How to find the derivative of a complex function with a degree. Derivatives of trigonometric functions

With this video I begin a long series of lessons on derivatives. This lesson consists of several parts.

First of all, I will tell you what derivatives are and how to calculate them, but not in sophisticated academic language, but the way I understand it myself and how I explain it to my students. Secondly, we will consider the simplest rule for solving problems in which we will look for derivatives of sums, derivatives of differences and derivatives of a power function.

We will look at more complex combined examples, from which you will, in particular, learn that similar problems involving roots and even fractions can be solved using the formula for the derivative of a power function. In addition, of course, there will be many problems and examples of solutions for the most different levels difficulties.

In general, initially I was going to record a short 5-minute video, but you can see how it turned out. So enough of the lyrics - let's get down to business.

What is a derivative?

So, let's start from afar. Many years ago, when the trees were greener and life was more fun, mathematicians thought about this: consider simple function, given by its graph, let's call it $y=f\left(x \right)$. Of course, the graph does not exist on its own, so you need to draw the $x$ axes as well as the $y$ axis. Now let's choose any point on this graph, absolutely any. Let's call the abscissa $((x)_(1))$, the ordinate, as you might guess, will be $f\left(((x)_(1)) \right)$.

Let's look at another point on the same graph. It doesn’t matter which one, the main thing is that it differs from the original one. It, again, has an abscissa, let's call it $((x)_(2))$, and also an ordinate - $f\left(((x)_(2)) \right)$.

So, we got two points: they have different abscissas and, therefore, different meanings functions, although the latter is optional. But what is really important is that we know from the planimetry course: through two points you can draw a straight line and, moreover, only one. So let's carry it out.

Now let’s draw a straight line through the very first of them, parallel to the abscissa axis. We get right triangle. Let's call it $ABC$, right angle $C$. This triangle has one very interesting property: the fact is that the angle $\alpha $, in fact, equal to angle, under which the straight line $AB$ intersects with the continuation of the abscissa axis. Judge for yourself:

the straight line $AC$ is parallel to the $Ox$ axis by construction,
line $AB$ intersects $AC$ under $\alpha $,
hence $AB$ intersects $Ox$ under the same $\alpha $.

What can we say about $\text( )\!\!\alpha\!\!\text( )$? Nothing specific, except that in the triangle $ABC$ the ratio of leg $BC$ to leg $AC$ is equal to the tangent of this very angle. So let's write it down:

Of course, $AC$ in in this case easy to calculate:

Likewise for $BC$:

In other words, we can write the following:

\[\operatorname(tg)\text( )\!\!\alpha\!\!\text( )=\frac(f\left(((x)_(2)) \right)-f\left( ((x)_(1)) \right))(((x)_(2))-((x)_(1)))\]

Now that we've got all that out of the way, let's go back to our graph and look at new point$B$. Let's erase the old values and take $B$ somewhere closer to $((x)_(1))$. Let us again denote its abscissa by $((x)_(2))$, and its ordinate by $f\left(((x)_(2)) \right)$.

Let's look again at our little triangle $ABC$ and $\text( )\!\!\alpha\!\!\text( )$ inside it. It is quite obvious that this will be a completely different angle, the tangent will also be different because the lengths of the segments $AC$ and $BC$ have changed significantly, but the formula for the tangent of the angle has not changed at all - this is still the relationship between a change in the function and a change in the argument .

Finally, we continue to move $B$ closer to the original point $A$, as a result the triangle will become even smaller, and the straight line containing the segment $AB$ will look more and more like a tangent to the graph of the function.

As a result, if we continue to bring the points closer together, i.e., reduce the distance to zero, then the straight line $AB$ will indeed turn into a tangent to the graph at a given point, and $\text( )\!\!\alpha\!\ !\text( )$ will transform from a regular triangle element to the angle between the tangent to the graph and the positive direction of the $Ox$ axis.

And here we smoothly move on to the definition of $f$, namely, the derivative of a function at the point $((x)_(1))$ is the tangent of the angle $\alpha $ between the tangent to the graph at the point $((x)_( 1))$ and the positive direction of the $Ox$ axis:

\[(f)"\left(((x)_(1)) \right)=\operatorname(tg)\text( )\!\!\alpha\!\!\text( )\]

Returning to our graph, it should be noted that any point on the graph can be chosen as $((x)_(1))$. For example, with the same success we could remove the stroke at the point shown in the figure.

Let's call the angle between the tangent and the positive direction of the axis $\beta$. Accordingly, $f$ in $((x)_(2))$ will be equal to the tangent of this angle $\beta $.

\[(f)"\left(((x)_(2)) \right)=tg\text( )\!\!\beta\!\!\text( )\]

Each point on the graph will have its own tangent, and, therefore, its own function value. In each of these cases, in addition to the point at which we are looking for the derivative of a difference or sum, or the derivative of a power function, it is necessary to take another point located at some distance from it, and then direct this point to the original one and, of course, find out how in the process Such movement will change the tangent of the angle of inclination.

