increment is the process when the abscissa changes, then the ordinate also changes. Denoted as?y.
Let's find the tangent of the angle in triangle MPN using points M and N for this.

tg? = NP/MP = ?у/?x.

As?x goes to 0. The intersecting MN is getting closer to the tangent MT and the angle? will?. Therefore, tg? maximum value for tg?.

tg? = lim from?x-0 tg ? = lim from?x-0 ?y/?x

Derivatives table

If you pronounce the wording of each derivative formulas. The table will be easier to remember.
1) The derivative of a constant value is 0.
2) X with a prime equals one.
3) If there is a constant factor, we simply take it out as a derivative.
4) To find a derived power, you need to multiply the exponent of a given power by a power with the same base, whose exponent is 1 less.
5) Finding a root is equal to one divided by 2 of these roots.
6) The derivative of one divided by X is equal to one divided by X squared, with a minus sign.
7) P sine equals cosine
8) P cosine equals sine with a minus sign.
9) P tangent equals one divided by cosine squared.
10) P cotangent is equal to one with a minus sign, divided by sine squared.

There are also rules in differentiation, which are also easier to learn by speaking them out loud.

1) Very simply, n. of terms equals their sum.
2) The derivative in multiplication is equal to the multiplication of the first value by the second, adding to itself the multiplication of the second value by the first.
3) The derivative in division is equal to the multiplication of the first value by the second, subtracting the multiplication of the second value by the first. Fraction divided by the second value squared.
4) The formulation is a special case of the third formula.

In this lesson, we continue to study derivatives of functions and move on to a more advanced topic, namely, derivatives of products and quotients. If you watched the previous lesson, you probably realized that we considered only the simplest constructions, namely, the derivative of a power function, sum and difference. In particular, we learned that the derivative of a sum is equal to their sum, and the derivative of a difference is equal, respectively, to their difference. Unfortunately, in the case of quotient and product derivatives, the formulas will be much more complicated. We will start with the formula for the derivative of a product of functions.

Derivatives of trigonometric functions

To begin with, let me make a small lyrical digression. The fact is that in addition to the standard power function - $y=((x)^(n))$, in this lesson we will also encounter other functions, namely, $y=\sin x$, as well as $y=\ cos x$ and other trigonometry - $y=tgx$ and, of course, $y=ctgx$.

If we all know perfectly well the derivative of a power function, namely $\left(((x)^(n)) \right)=n\cdot ((x)^(n-1))$, then as for trigonometric functions , needs to be mentioned separately. Let's write it down:

\[\begin(align)& ((\left(\sinx \right))^(\prime ))=\cosx \\& ((\left(\cos x \right))^(\prime ))= -\sin x \\& ((\left(tgx \right))^(\prime ))=\frac(1)(((\cos )^(2))x) \\& ((\left( ctgx \right))^(\prime ))=\frac(1)(((\cos )^(2))x) \\\end(align)\]

But you know these formulas very well, let's move on.

What is the derivative of a product?

First, the most important thing: if a function is the product of two other functions, for example, $f\cdot g$, then the derivative of this construction will be equal to the following expression:

As you can see, this formula is significantly different and more complex than the formulas we looked at earlier. For example, the derivative of a sum is calculated in an elementary way - $((\left(f+g \right))^(\prime ))=(f)"+(g)"$, or the derivative of a difference, which is also calculated in an elementary way - $(( \left(f-g \right))^(\prime ))=(f)"-(g)"$.

Let's try to apply the first formula to calculate the derivatives of the two functions that are given to us in the problem. Let's start with the first example:

Obviously, the following construction acts as a product, or more precisely, as a multiplier: $((x)^(3))$, we can consider it as $f$, and $\left(x-5 \right)$ we can consider as $g$. Then their product will be precisely the product of two functions. We decide:

\[\begin(align)& ((\left(((x)^(3))\cdot \left(x-5 \right) \right))^(\prime ))=((\left(( (x)^(3)) \right))^(\prime ))\cdot \left(x-5 \right)+((x)^(3))\cdot ((\left(x-5 \ right))^(\prime ))= \\& =3((x)^(2))\cdot \left(x-5 \right)+((x)^(3))\cdot 1 \\ \end(align)\].

