First derivative If the derivative of a function is positive (negative) in some interval, then the function in this interval is monotonically increasing (monotonically decreasing). If the derivative function is positive (negative) in some interval, then the function in this interval is monotonically increasing (monotonically decreasing). Further
Definition A curve is called convex at a point if in some neighborhood of this point it is located under its tangent at a point A curve is called convex at a point if in some neighborhood of this point it is located under its tangent at a point point, it is located above its tangent at a point A curve is called concave at a point if, in some neighborhood of this point, it is located above its tangent at a point
Sign of concavity and convexity If the second derivative of a function in a given interval is positive, then the curve is concave in this interval, and if it is negative, it is convex in this interval. If the second derivative of a function in a given interval is positive, then the curve is concave in this interval, and if it is negative, it is convex in this interval. Definition
Plan for studying the function and constructing its graph 1. Find the domain of the function and determine the break points, if any 1. Find the domain of the function and determine the break points, if any 2. Find out if the function is even or odd; check its periodicity 2. Find out if the function is even or odd; check its periodicity 3. Determine the intersection points of the function graph with the coordinate axes 3. Determine the intersection points of the function graph with the coordinate axes 4. Find the critical points of the 1st kind 4. Find the critical points of the 1st kind 5. Determine the intervals of monotonicity and extrema of the function 5. Determine monotonicity intervals and extrema of the function 6. Determine the intervals of convexity and concavity and find inflection points 6. Determine the intervals of convexity and concavity and find inflection points 7. Using the results of the study, connect the obtained points of a smooth curve 7. Using the results of the study, connect the obtained points of a smooth curve Exit
What is a derivative?
Definition and meaning of the derivative of a function
Many will be surprised at the unexpected location of this article in my author's course on the derivative of a function of one variable and its applications. After all, as it was from school: a standard textbook, first of all, gives a definition of a derivative, its geometric, mechanical meaning. Next, students find derivatives of functions by definition, and, in fact, only then the differentiation technique is perfected using derivative tables.
But from my point of view, the following approach is more pragmatic: first of all, it is advisable to UNDERSTAND WELL function limit, and especially infinitesimals. The fact is that the definition of the derivative is based on the concept of a limit, which is poorly considered in the school course. That is why a significant part of young consumers of granite knowledge poorly penetrate into the very essence of the derivative. Thus, if you are not well versed in differential calculus, or the wise brain has successfully rid itself of this baggage over the years, please start with function limits. At the same time master / remember their decision.
The same practical sense suggests that it is profitable first learn to find derivatives, including derivatives of complex functions. Theory is a theory, but, as they say, you always want to differentiate. In this regard, it is better to work out the listed basic lessons, and maybe become differentiation master without even realizing the essence of their actions.
I recommend starting the materials on this page after reading the article. The simplest problems with a derivative, where, in particular, the problem of the tangent to the graph of a function is considered. But it can be delayed. The fact is that many applications of the derivative do not require understanding it, and it is not surprising that the theoretical lesson appeared quite late - when I needed to explain finding intervals of increase/decrease and extremums functions. Moreover, he was in the subject for quite a long time " Functions and Graphs”, until I decided to put it in earlier.
Therefore, dear teapots, do not rush to absorb the essence of the derivative, like hungry animals, because the saturation will be tasteless and incomplete.
The concept of increasing, decreasing, maximum, minimum of a function
Many tutorials lead to the concept of a derivative with the help of some practical problems, and I also came up with an interesting example. Imagine that we have to travel to a city that can be reached in different ways. We immediately discard the curved winding paths, and we will consider only straight lines. However, straight-line directions are also different: you can get to the city along a flat autobahn. Or on a hilly highway - up and down, up and down. Another road goes only uphill, and another one goes downhill all the time. Thrill-seekers will choose a route through the gorge with a steep cliff and a steep ascent.
But whatever your preferences, it is desirable to know the area, or at least have a topographical map of it. What if there is no such information? After all, you can choose, for example, a flat path, but as a result, stumble upon a ski slope with funny Finns. Not the fact that the navigator and even a satellite image will give reliable data. Therefore, it would be nice to formalize the relief of the path by means of mathematics.
Consider some road (side view): 
Just in case, I remind you of an elementary fact: the journey takes place from left to right. For simplicity, we assume that the function continuous in the area under consideration.