Derivative of a power function

Unfortunately, similar definition We are not satisfied at all. All these formulas, pictures, angles do not give us the slightest idea of how to calculate the real derivative in real problems. Therefore, let's digress a little from the formal definition and consider more effective formulas and techniques with which you can already solve real problems.

Let's start with the simplest constructions, namely, functions of the form $y=((x)^(n))$, i.e. power functions. In this case, we can write the following: $(y)"=n\cdot ((x)^(n-1))$. In other words, the degree that was in the exponent is shown in the front multiplier, and the exponent itself is reduced by unit. For example:

\[\begin(align)& y=((x)^(2)) \\& (y)"=2\cdot ((x)^(2-1))=2x \\\end(align) \]

Here's another option:

\[\begin(align)& y=((x)^(1)) \\& (y)"=((\left(x \right))^(\prime ))=1\cdot ((x )^(0))=1\cdot 1=1 \\& ((\left(x \right))^(\prime ))=1 \\\end(align)\]

Using these simple rules, let's try to remove the touch of the following examples:

So we get:

\[((\left(((x)^(6)) \right))^(\prime ))=6\cdot ((x)^(5))=6((x)^(5)) \]

Now let's solve the second expression:

\[\begin(align)& f\left(x \right)=((x)^(100)) \\& ((\left(((x)^(100)) \right))^(\ prime ))=100\cdot ((x)^(99))=100((x)^(99)) \\\end(align)\]

Of course, these were very simple tasks. However real problems more complex and they are not limited to just degrees of function.

So, rule No. 1 - if a function is presented in the form of the other two, then the derivative of this sum is equal to the sum of the derivatives:

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

Similarly, the derivative of the difference of two functions is equal to the difference of the derivatives:

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

\[((\left(((x)^(2))+x \right))^(\prime ))=((\left(((x)^(2)) \right))^(\ prime ))+((\left(x \right))^(\prime ))=2x+1\]

In addition, there is one more important rule: if some $f$ is preceded by a constant $c$, by which this function is multiplied, then the $f$ of this entire construction is calculated as follows:

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

\[((\left(3((x)^(3)) \right))^(\prime ))=3((\left(((x)^(3)) \right))^(\ prime ))=3\cdot 3((x)^(2))=9((x)^(2))\]

Finally, one more very important rule: in problems there is often a separate term that does not contain $x$ at all. For example, we can observe this in our expressions today. The derivative of a constant, i.e., a number that does not depend in any way on $x$, is always equal to zero, and it does not matter at all what the constant $c$ is equal to:

\[((\left(c \right))^(\prime ))=0\]

Example solution:

\[((\left(1001 \right))^(\prime ))=((\left(\frac(1)(1000) \right))^(\prime ))=0\]

Key points again:

The derivative of the sum of two functions is always equal to the sum of the derivatives: $((\left(f+g \right))^(\prime ))=(f)"+(g)"$;
For similar reasons, the derivative of the difference of two functions is equal to the difference of two derivatives: $((\left(f-g \right))^(\prime ))=(f)"-(g)"$;
If a function has a constant factor, then this constant can be taken out as a derivative sign: $((\left(c\cdot f \right))^(\prime ))=c\cdot (f)"$;
If the entire function is a constant, then its derivative is always zero: $((\left(c \right))^(\prime ))=0$.

Let's see how it all works on real examples. So:

We write down:

\[\begin(align)& ((\left(((x)^(5))-3((x)^(2))+7 \right))^(\prime ))=((\left (((x)^(5)) \right))^(\prime ))-((\left(3((x)^(2)) \right))^(\prime ))+(7) "= \\& =5((x)^(4))-3((\left(((x)^(2)) \right))^(\prime ))+0=5((x) ^(4))-6x \\\end(align)\]

In this example we see both the derivative of the sum and the derivative of the difference. In total, the derivative is equal to $5((x)^(4))-6x$.

Let's move on to the second function:

Let's write down the solution:

\[\begin(align)& ((\left(3((x)^(2))-2x+2 \right))^(\prime ))=((\left(3((x)^( 2)) \right))^(\prime ))-((\left(2x \right))^(\prime ))+(2)"= \\& =3((\left(((x) ^(2)) \right))^(\prime ))-2(x)"+0=3\cdot 2x-2\cdot 1=6x-2 \\\end(align)\]

Here we have found the answer.

Let's move on to the third function - it is more serious:

\[\begin(align)& ((\left(2((x)^(3))-3((x)^(2))+\frac(1)(2)x-5 \right)) ^(\prime ))=((\left(2((x)^(3)) \right))^(\prime ))-((\left(3((x)^(2)) \right ))^(\prime ))+((\left(\frac(1)(2)x \right))^(\prime ))-(5)"= \\& =2((\left(( (x)^(3)) \right))^(\prime ))-3((\left(((x)^(2)) \right))^(\prime ))+\frac(1) (2)\cdot (x)"=2\cdot 3((x)^(2))-3\cdot 2x+\frac(1)(2)\cdot 1=6((x)^(2)) -6x+\frac(1)(2) \\\end(align)\]

We have found the answer.

