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Derivative of the product of functions at a given point. Find the derivative: algorithm and examples of solutions

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 both the first and second terms contain the degree $x$: 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 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 before us the 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, while 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 from the root to the power with a rational exponent and back several times, 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 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 got earlier when calculating in a different way, then 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.

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 tabulated. 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:

(f + g)’ = f ’ + g ’
(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, huh? 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:

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Solving physical problems or examples in mathematics is completely impossible without knowledge of the derivative and methods for calculating it. The derivative is one of the most important concepts in 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 of the 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 over 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 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.

Let the functions u be defined in a certain neighborhood of a point and have derivatives at the point. Then their product has a derivative at the point, which is determined by the formula:
(1) .

Proof

Let us introduce the following notation:
;
.
Here and are functions of the variables and . But for ease of notation, we will omit the designations of their arguments.

Next we notice that
;
.
By condition, the functions and have derivatives at the point, which are the following limits:
;
.
From the existence of derivatives it follows that the functions and are continuous at the point. That's why
;
.

Consider the function y of the variable x, which is the product of the functions and:
.
Let's consider the increment of this function at the point:

.
Now we find the derivative:

.

So,
.
The rule has been proven.

Instead of a variable, you can use any other variable. Let's denote it as x. Then if there are derivatives and , then the derivative of the product of two functions is determined by the formula:
.
Or in a shorter version
(1) .