What are the features of this chart?
At intervals 
function increases, that is, each of its next value more the previous one. Roughly speaking, the schedule goes down up(we climb the hill). And on the interval the function decreasing- each next value smaller the previous one, and our schedule goes top down(going down the slope).
Let's also pay attention to special points. At the point we reach maximum, i.e exist such a section of the path on which the value will be the largest (highest). At the same point, minimum, and exist such its neighborhood, in which the value is the smallest (lowest).
More rigorous terminology and definitions will be considered in the lesson. about the extrema of the function, but for now let's study one more important feature: on the intervals 
the function is increasing, but it is increasing at different speeds. And the first thing that catches your eye is that the chart soars up on the interval much more cool than on the interval. Is it possible to measure the steepness of the road using mathematical tools?
Function change rate
The idea is this: take some value (read "delta x"), which we will call argument increment, and let's start "trying it on" to various points of our path: 
1) Let's look at the leftmost point: bypassing the distance , we climb the slope to a height (green line). The value is called function increment, and in this case this increment is positive (the difference of values along the axis is greater than zero). Let's make the ratio , which will be the measure of the steepness of our road. Obviously, is a very specific number, and since both increments are positive, then .
Attention! Designation are ONE symbol, that is, you cannot “tear off” the “delta” from the “x” and consider these letters separately. Of course, the comment also applies to the function's increment symbol.
Let's explore the nature of the resulting fraction more meaningful. Suppose initially we are at a height of 20 meters (in the left black dot). Having overcome the distance of meters (left red line), we will be at a height of 60 meters. Then the increment of the function will be 
meters (green line) and: . Thus, on every meter this section of the road height increases average by 4 meters…did you forget your climbing equipment? =) In other words, the constructed ratio characterizes the AVERAGE RATE OF CHANGE (in this case, growth) of the function.
Note : The numerical values of the example in question correspond to the proportions of the drawing only approximately.
2) Now let's go the same distance from the rightmost black dot. Here the rise is more gentle, so the increment (crimson line) is relatively small, and the ratio compared to the previous case will be quite modest. Relatively speaking, 
meters and function growth rate is . That is, here for every meter of the road there is average half a meter up.
3) A little adventure on the mountainside. Let's look at the top black dot located on the y-axis. Let's assume that this is a mark of 50 meters. Again we overcome the distance, as a result of which we find ourselves lower - at the level of 30 meters. Since the movement has been made top down(in the "opposite" direction of the axis), then the final the increment of the function (height) will be negative: 
meters (brown line in the drawing). And in this case we are talking about decay rate features: 
, that is, for each meter of the path of this section, the height decreases average by 2 meters. Take care of clothes on the fifth point.
Now let's ask the question: what is the best value of "measuring standard" to use? It is clear that 10 meters is very rough. A good dozen bumps can easily fit on them. Why are there bumps, there may be a deep gorge below, and after a few meters - its other side with a further steep ascent. Thus, with a ten-meter one, we will not get an intelligible characteristic of such sections of the path through the ratio.
From the above discussion, the following conclusion follows: the smaller the value, the more accurately we will describe the relief of the road. Moreover, the following facts are true:
– For any lifting points 
you can choose a value (albeit a very small one) that fits within the boundaries of one or another rise. And this means that the corresponding height increment will be guaranteed to be positive, and the inequality will correctly indicate the growth of the function at each point of these intervals.
- Likewise, for any slope point, there is a value that will fit completely on this slope. Therefore, the corresponding increase in height is unequivocally negative, and the inequality will correctly show the decrease in the function at each point of the given interval.
– Of particular interest is the case when the rate of change of the function is zero: . First, a zero height increment () is a sign of an even path. And secondly, there are other curious situations, examples of which you see in the figure. Imagine that fate has taken us to the very top of a hill with soaring eagles or the bottom of a ravine with croaking frogs. If you take a small step in any direction, then the change in height will be negligible, and we can say that the rate of change of the function is actually zero. The same pattern is observed at points.
Thus, we have approached an amazing opportunity to perfectly accurately characterize the rate of change of a function. After all, mathematical analysis allows us to direct the increment of the argument to zero: that is, to make it infinitesimal.