Let's move on to the last expression - the most complex and longest:

So, we consider:

\[\begin(align)& ((\left(6((x)^(7))-14((x)^(3))+4x+5 \right))^(\prime ))=( (\left(6((x)^(7)) \right))^(\prime ))-((\left(14((x)^(3)) \right))^(\prime )) +((\left(4x \right))^(\prime ))+(5)"= \\& =6\cdot 7\cdot ((x)^(6))-14\cdot 3((x )^(2))+4\cdot 1+0=42((x)^(6))-42((x)^(2))+4 \\\end(align)\]

But the solution does not end there, because we are asked not just to remove a stroke, but to calculate its value at a specific point, so we substitute −1 instead of $x$ into the expression:

\[(y)"\left(-1 \right)=42\cdot 1-42\cdot 1+4=4\]

Let's go further and move on to even more complex and interesting examples. The fact is that the formula for solving the power derivative $((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))$ has an even wider scope than is usually believed. With its help, you can solve examples with fractions, roots, etc. This is what we will do now.

To begin with, let’s once again write down the formula that will help us find the derivative of a power function:

And now attention: so far we have considered as $n$ only integers, however, nothing prevents us from considering fractions and even negative numbers. For example, we can write the following:

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

Nothing complicated, so let's see how this formula will help us when solving more complex tasks. So, an example:

Let's write down the solution:

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

Let's go back to our example and write:

\[(y)"=\frac(1)(2\sqrt(x))+\frac(1)(3\sqrt(((x)^(2))))+\frac(1)(4 \sqrt(((x)^(3))))\]

This is such a difficult decision.

Let's move on to the second example - there are only two terms, but each of them contains both a classical degree and roots.

Now we will learn how to find the derivative of a power function, which, in addition, contains the root:

\[\begin(align)& ((\left(((x)^(3))\sqrt(((x)^(2)))+((x)^(7))\sqrt(x) \right))^(\prime ))=((\left(((x)^(3))\cdot \sqrt(((x)^(2))) \right))^(\prime )) =((\left(((x)^(3))\cdot ((x)^(\frac(2)(3))) \right))^(\prime ))= \\& =(( \left(((x)^(3+\frac(2)(3))) \right))^(\prime ))=((\left(((x)^(\frac(11)(3 ))) \right))^(\prime ))=\frac(11)(3)\cdot ((x)^(\frac(8)(3)))=\frac(11)(3)\ cdot ((x)^(2\frac(2)(3)))=\frac(11)(3)\cdot ((x)^(2))\cdot \sqrt(((x)^(2 ))) \\& ((\left(((x)^(7))\cdot \sqrt(x) \right))^(\prime ))=((\left(((x)^(7 ))\cdot ((x)^(\frac(1)(3))) \right))^(\prime ))=((\left(((x)^(7\frac(1)(3 ))) \right))^(\prime ))=7\frac(1)(3)\cdot ((x)^(6\frac(1)(3)))=\frac(22)(3 )\cdot ((x)^(6))\cdot \sqrt(x) \\\end(align)\]

Both terms have been calculated, all that remains is to write down the final answer:

\[(y)"=\frac(11)(3)\cdot ((x)^(2))\cdot \sqrt(((x)^(2)))+\frac(22)(3) \cdot ((x)^(6))\cdot \sqrt(x)\]

We have found the answer.

Derivative of a fraction through a power function

But the possibilities of the formula for solving the derivative of a power function do not end there. The fact is that with its help you can calculate not only examples with roots, but also with fractions. This is precisely the rare opportunity that greatly simplifies the solution of such examples, but is often ignored not only by students, but also by teachers.

So, now we will try to combine two formulas at once. On the one hand, the classical derivative of a power function

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

On the other hand, we know that an expression of the form $\frac(1)(((x)^(n)))$ can be represented as $((x)^(-n))$. Hence,

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

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

Thus, derivatives simple fractions, where the numerator is a constant and the denominator is a degree, are also calculated using classical formula. Let's see how this works in practice.

So, the first function:

\[((\left(\frac(1)(((x)^(2))) \right))^(\prime ))=((\left(((x)^(-2)) \ right))^(\prime ))=-2\cdot ((x)^(-3))=-\frac(2)(((x)^(3)))\]