Now let's take a closer look at each of our terms. We see that in both the first and second terms there is a power $x$: in the first case it is $((x)^(2))$, and in the second it is $((x)^(3))$. Let's take the smallest degree out of brackets, leaving in brackets:

\[\begin(align)& 3((x)^(2))\cdot \left(x-5 \right)+((x)^(3))\cdot 1=((x)^(2 ))\left(3\cdot 1\left(x-5 \right)+x \right)= \\& =((x)^(2))\left(3x-15+x \right)=( (x)^(2))(4x-15)\\\end(align)\]

That's it, we have found the answer.

Let's return to our problems and try to solve:

So, let's rewrite:

Again, we note that we are talking about the product of the product of two functions: $x$, which can be denoted by $f$, and $\left(\sqrt(x)-1 \right)$, which can be denoted by $g$.

Thus, we again have a product of two functions. To find the derivative of the function $f\left(x \right)$ we will again use our formula. We get:

\[\begin(align)& (f)"=\left(x \right)"\cdot \left(\sqrt(x)-1 \right)+x\cdot ((\left(\sqrt(x) -1 \right))^(\prime ))=1\cdot \left(\sqrt(x)-1 \right)+x\frac(1)(3\sqrt(x))= \\& =\ sqrt(x)-1+\sqrt(x)\cdot \frac(1)(3)=\frac(4)(3)\sqrt(x)-1 \\\end(align)\]

The answer has been found.

Why factor derivatives?

We have just used several very important mathematical facts, which in themselves are not related to derivatives, but without their knowledge, all further study of this topic simply does not make sense.

Firstly, solving the very first problem and having already gotten rid of all the signs of derivatives, for some reason we began to factor this expression.

Secondly, when solving the following problem, we passed several times from the root to the power with a rational exponent and back, while using the 8-9th grade formula, which would be worth repeating separately.

Regarding factorization - why are all these additional efforts and transformations needed? In fact, if the problem simply says “find the derivative of a function,” then these additional steps are not required. However, in real problems that await you in all kinds of exams and tests, simply finding the derivative is often not enough. The fact is that the derivative is only a tool with which you can find out, for example, the increase or decrease of a function, and for this you need to solve the equation and factor it. And this is where this technique will be very appropriate. And in general, it is much more convenient and pleasant to work with a function factorized in the future if any transformations are required. Therefore, rule No. 1: if the derivative can be factorized, that’s what you should do. And immediately rule No. 2 (essentially, this is 8th-9th grade material): if the problem contains a root n-th degree, and the root is clearly greater than two, then this root can be replaced by an ordinary degree with a rational exponent, and a fraction will appear in the exponent, where n― that very degree ― will be in the denominator of this fraction.

Of course, if there is some degree under the root (in our case this is the degree k), then it doesn’t go anywhere, but simply ends up in the numerator of this very degree.

Now that you understand all this, let's go back to the derivatives of the product and calculate a few more equations.

But before moving directly to the calculations, I would like to remind you of the following patterns:

\[\begin(align)& ((\left(\sin x \right))^(\prime ))=\cos x \\& ((\left(\cos x \right))^(\prime ) )=-\sin x \\& \left(tgx \right)"=\frac(1)(((\cos )^(2))x) \\& ((\left(ctgx \right))^ (\prime ))=-\frac(1)(((\sin )^(2))x) \\\end(align)\]

Let's consider the first example:

We again have a product of two functions: the first is $f$, the second is $g$. Let me remind you of the formula:

\[((\left(f\cdot g \right))^(\prime ))=(f)"\cdot g+f\cdot (g)"\]