As a result, another logical question arises: is it possible to find for the road and its schedule another function, which would tell us about all flats, uphills, downhills, peaks, lowlands, as well as the rate of increase / decrease at each point of the path?
What is a derivative? Definition of a derivative.
The geometric meaning of the derivative and differential
Please read thoughtfully and not too quickly - the material is simple and accessible to everyone! It's okay if in some places something seems not very clear, you can always return to the article later. I will say more, it is useful to study the theory several times in order to qualitatively understand all the points (the advice is especially relevant for “technical” students, for whom higher mathematics plays a significant role in the educational process).
Naturally, in the very definition of the derivative at a point, we will replace it with:
What have we come to? And we came to the conclusion that for a function according to the law 
is aligned other function, which is called derivative function(or simply derivative).
The derivative characterizes rate of change functions . How? The thought goes like a red thread from the very beginning of the article. Consider some point domains functions . Let the function be differentiable at a given point. Then:
1) If , then the function increases at the point . And obviously there is interval(even if very small) containing the point at which the function grows, and its graph goes “from bottom to top”.
2) If , then the function decreases at the point . And there is an interval containing a point at which the function decreases (the graph goes “from top to bottom”).
3) If , then infinitely close near the point, the function keeps its speed constant. This happens, as noted, for a function-constant and at critical points of the function, in particular at the minimum and maximum points.
Some semantics. What does the verb "differentiate" mean in a broad sense? To differentiate means to single out a feature. Differentiating the function , we "select" the rate of its change in the form of a derivative of the function . And what, by the way, is meant by the word "derivative"? Function happened from the function.
The terms very successfully interpret the mechanical meaning of the derivative
:
Let's consider the law of change of the body's coordinates, which depends on time, and the function of the speed of motion of the given body. The function characterizes the rate of change of the body coordinate, therefore it is the first derivative of the function with respect to time: . If the concept of “body motion” did not exist in nature, then there would not exist derivative concept of "velocity".
The acceleration of the body is the rate of change of speed, therefore: . If the original concepts of “body movement” and “body movement speed” did not exist in nature, then there would be no derivative the concept of acceleration of a body.
If we follow the definition, then the derivative of a function at a point is the limit of the increment ratio of the function Δ y to the increment of the argument Δ x:
Everything seems to be clear. But try to calculate by this formula, say, the derivative of the function f(x) = x 2 + (2x+ 3) · e x sin x. If you do everything by definition, then after a couple of pages of calculations you will simply fall asleep. Therefore, there are simpler and more effective ways.
To begin with, we note that the so-called elementary functions can be distinguished from the whole variety of functions. These are relatively simple expressions, the derivatives of which have long been calculated and entered in the table. Such functions are easy enough to remember, along with their derivatives.
Derivatives of elementary functions
Elementary functions are everything listed below. The derivatives of these functions must be known by heart. Moreover, it is not difficult to memorize them - that's why they are elementary.
So, the derivatives of elementary functions:
| Name | Function | Derivative |
| Constant | f(x) = C, C ∈ R | 0 (yes, yes, zero!) |
| Degree with rational exponent | f(x) = x n | n · x n − 1 |
| Sinus | f(x) = sin x | cos x |
| Cosine | f(x) = cos x | − sin x(minus sine) |
| Tangent | f(x) = tg x | 1/cos 2 x |
| Cotangent | f(x) = ctg x | − 1/sin2 x |
| natural logarithm | f(x) = log x | 1/x |
| Arbitrary logarithm | f(x) = log a x | 1/(x ln a) |
| Exponential function | f(x) = e x | e x(nothing changed) |
If an elementary function is multiplied by an arbitrary constant, then the derivative of the new function is also easily calculated:
(C · f)’ = C · f ’.
In general, constants can be taken out of the sign of the derivative. For example:
(2x 3)' = 2 ( x 3)' = 2 3 x 2 = 6x 2 .
Obviously, elementary functions can be added to each other, multiplied, divided, and much more. This is how new functions will appear, no longer very elementary, but also differentiable according to certain rules. These rules are discussed below.