The first example is solved, let's move on to the second:

\[\begin(align)& ((\left(\frac(7)(4((x)^(4)))-\frac(2)(3((x)^(3)))+\ frac(5)(2)((x)^(2))+2((x)^(3))-3((x)^(4)) \right))^(\prime ))= \ \& =((\left(\frac(7)(4((x)^(4))) \right))^(\prime ))-((\left(\frac(2)(3(( x)^(3))) \right))^(\prime ))+((\left(2((x)^(3)) \right))^(\prime ))-((\left( 3((x)^(4)) \right))^(\prime )) \\& ((\left(\frac(7)(4((x)^(4))) \right))^ (\prime ))=\frac(7)(4)((\left(\frac(1)(((x)^(4))) \right))^(\prime ))=\frac(7 )(4)\cdot ((\left(((x)^(-4)) \right))^(\prime ))=\frac(7)(4)\cdot \left(-4 \right) \cdot ((x)^(-5))=\frac(-7)(((x)^(5))) \\& ((\left(\frac(2)(3((x)^ (3))) \right))^(\prime ))=\frac(2)(3)\cdot ((\left(\frac(1)(((x)^(3))) \right) )^(\prime ))=\frac(2)(3)\cdot ((\left(((x)^(-3)) \right))^(\prime ))=\frac(2)( 3)\cdot \left(-3 \right)\cdot ((x)^(-4))=\frac(-2)(((x)^(4))) \\& ((\left( \frac(5)(2)((x)^(2)) \right))^(\prime ))=\frac(5)(2)\cdot 2x=5x \\& ((\left(2 ((x)^(3)) \right))^(\prime ))=2\cdot 3((x)^(2))=6((x)^(2)) \\& ((\ left(3((x)^(4)) \right))^(\prime ))=3\cdot 4((x)^(3))=12((x)^(3)) \\\ end(align)\]...

Now we collect all these terms into a single formula:

\[(y)"=-\frac(7)(((x)^(5)))+\frac(2)(((x)^(4)))+5x+6((x)^ (2))-12((x)^(3))\]

We have received an answer.

However, before moving on, I would like to draw your attention to the form of writing the original expressions themselves: in the first expression we wrote $f\left(x \right)=...$, in the second: $y=...$ Many students get lost when they see different shapes records. What is the difference between $f\left(x \right)$ and $y$? Nothing really. They are just different entries with the same meaning. It's just that when we say $f\left(x \right)$, then we're talking about, first of all, about a function, and when we talk about $y$, we most often mean the graph of a function. Otherwise, this is the same thing, i.e., the derivative in both cases is considered the same.

Complex problems with derivatives

In conclusion, I would like to consider a couple of complex combined problems that use everything we have considered today. They contain roots, fractions, and sums. However, these examples will only be complex in today’s video tutorial, because truly complex derivative functions will be waiting for you ahead.

So, the final part of today's video lesson, consisting of two combined tasks. Let's start with the first of them:

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

The derivative of the function is equal to:

\[(y)"=3((x)^(2))-\frac(3)(((x)^(4)))+\frac(1)(3\sqrt(((x)^ (2))))\]

The first example is solved. Let's consider the second problem:

In the second example we proceed similarly:

\[((\left(-\frac(2)(((x)^(4)))+\sqrt(x)+\frac(4)(x\sqrt(((x)^(3)) )) \right))^(\prime ))=((\left(-\frac(2)(((x)^(4))) \right))^(\prime ))+((\left (\sqrt(x) \right))^(\prime ))+((\left(\frac(4)(x\cdot \sqrt(((x)^(3)))) \right))^ (\prime ))\]

Let's calculate each term separately:

\[\begin(align)& ((\left(-\frac(2)(((x)^(4))) \right))^(\prime ))=-2\cdot ((\left( ((x)^(-4)) \right))^(\prime ))=-2\cdot \left(-4 \right)\cdot ((x)^(-5))=\frac(8 )(((x)^(5))) \\& ((\left(\sqrt(x) \right))^(\prime ))=((\left(((x)^(\frac( 1)(4))) \right))^(\prime ))=\frac(1)(4)\cdot ((x)^(-\frac(3)(4)))=\frac(1 )(4\cdot ((x)^(\frac(3)(4))))=\frac(1)(4\sqrt(((x)^(3)))) \\& ((\ left(\frac(4)(x\cdot \sqrt(((x)^(3)))) \right))^(\prime ))=((\left(\frac(4)(x\cdot ((x)^(\frac(3)(4)))) \right))^(\prime ))=((\left(\frac(4)(((x)^(1\frac(3 )(4)))) \right))^(\prime ))=4\cdot ((\left(((x)^(-1\frac(3)(4))) \right))^( \prime ))= \\& =4\cdot \left(-1\frac(3)(4) \right)\cdot ((x)^(-2\frac(3)(4)))=4 \cdot \left(-\frac(7)(4) \right)\cdot \frac(1)(((x)^(2\frac(3)(4))))=\frac(-7) (((x)^(2))\cdot ((x)^(\frac(3)(4))))=-\frac(7)(((x)^(2))\cdot \sqrt (((x)^(3)))) \\\end(align)\]

All terms have been calculated. Now we return to the original formula and add all three terms together. We get that the final answer will be like this:

\[(y)"=\frac(8)(((x)^(5)))+\frac(1)(4\sqrt(((x)^(3))))-\frac(7 )(((x)^(2))\cdot \sqrt(((x)^(3))))\]

And that is all. This was our first lesson. In the following lessons we will look at more complex constructions, and also find out why derivatives are needed in the first place.