Let's decide:

\[\begin(align)& (y)"=((\left(((x)^(4)) \right))^(\prime ))\cdot \sin x+((x)^(4) )\cdot ((\left(\sin x \right))^(\prime ))= \\& =3((x)^(3))\cdot \sin x+((x)^(4)) \cdot \cos x=((x)^(3))\left(3\sin x+x\cdot \cos x \right) \\\end(align)\]

Let's move on to the second function:

Again, $\left(3x-2 \right)$ is a function of $f$, $\cos x$ is a function of $g$. In total, the derivative of the product of two functions will be equal to:

\[\begin(align)& (y)"=((\left(3x-2 \right))^(\prime ))\cdot \cos x+\left(3x-2 \right)\cdot ((\ left(\cos x \right))^(\prime ))= \\& =3\cdot \cos x+\left(3x-2 \right)\cdot \left(-\sin x \right)=3\ cos x-\left(3x-2 \right)\cdot \sin x \\\end(align)\]

\[(y)"=((\left(((x)^(2))\cdot \cos x \right))^(\prime ))+((\left(4x\sin x \right)) ^(\prime ))\]

Let's write it down separately:

\[\begin(align)& ((\left(((x)^(2))\cdot \cos x \right))^(\prime ))=\left(((x)^(2)) \right)"\cos x+((x)^(2))\cdot ((\left(\cos x \right))^(\prime ))= \\& =2x\cdot \cos x+((x )^(2))\cdot \left(-\sin x \right)=2x\cdot \cos x-((x)^(2))\cdot \sin x \\\end(align)\]

We do not factorize this expression, because this is not the final answer yet. Now we have to solve the second part. Let's write it out:

\[\begin(align)& ((\left(4x\cdot \sin x \right))^(\prime ))=((\left(4x \right))^(\prime ))\cdot \sin x+4x\cdot ((\left(\sin x \right))^(\prime ))= \\& =4\cdot \sin x+4x\cdot \cos x \\\end(align)\]

Now let’s return to our original task and put everything together into a single structure:

\[\begin(align)& (y)"=2x\cdot \cos x-((x)^(2))\cdot \sin x+4\sin x+4x\cos x=6x\cdot \cos x= \\& =6x\cdot \cos x-((x)^(2))\cdot \sin x+4\sin x \\\end(align)\]

That's it, this is the final answer.

Let's move on to the last example - it will be the most complex and most voluminous in terms of calculations. So, an example:

\[(y)"=((\left(((x)^(2))\cdot tgx \right))^(\prime ))-((\left(2xctgx \right))^(\prime ) )\]

We count each part separately:

\[\begin(align)& ((\left(((x)^(2))\cdot tgx \right))^(\prime ))=((\left(((x)^(2)) \right))^(\prime ))\cdot tgx+((x)^(2))\cdot ((\left(tgx \right))^(\prime ))= \\& =2x\cdot tgx+( (x)^(2))\cdot \frac(1)(((\cos )^(2))x) \\\end(align)\]

\[\begin(align)& ((\left(2x\cdot ctgx \right))^(\prime ))=((\left(2x \right))^(\prime ))\cdot ctgx+2x\ cdot ((\left(ctgx \right))^(\prime ))= \\& =2\cdot ctgx+2x\left(-\frac(1)(((\sin )^(2))x) \right)=2\cdot ctgx-\frac(2x)(((\sin )^(2))x) \\\end(align)\]

Returning to the original function, let's calculate its derivative as a whole:

\[\begin(align)& (y)"=2x\cdot tgx+\frac(((x)^(2)))(((\cos )^(2))x)-\left(2ctgx-\ frac(2x)(((\sin )^(2))x) \right)= \\& =2x\cdot tgx+\frac(((x)^(2)))(((\cos )^( 2))x)-2ctgx+\frac(2x)(((\sin )^(2))x) \\\end(align)\]

That, in fact, is all I wanted to tell you about the derivative works. As you can see, the main problem with the formula is not in memorizing it, but in the fact that it involves a fairly large amount of calculations. But that's okay, because now we're moving on to the quotient derivative, where we're going to have to work really hard.