Derivative of sum and difference
Let the functions f(x) and g(x), whose derivatives are known to us. For example, you can take the elementary functions discussed above. Then you can find the derivative of the sum and difference of these functions:
- (f + g)’ = f ’ + g ’
- (f − g)’ = f ’ − g ’
So, the derivative of the sum (difference) of two functions is equal to the sum (difference) of the derivatives. There may be more terms. For example, ( f + g + h)’ = f ’ + g ’ + h ’.
Strictly speaking, there is no concept of "subtraction" in algebra. There is a concept of "negative element". Therefore, the difference f − g can be rewritten as a sum f+ (−1) g, and then only one formula remains - the derivative of the sum.
f(x) = x 2 + sinx; g(x) = x 4 + 2x 2 − 3.
Function f(x) is the sum of two elementary functions, so:
f ’(x) = (x 2+ sin x)’ = (x 2)' + (sin x)’ = 2x+ cosx;
We argue similarly for the function g(x). Only there are already three terms (from the point of view of algebra):
g ’(x) = (x 4 + 2x 2 − 3)’ = (x 4 + 2x 2 + (−3))’ = (x 4)’ + (2x 2)’ + (−3)’ = 4x 3 + 4x + 0 = 4x · ( x 2 + 1).
Answer:
f ’(x) = 2x+ cosx;
g ’(x) = 4x · ( x
2 + 1).
Derivative of a product
Mathematics is a logical science, so many people believe that if the derivative of the sum is equal to the sum of the derivatives, then the derivative of the product strike"\u003e equal to the product of derivatives. But figs to you! The derivative of the product is calculated using a completely different formula. Namely:
(f · g) ’ = f ’ · g + f · g ’
The formula is simple, but often forgotten. And not only schoolchildren, but also students. The result is incorrectly solved problems.
Task. Find derivatives of functions: f(x) = x 3 cosx; g(x) = (x 2 + 7x− 7) · e x .
Function f(x) is a product of two elementary functions, so everything is simple:
f ’(x) = (x 3 cos x)’ = (x 3)' cos x + x 3 (cos x)’ = 3x 2 cos x + x 3 (−sin x) = x 2 (3cos x − x sin x)
Function g(x) the first multiplier is a little more complicated, but the general scheme does not change from this. Obviously, the first multiplier of the function g(x) is a polynomial, and its derivative is the derivative of the sum. We have:
g ’(x) = ((x 2 + 7x− 7) · e x)’ = (x 2 + 7x− 7)' · e x + (x 2 + 7x− 7) ( e x)’ = (2x+ 7) · e x + (x 2 + 7x− 7) · e x = e x(2 x + 7 + x 2 + 7x −7) = (x 2 + 9x) · e x = x(x+ 9) · e x .
Answer:
f ’(x) = x 2 (3cos x − x sin x);
g ’(x) = x(x+ 9) · e
x
.
Note that in the last step, the derivative is factorized. Formally, this is not necessary, but most derivatives are not calculated on their own, but to explore the function. This means that further the derivative will be equated to zero, its signs will be found out, and so on. For such a case, it is better to have an expression decomposed into factors.
If there are two functions f(x) and g(x), and g(x) ≠ 0 on the set of interest to us, we can define a new function h(x) = f(x)/g(x). For such a function, you can also find the derivative:
Not weak, right? Where did the minus come from? Why g 2? But like this! This is one of the most complex formulas - you can’t figure it out without a bottle. Therefore, it is better to study it with specific examples.
Task. Find derivatives of functions:
There are elementary functions in the numerator and denominator of each fraction, so all we need is the formula for the derivative of the quotient:
By tradition, we factor the numerator into factors - this will greatly simplify the answer:
A complex function is not necessarily a formula half a kilometer long. For example, it suffices to take the function f(x) = sin x and replace the variable x, say, on x 2+ln x. It turns out f(x) = sin ( x 2+ln x) is a complex function. She also has a derivative, but it will not work to find it according to the rules discussed above.
How to be? In such cases, the replacement of a variable and the formula for the derivative of a complex function help:
f ’(x) = f ’(t) · t', if x is replaced by t(x).
As a rule, the situation with the understanding of this formula is even more sad than with the derivative of the quotient. Therefore, it is also better to explain it with specific examples, with a detailed description of each step.