First level

Derivative of a function. Comprehensive Guide (2019)

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

The axis is a certain level of zero altitude; in life we use sea level as it.

As we move forward along such a road, we also move up or down. We can also say: when the argument changes (movement along the abscissa axis), the value of the function changes (movement along the ordinate axis). Now let's think about how to determine the “steepness” of our road? What kind of value could this be? It’s very simple: how much does the height change when moving forward by certain distance. Indeed, on different sections of the road, moving forward (along the x-axis) by one kilometer, we will rise or fall by different quantities meters relative to sea level (along the ordinate axis).

Let’s denote progress (read “delta x”).

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

Important: an expression is a single whole, one variable. Never separate the “delta” from the “x” or any other letter! That is, for example, .

So, we have moved forward, horizontally, by. If we compare the line of the road with the graph of the function, then how do we denote the rise? Certainly, . That is, as we move forward, we rise higher.

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

Let's return to "steepness": this is a value that shows how much (steeply) the height increases when moving forward one unit of distance:

Let us assume that on some section of the road, when moving forward by a kilometer, the road rises up by a kilometer. Then the slope at this place is equal. And if the road, while moving forward by m, dropped by km? Then the slope is equal.

Now let's look at the top of a hill. If you take the beginning of the section half a kilometer before the summit, and the end half a kilometer after it, you can see that the height is almost the same.

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

IN real life Measuring distances to the nearest millimeter is more than enough. But mathematicians always strive for perfection. Therefore, the concept was invented infinitesimal, that is, the absolute value is less than any number that we can name. For example, you say: one trillionth! How much less? And you divide this number by - and it will be even less. And so on. If we want to write that a quantity is infinitesimal, we write like this: (we read “x tends to zero”). It is very important to understand that this number is not zero! But very close to it. This means that you can divide by it.

The concept opposite to infinitesimal is infinitely large (). You've probably already come across it when you were working on inequalities: this number is modulo greater than any number you can think of. If you come up with the biggest number possible, just multiply it by two and you'll get an even bigger number. And infinity still Furthermore what will happen. In fact, the infinitely large and the infinitely small are the inverse of each other, that is, at, and vice versa: at.

Now let's get back to our road. The ideally calculated slope is the slope calculated for an infinitesimal segment of the path, that is:

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

What is all this for? The road, the steepness... We’re not going on a car rally, but we’re teaching mathematics. And in mathematics everything is exactly the same, only called differently.

Concept of derivative

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

Incrementally in mathematics they call change. The extent to which the argument () changes as it moves along the axis is called argument increment and is designated. How much the function (height) has changed when moving forward along the axis by a distance is called function increment and is designated.

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

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

Can the derivative be equal to zero? Certainly. For example, if we are driving on a flat horizontal road, the steepness is zero. And it’s true, the height doesn’t change at all. Same with the derivative: derivative constant function(constants) is equal to zero:

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

Let's remember the hilltop example. It turned out that it was possible to arrange the ends of the segment along different sides from the top, so that the height at the ends is the same, that is, the segment is parallel to the axis:

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

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

This can be understood this way: when we stand at the very top, a small shift to the left or right changes our height negligibly.

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

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

A little more about increments.

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

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

Practice finding increments:

Find the increment of the function at a point when the increment of the argument is equal to.
The same goes for the function at a point.

Solutions:

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

Power function.

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

Moreover - to any extent: .

The simplest case- this is when the exponent:

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

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

Increment is this. But a function at any point is equal to its argument. That's why:

The derivative is equal to:

The derivative of is equal to:

b) Now consider quadratic function (): .

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

So, we came up with another rule:

c) We continue the logical series: .

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

So, I got the following:

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

We get: .

d) Similar rules can be obtained for large powers:

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

(2)

The rule can be formulated in the words: “the degree is brought forward as a coefficient, and then reduced by .”

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

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

. You won't believe it, but this power function. If you have questions like “How is this? Where is the degree?”, remember the topic “”!
Yes, yes, the root is also a degree, only fractional: .
So ours Square root- this is just a degree with an indicator:
.
We look for the derivative using the recently learned formula:
If at this point it becomes unclear again, repeat the topic “”!!! (about degree with negative indicator)
. Now the exponent:
And now through the definition (have you forgotten yet?):
;
.
Now, as usual, we neglect the term containing:
.
. Combination of previous cases: .

Trigonometric functions.

Here we will use one fact from higher mathematics:

With expression.

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

We see that when the function does not exist - the point on the graph is cut out. But the closer to the value, the closer the function is to. This is what “aims.”

Additionally, you can check this rule using a calculator. Yes, yes, don’t be shy, take a calculator, we’re not at the Unified State Exam yet.

So, let's try: ;

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

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

a) Consider the function. As usual, let's find its increment:

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

Now the derivative:

Let's make a replacement: . Then for infinitesimal it is also infinitesimal: . The expression for takes the form:

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

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

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

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

Practice:

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

Solutions:

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

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

Exponent and natural logarithm.