What is the derivative of a quotient?

So, the formula for the derivative of the quotient. This is perhaps the most complex formula in the school course on derivatives. Let's say we have a function of the form $\frac(f)(g)$, where $f$ and $g$ are also functions from which we can also remove the prime. Then it will be calculated according to the following formula:

The numerator somewhat reminds us of the formula for the derivative of a product, but there is a minus sign between the terms and the square of the original denominator has also been added to the denominator. Let's see how this works in practice:

Let's try to solve:

\[(f)"=((\left(\frac(((x)^(2))-1)(x+2) \right))^(\prime ))=\frac(((\left (((x)^(2))-1 \right))^(\prime ))\cdot \left(x+2 \right)-\left(((x)^(2))-1 \right )\cdot ((\left(x+2 \right))^(\prime )))(((\left(x+2 \right))^(2)))\]

I suggest writing out each part separately and writing down:

\[\begin(align)& ((\left(((x)^(2))-1 \right))^(\prime ))=((\left(((x)^(2)) \ right))^(\prime ))-(1)"=2x \\& ((\left(x+2 \right))^(\prime ))=(x)"+(2)"=1 \ \\end(align)\]

Let's rewrite our expression:

\[\begin(align)& (f)"=\frac(2x\cdot \left(x+2 \right)-\left(((x)^(2))-1 \right)\cdot 1) (((\left(x+2 \right))^(2)))= \\& =\frac(2((x)^(2))+4x-((x)^(2))+ 1)(((\left(x+2 \right))^(2)))=\frac(((x)^(2))+4x+1)(((\left(x+2 \right ))^(2))) \\\end(align)\]

We have found the answer. Let's move on to the second function:

Judging by the fact that its numerator is simply one, the calculations here will be a little simpler. So, let's write:

\[(y)"=((\left(\frac(1)(((x)^(2))+4) \right))^(\prime ))=\frac((1)"\cdot \left(((x)^(2))+4 \right)-1\cdot ((\left(((x)^(2))+4 \right))^(\prime )))(( (\left(((x)^(2))+4 \right))^(2)))\]

Let's calculate each part of the example separately:

\[\begin(align)& (1)"=0 \\& ((\left(((x)^(2))+4 \right))^(\prime ))=((\left(( (x)^(2)) \right))^(\prime ))+(4)"=2x \\\end(align)\]

Let's rewrite our expression:

\[(y)"=\frac(0\cdot \left(((x)^(2))+4 \right)-1\cdot 2x)(((\left(((x)^(2) )+4 \right))^(2)))=-\frac(2x)(((\left(((x)^(2))+4 \right))^(2)))\]

We have found the answer. As expected, the amount of computation turned out to be significantly less than for the first function.

What is the difference between the designations?

Attentive students probably already have a question: why in some cases do we denote the function as $f\left(x \right)$, and in other cases we simply write $y$? In fact, from the point of view of mathematics, there is absolutely no difference - you have the right to use both the first designation and the second, and there will be no penalties in exams or tests. For those who are still interested, I will explain why the authors of textbooks and problems in some cases write $f\left(x \right)$, and in others (much more frequent) - simply $y$. The fact is that by writing a function in the form \, we implicitly hint to those who read our calculations that we are talking specifically about the algebraic interpretation of functional dependence. That is, there is a certain variable $x$, we consider the dependence on this variable and denote it $f\left(x \right)$. At the same time, having seen such a designation, the one who reads your calculations, for example, the inspector, will subconsciously expect that in the future only algebraic transformations await him - no graphs and no geometry.