Task. Find derivatives of functions: f(x) = e 2x + 3 ; g(x) = sin ( x 2+ln x)
Note that if in the function f(x) instead of expression 2 x+ 3 will be easy x, then we get an elementary function f(x) = e x. Therefore, we make a substitution: let 2 x + 3 = t, f(x) = f(t) = e t. We are looking for the derivative of a complex function by the formula:
f ’(x) = f ’(t) · t ’ = (e t)’ · t ’ = e t · t ’
And now - attention! Performing a reverse substitution: t = 2x+ 3. We get:
f ’(x) = e t · t ’ = e 2x+ 3 (2 x + 3)’ = e 2x+ 3 2 = 2 e 2x + 3
Now let's look at the function g(x). Obviously needs to be replaced. x 2+ln x = t. We have:
g ’(x) = g ’(t) · t' = (sin t)’ · t' = cos t · t ’
Reverse replacement: t = x 2+ln x. Then:
g ’(x) = cos( x 2+ln x) · ( x 2+ln x)' = cos ( x 2+ln x) · (2 x + 1/x).
That's all! As can be seen from the last expression, the whole problem has been reduced to calculating the derivative of the sum.
Answer:
f ’(x) = 2 e
2x + 3 ;
g ’(x) = (2x + 1/x) cos ( x 2+ln x).
Very often in my lessons, instead of the term “derivative”, I use the word “stroke”. For example, the stroke of the sum is equal to the sum of the strokes. Is that clearer? Well, that's good.
Thus, the calculation of the derivative comes down to getting rid of these very strokes according to the rules discussed above. As a final example, let's return to the derivative power with a rational exponent:
(x n)’ = n · x n − 1
Few know that in the role n may well be a fractional number. For example, the root is x 0.5 . But what if there is something tricky under the root? Again, a complex function will turn out - they like to give such constructions in tests and exams.
Task. Find the derivative of a function:
First, let's rewrite the root as a power with a rational exponent:
f(x) = (x 2 + 8x − 7) 0,5 .
Now we make a substitution: let x 2 + 8x − 7 = t. We find the derivative by the formula:
f ’(x) = f ’(t) · t ’ = (t 0.5)' t' = 0.5 t−0.5 t ’.
We make a reverse substitution: t = x 2 + 8x− 7. We have:
f ’(x) = 0.5 ( x 2 + 8x− 7) −0.5 ( x 2 + 8x− 7)' = 0.5 (2 x+ 8) ( x 2 + 8x − 7) −0,5 .
Finally, back to the roots:
Definition. Let the function \(y = f(x) \) be defined in some interval containing the point \(x_0 \) inside. Let's increment \(\Delta x \) to the argument so as not to leave this interval. Find the corresponding increment of the function \(\Delta y \) (when passing from the point \(x_0 \) to the point \(x_0 + \Delta x \)) and compose the relation \(\frac(\Delta y)(\Delta x) \). If there is a limit of this relation at \(\Delta x \rightarrow 0 \), then the indicated limit is called derivative function\(y=f(x) \) at the point \(x_0 \) and denote \(f"(x_0) \).
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x_0) $$
The symbol y is often used to denote the derivative. Note that y" = f(x) is a new function, but naturally associated with the function y = f(x), defined at all points x at which the above limit exists . This function is called like this: derivative of the function y \u003d f (x).
The geometric meaning of the derivative consists of the following. If a tangent that is not parallel to the y axis can be drawn to the graph of the function y \u003d f (x) at a point with the abscissa x \u003d a, then f (a) expresses the slope of the tangent:
\(k = f"(a)\)
Since \(k = tg(a) \), the equality \(f"(a) = tg(a) \) is true.
And now we interpret the definition of the derivative in terms of approximate equalities. Let the function \(y = f(x) \) have a derivative at a particular point \(x \):
$$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) = f"(x) $$
This means that near the point x, the approximate equality \(\frac(\Delta y)(\Delta x) \approx f"(x) \), i.e. \(\Delta y \approx f"(x) \cdot\Deltax\). The meaningful meaning of the obtained approximate equality is as follows: the increment of the function is “almost proportional” to the increment of the argument, and the coefficient of proportionality is the value of the derivative at a given point x. For example, for the function \(y = x^2 \) the approximate equality \(\Delta y \approx 2x \cdot \Delta x \) is true. If we carefully analyze the definition of the derivative, we will find that it contains an algorithm for finding it.