There is a function in mathematics whose derivative for any value is equal to the value of the function itself at the same time. It is called “exponent”, and is an exponential function

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

So, the rule:

Very easy to remember.

Well, let’s not go far, let’s look at it right away inverse function. Which function is the inverse of exponential function? Logarithm:

In our case, the base is the number:

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

What is it equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

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

Answers: Exhibitor and natural logarithm- functions are uniquely simple in terms of derivatives. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after let's go through the rules differentiation.

Rules of differentiation

Rules of what? Again new term, again?!...

Differentiation is the process of finding the derivative.

That's all. What else can you call this process in one word? Not derivative... Mathematicians call the differential the same increment of a function at. This term comes from the Latin differentia - difference. Here.

When deriving all these rules, we will use two functions, for example, and. We will also need formulas for their increments:

There are 5 rules in total.

The constant is taken out of the derivative sign.

If - some constant number(constant), then.

Obviously, this rule also works for the difference: .

Let's prove it. Let it be, or simpler.

Examples.

Find the derivatives of the functions:

at a point;
at a point;
at a point;
at the point.

Solutions:

(the derivative is the same at all points, since this linear function, remember?);

Derivative of the product

Everything is similar here: let’s enter new feature and find its increment:

Derivative:

Examples:

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

Solutions:

Derivative of an exponential function

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

So, where is some number.

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

For this we will use simple rule: . Then:

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

Happened?

Here, check yourself:

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

Examples:
Find the derivatives of the functions:

Answers:

This is just a number that cannot be calculated without a calculator, that is, it cannot be written down in any more in simple form. Therefore, we leave it in this form in the answer.

Derivative of a logarithmic function

It’s similar here: you already know the derivative of the natural logarithm:

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

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

Only now we will write instead:

The denominator is simply a constant (a constant number, without a variable). The derivative is obtained very simply:

Derivatives of exponential and logarithmic functions almost never appear in the Unified State Examination, but it wouldn’t hurt to know them.

Derivative of a complex function.

What is a "complex function"? No, this is not a logarithm, and not an arctangent. These functions can be difficult to understand (although if you find the logarithm difficult, read the topic “Logarithms” and you will be fine), but from a mathematical point of view, the word “complex” does not mean “difficult”.

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

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then square the resulting number. So, we are given a number (chocolate), I find its cosine (wrapper), and then you square what I got (tie it with a ribbon). What happened? Function. This is an example complex function: when, to find its value, we perform the first action directly with the variable, and then a second action with what resulted from the first.

We can easily do the same steps in reverse order: first you square it, and I then look for the cosine of the resulting number: . It’s easy to guess that the result will almost always be different. Important Feature complex functions: when the order of actions changes, the function changes.

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

For the first example, .

Second example: (same thing). .

The action we do last will be called "external" function, and the action performed first - accordingly "internal" function(these are informal names, I use them only to explain the material in simple language).

Try to determine for yourself which function is external and which internal:

Answers: Separating inner and outer functions is very similar to changing variables: for example, in a function

What action will we perform first? First, let's calculate the sine, and only then cube it. This means that it is an internal function, but an external one.
And the original function is their composition: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .
Internal: ; external: .
Examination: .

We change variables and get a function.

Well, now we will extract our chocolate bar and look for the derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. In relation to the original example, it looks like this:

Another example:

So, let's finally formulate the official rule:

Algorithm for finding the derivative of a complex function:

It seems simple, right?

Let's check with examples:

Solutions:

1) Internal: ;

External: ;

2) Internal: ;

(Just don’t try to cut it by now! Nothing comes out from under the cosine, remember?)

3) Internal: ;

External: ;

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

That is, first we differentiate the root, then the cosine, and only then the expression in brackets. And then we multiply it all.

In such cases, it is convenient to number the actions. That is, let's imagine what we know. In what order will we perform actions to calculate the value of this expression? Let's look at an example:

The later the action is performed, the more “external” the corresponding function will be. The sequence of actions is the same as before:

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

1. Radical expression. .

2. Root. .

3. Sine. .

4. Square. .

5. Putting it all together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN THINGS

Derivative of a function- the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument:

Basic derivatives:

Rules of differentiation:

The constant is taken out of the derivative sign:

Derivative of the sum:

Derivative of the product:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

We define the “internal” function and find its derivative.
We define the “external” function and find its derivative.
We multiply the results of the first and second points.

On which we analyzed the simplest derivatives, and also got acquainted with the rules of differentiation and some technical methods finding derivatives. Thus, if you are not very good with derivatives of functions or some points in this article are not entirely clear, then first read the above lesson. Please get in a serious mood - the material is not simple, but I will still try to present it simply and clearly.

In practice, you have to deal with the derivative of a complex function very often, I would even say, almost always, when you are given tasks to find derivatives.

