It is absolutely impossible to solve physical problems or examples in mathematics without knowledge about the derivative and methods for calculating it. The derivative is one of the most important concepts of 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 the derivative
Let there be a function f(x) , given in some interval (a,b) . The points x and x0 belong to this interval. When x changes, the function itself changes. Argument change - difference of its values x-x0 . This difference is written as delta x and is called argument increment. The change or increment of a function is the difference between the values of the function at two points. Derivative definition:
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 is the point in finding such a limit? But which one:
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.
The physical meaning of the derivative: the time derivative of the path is equal to the speed of the rectilinear motion.
Indeed, since school days, everyone knows that speed is a private path. x=f(t) and time t . Average speed over a certain period of time:
To find out the speed of movement at a time t0 you need to calculate the limit:
Rule one: take out the constant
The constant can be taken out of the sign of the derivative. Moreover, it must be done. When solving examples in mathematics, take as a rule - if you can simplify the expression, be sure to simplify .
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 a function:
Rule three: the 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:
Decision: 
Here it is important to say about the calculation of derivatives of complex functions. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument by the derivative of the intermediate argument with respect to the independent variable.
In the above example, we encounter 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 consider 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: The derivative of the quotient of two functions
Formula for determining the derivative of a 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 question on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult control and deal with tasks, even if you have never dealt with the calculation of derivatives before.
With proofreading materials on the topic "derivative". Basic school level.
Theoretical information for students, teachers and tutors in mathematics. To help with the lessons.
Definition: the derivative of a function at a point is called the limit of the ratio of the increment of the function to the increment of the variable, that is
Table of derivatives of basic mathematical functions:
Rules for calculating derivatives
Derivative of sum of any two expressions is equal to the sum of the derivatives of these expressions (the derivative of the sum is equal to the sum of the derivatives)
Difference derivative of any two expressions is equal to the difference of the derivatives of these terms (the derivative of the difference is equal to the difference of the derivatives).
Derivative of the product two factors is equal to the product of the derivative of the first factor by the second plus the product of the first factor by the derivative of the second (the sum of the derivatives of the factors taken in turn).
Math tutor's comment: when I remind the student in short phrases about the rule for calculating the derivative of a product, I say this: the derivative of the first factor by the second plus stroke exchange!
Derivative of quotient of two expressions is equal to the quotient of the difference of alternately taken derivatives of the factors and the square of the denominator.
Derivative of the product of a number and a function. To find the derivative of the product of a number and a literal expression (a function), you need to multiply this number by the derivative of this literal expression.
Derivative of a complex function:
To calculate the derivative of a complex function, you need to find the derivative of the outer function and multiply it by the derivative of the inner function.
Your comments and feedback on the page with derivatives:
Alexander S.
I really needed a table. One of the most on the internet. Thank you very much for the explanations and rules. At least one more example to them and in general it would be great. Thanks again.
Kolpakov A.N., tutor in mathematics: ok, I'll try to update the page with examples soon.
Virtual mathematical reference book.
Kolpakov Alexander Nikolaevich, tutor in mathematics.
What is a derivative function - this is the main mathematical concept, is on the same level with integrals, in the analysis. This function at a certain point gives a characteristic of the rate of change of the function at a given point.
Such concepts as differentiation and integration, the first stands for the action of finding a derivative, the second, on the contrary, restores the function starting from this derivative.
Derivative calculations play an important part in differential calculations.
For an illustrative example, we will depict the derivative on the coordinate plane.
in the function y \u003d f (x), we fix the points M at which (x0; f (X0)) and N f (x0 +? x) to each abscissa there is an increment in the form? x. An increment is the process when the abscissa changes, then the ordinate also changes. Designated as?
Let's find the tangent of the angle in triangle MPN using points M and N for this.
tg? = NP/MP = ?y/?x.
With? x going to 0. 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
Derivative 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 stroke equals one.
3) If there is a constant factor, we simply take out eo for the derivative.
4) To find the derivative power, you need to multiply the exponent of this degree by the exponent with the same base, in which the exponent is 1 less.
5) Finding a root is 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 is equal to the sine with a minus sign.
9) P tangent equals one divided by the squared cosine.
10) P cotangent equals one with a minus sign, divided by the sine squared.
In differentiation, there are also rules that are also easier to learn by pronouncing them out loud.
1) Very simply, the number of terms is equal to 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 from itself the multiplication of the second value by the first. The fraction divided by the second value squared.
4) The formulation is a special case of the third formula.
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, the average student no longer makes this mistake.
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:
And you can check the solution of the problem on the derivative on .