Let's formulate it.
How to find the derivative of the function y \u003d f (x) ?
1. Fix value \(x \), find \(f(x) \)
2. Increment \(x \) argument \(\Delta x \), move to a new point \(x+ \Delta x \), find \(f(x+ \Delta x) \)
3. Find the function increment: \(\Delta y = f(x + \Delta x) - f(x) \)
4. Compose the relation \(\frac(\Delta y)(\Delta x) \)
5. Calculate $$ \lim_(\Delta x \to 0) \frac(\Delta y)(\Delta x) $$
This limit is the derivative of the function at x.
If the function y = f(x) has a derivative at the point x, then it is called differentiable at the point x. The procedure for finding the derivative of the function y \u003d f (x) is called differentiation functions y = f(x).
Let us discuss the following question: how are the continuity and differentiability of a function at a point related?
Let the function y = f(x) be differentiable at the point x. Then a tangent can be drawn to the graph of the function at the point M (x; f (x)) and, recall, the slope of the tangent is equal to f "(x). Such a graph cannot "break" at the point M, i.e., the function must be continuous at x.
It was reasoning "on the fingers". Let us present a more rigorous argument. If the function y = f(x) is differentiable at the point x, then the approximate equality \(\Delta y \approx f"(x) \cdot \Delta x \) holds. zero, then \(\Delta y \) will also tend to zero, and this is the condition for the continuity of the function at a point.
So, if a function is differentiable at a point x, then it is also continuous at that point.
The converse is not true. For example: function y = |x| is continuous everywhere, in particular at the point x = 0, but the tangent to the graph of the function at the “joint point” (0; 0) does not exist. If at some point it is impossible to draw a tangent to the function graph, then there is no derivative at this point.
One more example. The function \(y=\sqrt(x) \) is continuous on the entire number line, including at the point x = 0. And the tangent to the graph of the function exists at any point, including at the point x = 0. But at this point the tangent coincides with the y-axis, that is, it is perpendicular to the abscissa axis, its equation has the form x \u003d 0. There is no slope for such a straight line, which means that \ (f "(0) \) does not exist either
So, we got acquainted with a new property of a function - differentiability. How can you tell if a function is differentiable from the graph of a function?
The answer is actually given above. If at some point a tangent can be drawn to the graph of a function that is not perpendicular to the x-axis, then at this point the function is differentiable. If at some point the tangent to the graph of the function does not exist or it is perpendicular to the x-axis, then at this point the function is not differentiable.
Differentiation rules
The operation of finding the derivative is called differentiation. When performing this operation, you often have to work with quotients, sums, products of functions, as well as with "functions of functions", that is, complex functions. Based on the definition of the derivative, we can derive differentiation rules that facilitate this work. If C is a constant number and f=f(x), g=g(x) are some differentiable functions, then the following are true differentiation rules:
$$ f"_x(g(x)) = f"_g \cdot g"_x $$
Table of derivatives of some functions
$$ \left(\frac(1)(x) \right) " = -\frac(1)(x^2) $$ $$ (\sqrt(x)) " = \frac(1)(2\ sqrt(x)) $$ $$ \left(x^a \right) " = a x^(a-1) $$ $$ \left(a^x \right) " = a^x \cdot \ln a $$ $$ \left(e^x \right) " = e^x $$ $$ (\ln x)" = \frac(1)(x) $$ $$ (\log_a x)" = \frac (1)(x\ln a) $$ $$ (\sin x)" = \cos x $$ $$ (\cos x)" = -\sin x $$ $$ (\text(tg) x) " = \frac(1)(\cos^2 x) $$ $$ (\text(ctg) x)" = -\frac(1)(\sin^2 x) $$ $$ (\arcsin x) " = \frac(1)(\sqrt(1-x^2)) $$ $$ (\arccos x)" = \frac(-1)(\sqrt(1-x^2)) $$ $$ (\text(arctg) x)" = \frac(1)(1+x^2) $$ $$ (\text(arctg) x)" = \frac(-1)(1+x^2) $ $
The derivative of a function is one of the most difficult topics in the school curriculum. Not every graduate will answer the question of what a derivative is.
This article simply and clearly explains what a derivative is and why it is needed.. We will not now strive for mathematical rigor of presentation. The most important thing is to understand the meaning.