On the other hand, using notations of the form \, i.e., denoting a variable with one single letter, we immediately make it clear that in the future we are interested in the geometric interpretation of the function, i.e., we are interested, first of all, in its graph. Accordingly, when faced with a record of the form\, the reader has the right to expect graphic calculations, i.e., graphs, constructions, etc., but, in no case, analytical transformations.

I would also like to draw your attention to one feature of the design of the tasks that we are considering today. Many students think that I give too detailed calculations, and many of them could be skipped or simply solved in their heads. However, it is precisely such a detailed record that will allow you to get rid of offensive mistakes and significantly increase the percentage of correctly solved problems, for example, in the case of self-preparation for tests or exams. Therefore, if you are still unsure of your abilities, if you are just starting to study this topic, do not rush - describe every step in detail, write down every factor, every stroke, and very soon you will learn to solve such examples better than many school teachers. I hope this is clear. Let's count a few more examples.

Several interesting tasks

This time, as we see, trigonometry is present in the derivatives being calculated. Therefore, let me remind you the following:

\[\begin(align)& (sinx())"=\cos x \\& ((\left(\cos x \right))^(\prime ))=-\sin x \\\end(align )\]

Of course, we cannot do without the derivative of the quotient, namely:

\[((\left(\frac(f)(g) \right))^(\prime ))=\frac((f)"\cdot g-f\cdot (g)")(((g)^( 2)))\]

Let's consider the first function:

\[\begin(align)& (f)"=((\left(\frac(\sin x)(x) \right))^(\prime ))=\frac(((\left(\sin x \right))^(\prime ))\cdot x-\sin x\cdot \left(((x)") \right))(((x)^(2)))= \\& =\frac (x\cdot \cos x-1\cdot \sin x)(((x)^(2)))=\frac(x\cos x-\sin x)(((x)^(2))) \\\end(align)\]

So we have found a solution to this expression.

Let's move on to the second example:

Obviously, its derivative will be more complex, if only because trigonometry is present in both the numerator and denominator of this function. We decide:

\[(y)"=((\left(\frac(x\sin x)(\cos x) \right))^(\prime ))=\frac(((\left(x\sin x \right ))^(\prime ))\cdot \cos x-x\sin x\cdot ((\left(\cos x \right))^(\prime )))(((\left(\cos x \right)) ^(2)))\]

Note that we have a derivative of the product. In this case it will be equal to:

\[\begin(align)& ((\left(x\cdot \sin x \right))^(\prime ))=(x)"\cdot \sin x+x((\left(\sin x \ right))^(\prime ))= \\& =\sin x+x\cos x \\\end(align)\]

Let's return to our calculations. We write down:

\[\begin(align)& (y)"=\frac(\left(\sin x+x\cos x \right)\cos x-x\cdot \sin x\cdot \left(-\sin x \right) )(((\cos )^(2))x)= \\& =\frac(\sin x\cdot \cos x+x((\cos )^(2))x+x((\sin ) ^(2))x)(((\cos )^(2))x)= \\& =\frac(\sin x\cdot \cos x+x\left(((\sin )^(2) )x+((\cos )^(2))x \right))(((\cos )^(2))x)=\frac(\sin x\cdot \cos x+x)(((\cos )^(2))x) \\\end(align)\]

That's all! We did the math.

How to reduce the derivative of a quotient to a simple formula for the derivative of a product?

And here I would like to make one very important remark concerning trigonometric functions. The fact is that our original construction contains an expression of the form $\frac(\sin x)(\cos x)$, which can easily be replaced simply by $tgx$. Thus, we reduce the derivative of a quotient to a simpler formula for the derivative of a product. Let's calculate this example again and compare the results.