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 in the current example, is taken with a minus sign:
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:
You can check the solution of the derivative problem on derivative calculator online .
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 .
In this lesson, we continue to study the derivatives of functions and move on to a more complex topic, namely, the derivatives of the product and the quotient. If you watched the previous lesson, you probably understood that we considered only the simplest constructions, namely, the derivative of a power function, sums and differences. In particular, we learned that the derivative of the sum is equal to their sum, and the derivative of the difference is equal, respectively, to their difference. Unfortunately, in the case of derivatives of the quotient and the product, the formulas will be much more complicated. Let's start with the formula for the derivative of a product of functions.
Derivatives of trigonometric functions
To begin with, I will allow myself a small lyrical digression. The fact is that in addition to the standard power function - $y=((x)^(n))$, in this lesson there will be 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 must be mentioned separately. Let's write:
\[\begin(align)& ((\left(\sinx \right))^(\prime ))=\cosx \\& ((\left(\cos x \right))^(\prime ))= -\sin x \\& ((\left(tgx \right))^(\prime ))=\frac(1)(((\cos )^(2))x) \\& ((\left( ctgx \right))^(\prime ))=\frac(1)(((\cos )^(2))x) \\\end(align)\]
But you know these formulas very well, let's go further.
What is a derivative of a product?
First, the most important thing: if a function is a 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 that we considered earlier. For example, the derivative of the sum is considered elementary — $((\left(f+g \right))^(\prime ))=(f)"+(g)"$, or the derivative of the difference, which is also considered elementary — $(( \left(f-g \right))^(\prime ))=(f)"-(g)"$.
Let's try to apply the first formula to calculate the derivatives of two functions that we are given in the problem. Let's start with the first example:
It is obvious that the following construction acts as a product, more precisely, as a factor: $((x)^(3))$, we can consider as $f$, and $\left(x-5 \right)$ we can consider as $g$. Then their product will just be 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 power of $x$: 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, it will remain in the bracket:
\[\begin(align)& 3((x)^(2))\cdot \left(x-5 \right)+((x)^(3))\cdot 1=((x)^(2 ))\left(3\cdot 1\left(x-5 \right)+x \right)= \\& =((x)^(2))\left(3x-15+x \right)=( (x)^(2))(4x-15) \\\end(align)\]
All we found the answer.
We return to our tasks 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 the product of two functions. To find the derivative of the function $f\left(x \right)$, we again use our formula. We get:
\[\begin(align)& (f)"=\left(x \right)"\cdot \left(\sqrt(x)-1 \right)+x\cdot ((\left(\sqrt(x) -1 \right))^(\prime ))=1\cdot \left(\sqrt(x)-1 \right)+x\frac(1)(3\sqrt(x))= \\& =\ sqrt(x)-1+\sqrt(x)\cdot \frac(1)(3)=\frac(4)(3)\sqrt(x)-1 \\\end(align)\]
Answer found.
Why factorize derivatives?
We have just used some 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 got rid of all the signs of the 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 degree with a rational exponent and vice versa, while using the formula of the 8th-9th grade, which should be repeated separately.
Regarding factorization - why do we need all these additional efforts and transformations? 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 at various exams and tests, just 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, an increase or decrease in a function, and for this you need to solve the equation, factor it. And here this technique will be very appropriate. And in general, with a function decomposed into factors, it is much more convenient and pleasant to work in the future if any transformations are required. Therefore, rule number 1: if the derivative can be factored, that's exactly what you should do. And immediately rule number 2 (in fact, this is the material of the 8th-9th grade): if the root occurs in the problem n-th degree, moreover, 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- the same 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 does not go anywhere, but simply appears in the numerator of this very degree.
And now that you understand all this, let's go back to the derivatives of the product and calculate a few more equations.
But before proceeding directly to the calculations, I would like to recall the following patterns:
\[\begin(align)& ((\left(\sin x \right))^(\prime ))=\cos x \\& ((\left(\cos x \right))^(\prime ) )=-\sin x \\& \left(tgx \right)"=\frac(1)(((\cos )^(2))x) \\& ((\left(ctgx \right))^ (\prime ))=-\frac(1)(((\sin )^(2))x) \\\end(align)\]
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 $g$. The total 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 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 factor this expression into factors, because this is not yet the final answer. Now we have to solve the second part. Let's write it out:
\[\begin(align)& ((\left(4x\cdot \sin x \right))^(\prime ))=((\left(4x \right))^(\prime ))\cdot \sin x+4x\cdot ((\left(\sin x \right))^(\prime ))= \\& =4\cdot \sin x+4x\cdot \cos x \\\end(align)\]
And now we return to our original task and collect everything 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 the 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, we 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 that I wanted to tell about the derivatives of the work. As you can see, the main problem of the formula is not to memorize it, but that a rather large amount of calculations is obtained. But that's okay, because now we're moving on to the derivative of the quotient, where we have to work very hard.