We look at the table at the rule (No. 5) for differentiating a complex function:

Let's figure it out. First of all, let's pay attention to the entry. Here we have two functions - and , and the function, figuratively speaking, is nested within the function . A function of this type (when one function is nested within another) is called a complex function.

I will call the function external function, and the function – internal (or nested) function.

! These definitions are not theoretical and should not appear in the final design of assignments. I use informal expressions “external function”, “internal” function only to make it easier for you to understand the material.

To clarify the situation, consider:

Example 1

Find the derivative of a function

Under the sine we have not just the letter “X”, but an entire expression, so finding the derivative right away from the table will not work. We also notice that it is impossible to apply the first four rules here, there seems to be a difference, but the fact is that the sine cannot be “torn into pieces”:

IN in this example It is already intuitively clear from my explanations that a function is a complex function, and the polynomial is an internal function (embedding), and an external function.

First step what you need to do when finding the derivative of a complex function is to understand which function is internal and which is external.

When simple examples It seems clear that a polynomial is embedded under the sine. But what if everything is not obvious? How to accurately determine which function is external and which is internal? To do this, I suggest using the following technique, which can be done mentally or in a draft.

Let's imagine that we need to calculate the value of the expression at on a calculator (instead of one there can be any number).

What will we calculate first? First of all you will need to perform the following action: , therefore the polynomial will be an internal function:

Secondly will need to be found, so sine – will be an external function:

After we SOLD OUT with internal and external functions, it’s time to apply the rule of differentiation of complex functions .

Let's start deciding. From the lesson How to find the derivative? we remember that the design of a solution to any derivative always begins like this - we enclose the expression in brackets and put a stroke at the top right:

At first find the derivative of the external function (sine), look at the table of derivatives elementary functions and we notice that . All table formulas are also applicable if “x” is replaced with a complex expression, in this case:

Please note that the inner function hasn't changed, we don't touch it.

Well, it's quite obvious that

The result of applying the formula in its final form it looks like this:

The constant factor is usually placed at the beginning of the expression:

If there is any misunderstanding, write the solution down on paper and read the explanations again.

Example 2

Find the derivative of a function

Example 3

Find the derivative of a function

As always, we write down:

Let's figure out where we have an external function and where we have an internal one. To do this, we try (mentally or in a draft) to calculate the value of the expression at . What should you do first? First of all, you need to calculate what the base is equal to: therefore, the polynomial is the internal function:

And, only then is the exponentiation performed, therefore, the power function is an external function:

According to the formula , first you need to find the derivative of the external function, in this case, the degree. Looking for in the table the required formula: . We repeat again: any tabular formula is valid not only for “X”, but also for a complex expression. Thus, the result of applying the rule for differentiating a complex function next:

I emphasize again that when we take the derivative of the external function, our internal function does not change:

Now all that remains is to find a very simple derivative of the internal function and tweak the result a little:

Example 4

Find the derivative of a function

This is an example for independent decision(answer at the end of the lesson).

To consolidate your understanding of the derivative of a complex function, I will give an example without comments, try to figure it out on your own, reason where the external and where the internal function is, why the tasks are solved this way?

Example 5

a) Find the derivative of the function

b) Find the derivative of the function

Example 6

Find the derivative of a function

Here we have a root, and in order to differentiate the root, it must be represented as a power. Thus, first we bring the function into the form appropriate for differentiation:

Analyzing the function, we come to the conclusion that the sum of the three terms is an internal function, and raising to a power is an external function. We apply the rule of differentiation of complex functions :

We again represent the degree as a radical (root), and for the derivative of the internal function we apply a simple rule for differentiating the sum:

Ready. You can also give the expression in parentheses to common denominator and write everything down as one fraction. It’s beautiful, of course, but when you get cumbersome long derivatives, it’s better not to do this (it’s easy to get confused, make an unnecessary mistake, and it will be inconvenient for the teacher to check).

Example 7

Find the derivative of a function

This is an example for you to solve on your own (answer at the end of the lesson).

It is interesting to note that sometimes instead of the rule for differentiating a complex function, you can use the rule for differentiating a quotient , but such a solution will look like an unusual perversion. Here is a typical example:

Example 8

Find the derivative of a function

Here you can use the rule of differentiation of the quotient , but it is much more profitable to find the derivative through the rule of differentiation of a complex function:

We prepare the function for differentiation - we move the minus out of the derivative sign, and raise the cosine into the numerator:

Cosine is an internal function, exponentiation is an external function.
Let's use our rule :

We find the derivative of the internal function and reset the cosine back down:

Ready. In the example considered, it is important not to get confused in the signs. By the way, try to solve it using the rule , the answers must match.

Example 9

Find the derivative of a function

This is an example for you to solve on your own (answer at the end of the lesson).

So far we have looked at cases where we had only one nesting in a complex function. In practical tasks, you can often find derivatives, where, like nesting dolls, one inside the other, 3 or even 4-5 functions are nested at once.

Example 10

Find the derivative of a function

Let's understand the attachments of this function. Let's try to calculate the expression using the experimental value. How would we count on a calculator?

First you need to find , which means the arcsine is the deepest embedding:

This arcsine of one should then be squared:

And finally, we raise seven to a power:

That is, in this example we have three different functions and two embeddings, with the innermost function being the arcsine and the outermost function being the exponential function.

Let's start deciding

According to the rule First you need to take the derivative of the outer function. We look at the table of derivatives and find the derivative of the exponential function: The only difference is that instead of “x” we have complex expression, which does not negate the validity of this formula. So, the result of applying the rule for differentiating a complex function next.

Proof and derivation of the formulas for the derivative of the exponential (e to the x power) and the exponential function (a to the x power). Examples of calculating derivatives of e^2x, e^3x and e^nx. Formulas for derivatives of higher orders.

The derivative of an exponent is equal to the exponent itself (the derivative of e to the x power is equal to e to the x power):
(1) (e x )′ = e x.

The derivative of an exponential function with a base a is equal to the function itself multiplied by the natural logarithm of a:
(2) .

Derivation of the formula for the derivative of the exponential, e to the x power

An exponential is an exponential function whose base is equal to the number e, which is the following limit:
.
Here it can be both natural and real number. Next, we derive formula (1) for the derivative of the exponential.

Derivation of the exponential derivative formula

Consider the exponential, e to the x power:
y = e x .
This function is defined for everyone. Let's find its derivative with respect to the variable x. By definition, the derivative is the following limit:
(3) .

Let's transform this expression to reduce it to the known ones mathematical properties and rules. To do this we need the following facts:
A) Exponent property:
(4) ;
B) Property of logarithm:
(5) ;
IN) Continuity of the logarithm and the property of limits for a continuous function:
(6) .
Here is a function that has a limit and this limit is positive.
G) The meaning of the second remarkable limit:
(7) .

Let's apply these facts to our limit (3). We use property (4):
;
.

Let's make a substitution. Then ; .
Due to the continuity of the exponential,
.
Therefore, when , . As a result we get:
.

Let's make a substitution. Then . At , . And we have:
.

Let's apply the logarithm property (5):
. Then
.

Let us apply property (6). Since there is a positive limit and the logarithm is continuous, then:
.
Here we also used the second remarkable limit(7). Then
.

Thus, we obtained formula (1) for the derivative of the exponential.

Derivation of the formula for the derivative of an exponential function

Now we derive formula (2) for the derivative of the exponential function with a base of degree a. We believe that and . Then the exponential function
(8)
Defined for everyone.

Let's transform formula (8). For this we will use properties of the exponential function and logarithm.
;
.
So, we transformed formula (8) to the following form:
.

Higher order derivatives of e to the x power

Now let's find derivatives of higher orders. Let's look at the exponent first:
(14) .
(1) .

We see that the derivative of function (14) is equal to function (14) itself. Differentiating (1), we obtain derivatives of the second and third order:
;
.

This shows that the nth order derivative is also equal to the original function:
.

Higher order derivatives of the exponential function

Now consider an exponential function with a base of degree a:
.
We found its first-order derivative:
(15) .

Differentiating (15), we obtain derivatives of the second and third order:
;
.

We see that each differentiation leads to the multiplication of the original function by . Therefore, the nth order derivative has the following form:
.

Definition of power-exponential function. Deriving a formula for calculating its derivative. Examples of calculating derivatives of power-exponential functions are analyzed in detail.

Power-exponential function is a function that has the form of a power function
y = u v ,
in which the base u and the exponent v are some functions of the variable x:
u = u (x); v = v (x).
This function is also called exponential or .

Note that the power-exponential function can be represented in exponential form:
.
Therefore it is also called complex exponential function.

Calculation using logarithmic derivative

Let's find the derivative of the power-exponential function
(2) ,
where and are functions of the variable.
To do this, we logarithm equation (2), using the property of the logarithm:
.
Differentiate with respect to the variable x:
(3) .
We apply rules for differentiating complex functions and works:
;
.

We substitute in (3):
.
From here
.

So, we found the derivative of the power-exponential function:
(1) .
If the exponent is constant, then . Then the derivative is equal to the derivative of a complex power function:
.
If the base of the degree is constant, then . Then the derivative is equal to the derivative of a complex exponential function:
.
When and are functions of x, then the derivative of the power-exponential function is equal to the sum of the derivatives of the complex power and exponential functions.

Calculation of the derivative by reduction to a complex exponential function

Now let's find the derivative of the power-exponential function
(2) ,
presenting it as a complex exponential function:
(4) .

Let's differentiate the product:
.
We apply the rule for finding the derivative of a complex function:

.
And we again got formula (1).

Example 1

Find the derivative of the following function:
.

Solution

We calculate using the logarithmic derivative. Let's logarithm the original function:
(A1.1) .

From the table of derivatives we find:
;
.
Using the product derivative formula, we have:
.
We differentiate (A1.1):
.
Because the
,
That
.