So now we need to consider the following:

\[\frac(\sin x)(\cos x)=tgx\]

Let's rewrite our original function $y=\frac(x\sin x)(\cos x)$ taking this fact into account. We get:

Let's count:

\[\begin(align)& (y)"=((\left(x\cdot tgx \right))^(\prime ))(x)"\cdot tgx+x((\left(tgx \right) )^(\prime ))=tgx+x\frac(1)(((\cos )^(2))x)= \\& =\frac(\sin x)(\cos x)+\frac( x)(((\cos )^(2))x)=\frac(\sin x\cdot \cos x+x)(((\cos )^(2))x) \\\end(align) \]

Now, if we compare the result obtained with what we received earlier when calculating in a different way, we will be convinced that we have received the same expression. Thus, no matter which way we go when calculating the derivative, if everything is calculated correctly, then the answer will be the same.

Important nuances when solving problems

In conclusion, I would like to tell you one more subtlety related to calculating the derivative of a quotient. What I’m going to tell you now was not in the original script of the video lesson. However, a couple of hours before filming, I was studying with one of my students, and we were just discussing the topic of quotient derivatives. And, as it turned out, many students do not understand this point. So, let's say we need to calculate the remove stroke of the following function:

In principle, at first glance there is nothing supernatural about it. However, in the calculation process we can make many stupid and offensive mistakes, which I would like to discuss now.

So, we calculate this derivative. First of all, we note that we have the term $3((x)^(2))$, so it is appropriate to recall the following formula:

\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]

In addition, we have the term $\frac(48)(x)$ - we will deal with it through the derivative of the quotient, namely:

\[((\left(\frac(f)(g) \right))^(\prime ))=\frac((f)"\cdot g-f\cdot (g)")(((g)^( 2)))\]

So, let's decide:

\[(y)"=((\left(\frac(48)(x) \right))^(\prime ))+((\left(3((x)^(2)) \right)) ^(\prime ))+10(0)"\]

There are no problems with the first term, see:

\[((\left(3((x)^(2)) \right))^(\prime ))=3\cdot ((\left(((x)^(2)) \right))^ (\prime ))=3k.2x=6x\]

But with the first term, $\frac(48)(x)$, you need to work separately. The fact is that many students confuse the situation when they need to find $((\left(\frac(x)(48) \right))^(\prime ))$ and when they need to find $((\left(\frac (48)(x) \right))^(\prime ))$. That is, they get confused when the constant is in the denominator and when the constant is in the numerator, respectively, when the variable is in the numerator or in the denominator.

Let's start with the first option:

\[((\left(\frac(x)(48) \right))^(\prime ))=((\left(\frac(1)(48)\cdot x \right))^(\prime ))=\frac(1)(48)\cdot (x)"=\frac(1)(48)\cdot 1=\frac(1)(48)\]

On the other hand, if we try to do the same with the second fraction, we will get the following:

\[\begin(align)& ((\left(\frac(48)(x) \right))^(\prime ))=((\left(48\cdot \frac(1)(x) \right ))^(\prime ))=48\cdot ((\left(\frac(1)(x) \right))^(\prime ))= \\& =48\cdot \frac((1)" \cdot x-1\cdot (x)")(((x)^(2)))=48\cdot \frac(-1)(((x)^(2)))=-\frac(48 )(((x)^(2))) \\\end(align)\]

However, the same example could be calculated differently: at the stage where we passed to the derivative of the quotient, we can consider $\frac(1)(x)$ as a power with a negative exponent, i.e., we get the following:

\[\begin(align)& 48\cdot ((\left(\frac(1)(x) \right))^(\prime ))=48\cdot ((\left(((x)^(- 1)) \right))^(\prime ))=48\cdot \left(-1 \right)\cdot ((x)^(-2))= \\& =-48\cdot \frac(1 )(((x)^(2)))=-\frac(48)(((x)^(2))) \\\end(align)\]

And so, and so we received the same answer.