Let's remember the definition:
The derivative is the rate of change of the function.
The figure shows graphs of three functions. Which one do you think grows the fastest?
The answer is obvious - the third. It has the highest rate of change, that is, the largest derivative.
Here is another example.
Kostya, Grisha and Matvey got jobs at the same time. Let's see how their income changed during the year:
You can see everything on the chart right away, right? Kostya's income has more than doubled in six months. And Grisha's income also increased, but just a little bit. And Matthew's income decreased to zero. The starting conditions are the same, but the rate of change of the function, i.e. derivative, - different. As for Matvey, the derivative of his income is generally negative.
Intuitively, we can easily estimate the rate of change of a function. But how do we do it?
What we are really looking at is how steeply the graph of the function goes up (or down). In other words, how fast y changes with x. Obviously, the same function at different points can have a different value of the derivative - that is, it can change faster or slower.
The derivative of a function is denoted by .
Let's show how to find using the graph.
A graph of some function is drawn. Take a point on it with an abscissa. Draw a tangent to the graph of the function at this point. We want to evaluate how steeply the graph of the function goes up. A handy value for this is tangent of the slope of the tangent.
The derivative of a function at a point is equal to the tangent of the slope of the tangent drawn to the graph of the function at that point.
Please note - as the angle of inclination of the tangent, we take the angle between the tangent and the positive direction of the axis.
Sometimes students ask what is the tangent to the graph of a function. This is a straight line that has the only common point with the graph in this section, moreover, as shown in our figure. It looks like a tangent to a circle.
Let's find . We remember that the tangent of an acute angle in a right triangle is equal to the ratio of the opposite leg to the adjacent one. From triangle:
We found the derivative using the graph without even knowing the formula of the function. Such tasks are often found in the exam in mathematics under the number.
There is another important correlation. Recall that the straight line is given by the equation
The quantity in this equation is called slope of a straight line. It is equal to the tangent of the angle of inclination of the straight line to the axis.
.
We get that
Let's remember this formula. It expresses the geometric meaning of the derivative.
The derivative of a function at a point is equal to the slope of the tangent drawn to the graph of the function at that point.
In other words, the derivative is equal to the tangent of the slope of the tangent.
We have already said that the same function can have different derivatives at different points. Let's see how the derivative is related to the behavior of the function.
Let's draw a graph of some function. Let this function increase in some areas, and decrease in others, and at different rates. And let this function have maximum and minimum points.
At a point, the function is increasing. The tangent to the graph, drawn at the point, forms an acute angle; with positive axis direction. So the derivative is positive at the point.
At the point, our function is decreasing. The tangent at this point forms an obtuse angle; with positive axis direction. Since the tangent of an obtuse angle is negative, the derivative at the point is negative.
Here's what happens:
If a function is increasing, its derivative is positive.
If it decreases, its derivative is negative.
And what will happen at the maximum and minimum points? We see that at (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the slope of the tangent at these points is zero, and the derivative is also zero.
The point is the maximum point. At this point, the increase of the function is replaced by a decrease. Consequently, the sign of the derivative changes at the point from "plus" to "minus".
At the point - the minimum point - the derivative is also equal to zero, but its sign changes from "minus" to "plus".
Conclusion: with the help of the derivative, you can find out everything that interests us about the behavior of the function.
If the derivative is positive, then the function is increasing.
If the derivative is negative, then the function is decreasing.
At the maximum point, the derivative is zero and changes sign from plus to minus.
At the minimum point, the derivative is also zero and changes sign from minus to plus.
We write these findings in the form of a table:
| increases | maximum point | decreasing | minimum point | increases | |
| + | 0 | - | 0 | + |
Let's make two small clarifications. You will need one of them when solving the problem. Another - in the first year, with a more serious study of functions and derivatives.
A case is possible when the derivative of a function at some point is equal to zero, but the function has neither a maximum nor a minimum at this point. This so-called :
At a point, the tangent to the graph is horizontal and the derivative is zero. However, before the point the function increased - and after the point it continues to increase. The sign of the derivative does not change - it has remained positive as it was.
It also happens that at the point of maximum or minimum, the derivative does not exist. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point.
But how to find the derivative if the function is given not by a graph, but by a formula? In this case, it applies