What is the derivative of a quotient?
So, the formula for the derivative of a quotient. Perhaps this is the most difficult formula in the school derivatives course. Suppose we have a function of the form $\frac(f)(g)$, where $f$ and $g$ are also functions that can also be unfinished. Then it will be calculated according to the following formula:
The numerator somewhat reminds us of the formula for the derivative of the product, however, 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 propose to write out each part separately and write down:
\[\begin(align)& ((\left(((x)^(2))-1 \right))^(\prime ))=((\left(((x)^(2)) \ right))^(\prime ))-(1)"=2x \\& ((\left(x+2 \right))^(\prime ))=(x)"+(2)"=1 \ \\end(align)\]
We 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 just one, here the calculations 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 count each part of the example separately:
\[\begin(align)& (1)"=0 \\& ((\left(((x)^(2))+4 \right))^(\prime ))=((\left(( (x)^(2)) \right))^(\prime ))+(4)"=2x \\\end(align)\]
We 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 notations?
Attentive students probably already have a question: why in some cases we denote the function as $f\left(x \right)$, while in other cases we just 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 for exams and 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) just $y$. The thing is that by writing a function in the form \, we implicitly hint to the one who will read our calculations that we are talking about the algebraic interpretation of the functional dependence. That is, there is some 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 will read your calculations, for example, the verifier, will subconsciously expect that in the future only algebraic transformations await him - no graphs and no geometry.
On the other hand, using the notation of the form \, i.e., denoting the variable with one single letter, we immediately make it clear that in the future we are interested in precisely the geometric interpretation of the function, i.e., we are primarily interested in its graph. Accordingly, 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, not 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 my head. 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 in detail each step, write down each multiplier, each stroke, and very soon you will learn how to solve such examples better than many school teachers. I hope this is understandable. Let's count some more examples.
Several interesting challenges
This time, as we see, trigonometry is present in the composition of the calculated derivatives. So 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:
\[((\left(\frac(f)(g) \right))^(\prime ))=\frac((f)"\cdot g-f\cdot (g)")(((g)^( 2)))\]
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 the solution to this expression.
Let's move on to the second example:
It is obvious that its derivative will be more complex if only because trigonometry is present in both the numerator and the 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:
\[\begin(align)& ((\left(x\cdot \sin x \right))^(\prime ))=(x)"\cdot \sin x+x((\left(\sin x \ right))^(\prime ))= \\& =\sin x+x\cos x \\\end(align)\]
We 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 counted.
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 specifically trigonometric functions. The point is that our original construction contains an expression of the form $\frac(\sin x)(\cos x)$, which can be easily replaced by just $tgx$. Thus, we will reduce the derivative of the quotient to a simpler formula for the derivative of the 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 $y=\frac(x\sin x)(\cos x)$ with this fact in mind. 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 with what we got earlier, when calculating in a different way, then we will make sure that we got 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 in solving problems
In conclusion, I would like to tell you one more subtlety related to the calculation of the derivative of a quotient. What I am going to tell you now was not in the original script of the video tutorial. However, a couple of hours before filming, I was studying with one of my students, and we were just sorting out the topic of derivatives of the quotient. And, as it turned out, many students do not understand this point. So, let's say we need to count the unprime of the following function:
In principle, there is nothing supernatural in it at first glance. However, in the process of calculation, we can make many stupid and offensive mistakes, which I would like to analyze now.
So, we consider this derivative. First of all, 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 you need to find $((\left(\frac(x)(48) \right))^(\prime ))$ and when you 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 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 got the same answer.
Thus, we are once again convinced of two important facts. First, the same derivative can be calculated in completely different ways. For example, $((\left(\frac(48)(x) \right))^(\prime ))$ can be considered both as a derivative of a quotient and as a 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 simply counts. And if the variable is in the denominator, then we get a more complex expression with the accompanying calculations given earlier.
This lesson can be considered complete, so if you don’t understand something about the derivatives of a private or product, and indeed, if you have any questions on this topic, do not hesitate - visit my website, write, call, and I will definitely try can I help you.
Derivatives themselves are by no means a difficult topic, but very voluminous, 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 calculations of derivatives of a quotient or a product immediately, right now. Not when they are a huge snowball of misunderstandings, but when they are a small tennis ball that is easy to deal with.