Thus, we are once again convinced of two important facts. Firstly, the same derivative can be calculated in completely different ways. For example, $((\left(\frac(48)(x) \right))^(\prime ))$ can be considered both as the derivative of a quotient and as the derivative of a power function. Moreover, if all calculations are performed correctly, then the answer will always be the same. Secondly, when calculating derivatives containing both a variable and a constant, it is fundamentally important where the variable is located - in the numerator or in the denominator. In the first case, when the variable is in the numerator, we get a simple linear function that can be easily calculated. And if the variable is in the denominator, then we get a more complex expression with the accompanying calculations given earlier.

At this point, the lesson can be considered complete, so if you don’t understand anything about the derivatives of a quotient or a product, and in general, if you have any questions on this topic, do not hesitate - go to my website, write, call, and I will definitely try can I help you.

Derivatives themselves are not a complex topic, but they are very extensive, and what we are studying now will be used in the future when solving more complex problems. That is why it is better to identify all misunderstandings related to the calculation of derivatives of a quotient or a product immediately, right now. Not when they are a huge snowball of misunderstanding, but when they are a small tennis ball that is easy to deal with.

The operation of finding the derivative is called differentiation.

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

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

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

Example 1. Find the derivative of a function

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

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

Example 2. Find the derivative of a function

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

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

Table of derivatives of simple functions

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

Rules of differentiation

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

Rule 1.If the functions

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

and

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

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

Rule 2.If the functions

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

and

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

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

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

For example, for three multipliers:

Rule 3.If the functions

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

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

Where to look for things on other pages

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

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

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

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

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

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

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

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

Example 3. Find the derivative of a function

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

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

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

Example 4. Find the derivative of a function

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

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

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

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

Example 5. Find the derivative of a function

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

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

Example 6. Find the derivative of a function

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

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

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

Everything seems to be clear. But try using this formula to calculate, 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 from the entire variety of functions we can distinguish the so-called elementary functions. These are relatively simple expressions, the derivatives of which have long been calculated and entered into the table. Such functions are quite easy to remember - along with their derivatives.

Derivatives of elementary functions

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

So, derivatives of elementary functions:

Name	Function	Derivative
Constant	f(x) = C, C ∈ R	0 (yes, zero!)
Power 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/sin 2 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 particularly elementary, but also differentiated according to certain rules. These rules are discussed below.

Derivative of sum and difference

Let the functions be given f(x) And g(x), the derivatives of which 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 + sin x; g(x) = x 4 + 2x 2 − 3.

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

f ’(x) = (x 2 + sin x)’ = (x 2)’ + (sin x)’ = 2x+ cos x;

We reason 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+ cos x;
g ’(x) = 4x · ( x 2 + 1).

Derivative of the product

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

(f · g) ’ = f ’ · g + f · g ’

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

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

Function f(x) is the 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. Obviously, the first factor 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 .

Please note that in the last step the derivative is factorized. Formally, this does not need to be done, but most derivatives are not calculated on their own, but to examine the function. This means that further the derivative will be equated to zero, its signs will be determined, and so on. For such a case, it is better to have an expression factorized.

If there are two functions f(x) And g(x), and g(x) ≠ 0 on the set we are interested in, 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? And 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:

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

According to tradition, let's factorize the numerator - this will greatly simplify the answer:

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

What should I do? In such cases, replacing a variable and formula for the derivative of a complex function helps:

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

As a rule, the situation with understanding this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it using 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 replacement: let 2 x + 3 = t, f(x) = f(t) = e t. We look for the derivative of a complex function using the formula:

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

And now - attention! We perform the reverse replacement: 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 it 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 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 “prime.” For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.

Thus, calculating the derivative comes down to getting rid of these same 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 people know that in the role n may well be a fractional number. For example, the root is x 0.5. What if there is something fancy under the root? Again, the result will be a complex function - they like to give such constructions in tests and exams.

Task. Find the derivative of the 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 replacement: let x 2 + 8x − 7 = t. We find the derivative using the formula:

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

Let's do the reverse replacement: 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: