Table of derivatives of elementary algebraic functions with conclusions. Find the derivative: algorithm and examples of solutions

Very easy to remember.

Well, let’s not go far, let’s immediately consider the inverse function. Which function is the inverse of the 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:

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

Answers: The exponential and natural logarithm are uniquely simple functions from a derivative perspective. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.

Rules of differentiation

Rules of what? Again a 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:

1. at a point;
2. at a point;
3. at a point;
4. at the point.

Solutions:

1. (the derivative is the same at all points, since it is a linear function, remember?);

Derivative of the product

Everything is similar here: let’s introduce a new function and find its increment:

Derivative:

Examples:

1. Find the derivatives of the functions and;
2. 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:

To do this, we will use a simple rule: . Then:

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

Did it work?

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 a simpler form. Therefore, we leave it in this form in the answer.

Note that here is the quotient of two functions, so we apply the corresponding differentiation rule:

In this example, the product of two functions:

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 are almost never found in the Unified State Examination, but it will not be superfluous 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 of a 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.

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

For our example, .

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. An important feature of complex functions: when the order of actions changes, the function changes.

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

1. 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: .
2. Internal: ; external: .
   Examination: .
3. Internal: ; external: .
   Examination: .
4. Internal: ; external: .
   Examination: .
5. 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 (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 order 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:

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

Without proof, we present the formulas for the derivatives of the basic elementary functions:

1. Power function: (x n)` =nx n -1 .

2. Exponential function: (a x)` =a x lna(in particular, (e x)` = e x).

3. Logarithmic function: (in particular, (lnx)` = 1/x).

4. Trigonometric functions:

(cosх)` = -sinx

(tgх)` = 1/cos 2 x

(ctgх)` = -1/sin 2 x

5. Inverse trigonometric functions:

It can be proven that to differentiate a power-exponential function, it is necessary to use the formula for the derivative of a complex function twice, namely, differentiate it both as a complex power function and as a complex exponential function, and add the results: (f(x)  (x))` =(x)*f(x)  (x)-1 *f(x)` +f(x)  (x) *lnf(x)*(x)`.

Higher order derivatives

Since the derivative of a function is itself a function, it can also have a derivative. The concept of a derivative, which was discussed above, refers to a first-order derivative.

Derivativen-th order is called the derivative of the (n- 1)th order derivative. For example, f``(x) = (f`(x))` - second order derivative (or second derivative), f```(x) = (f``(x))` - third order derivative (or third derivative), etc. Sometimes Roman Arabic numerals in parentheses are used to denote higher-order derivatives, for example, f (5) (x) or f (V) (x) for a fifth-order derivative.

The physical meaning of derivatives of higher orders is determined in the same way as for the first derivative: each of them represents the rate of change of the derivative of the previous order. For example, the second derivative represents the rate of change of the first, i.e. speed speed. For rectilinear motion, it means the acceleration of a point at a moment in time.

Elasticity function

Elasticity function E x (y) is the limit of the ratio of the relative increment of the function y to the relative increment of the argument x as the latter tends to zero:
.

The elasticity of a function shows approximately how many percent the function y = f(x) will change when the independent variable x changes by 1%.

In an economic sense, the difference between this indicator and the derivative is that the derivative has units of measurement, and therefore its value depends on the units in which the variables are measured. For example, if the dependence of production volume on time is expressed in tons and months, respectively, then the derivative will show the marginal increase in volume in tons per month; if we measure these indicators, say, in kilograms and days, then both the function itself and its derivative will be different. Elasticity is essentially a dimensionless quantity (measured in percentages or shares) and therefore does not depend on the scale of indicators.

Basic theorems on differentiable functions and their applications

Fermat's theorem. If a function differentiable on an interval reaches its greatest or minimum value at an internal point of this interval, then the derivative of the function at this point is zero.

No proof.

The geometric meaning of Fermat's theorem is that at the point of the largest or smallest value achieved inside the interval, the tangent to the graph of the function is parallel to the abscissa axis (Figure 3.3).

Rolle's theorem. Let the function y =f(x) satisfy the following conditions:

2) differentiable on the interval (a, b);

3) at the ends of the segment takes equal values, i.e. f(a) =f(b).

Then there is at least one point inside the segment at which the derivative of the function is equal to zero.

No proof.

The geometric meaning of Rolle's theorem is that there is at least one point at which the tangent to the graph of the function will be parallel to the abscissa axis (for example, in Figure 3.4 there are two such points).

If f(a) =f(b) = 0, then Rolle’s theorem can be formulated differently: between two consecutive zeros of the differentiable function there is at least one zero of the derivative.

Rolle's theorem is a special case of Lagrange's theorem.

Lagrange's theorem. Let the function y =f(x) satisfy the following conditions:

1) continuous on the interval [a, b];

2) differentiable on the interval (a, b).

Then inside the segment there is at least one such point c, at which the derivative is equal to the quotient of the function increment divided by the argument increment on this segment:
.

No proof.

To understand the physical meaning of Lagrange’s theorem, we note that
is nothing more than the average rate of change of the function over the entire interval [a, b]. Thus, the theorem states that inside the segment there is at least one point at which the “instantaneous” rate of change of the function is equal to the average rate of its change over the entire segment.

The geometric meaning of Lagrange's theorem is illustrated in Figure 3.5. Note that the expression
represents the angular coefficient of the straight line on which the chord AB lies. The theorem states that on the graph of a function there will be at least one point at which the tangent to it will be parallel to this chord (that is, the slope of the tangent - the derivative - will be the same).

Corollary: if the derivative of a function is equal to zero on a certain interval, then the function is identically constant on this interval.

In fact, let us take the interval . According to Lagrange's theorem, in this interval there is a point c for which
. Hence f(a) – f(x) = f `(с)(a – x) = 0; f(x) = f(a) = const.

L'Hopital's rule. The limit of the ratio of two infinitesimal or infinitely large functions is equal to the limit of the ratio of their derivatives (finite or infinite), if the latter exists in the indicated sense.

In other words, if there is uncertainty of the form
, That
.

No proof.

The application of L'Hopital's rule to find limits will be discussed in practical classes.

Sufficient condition for an increase (decrease) of a function. If the derivative of a differentiable function is positive (negative) within a certain interval, then the function increases (decreases) on this interval.

Proof. Consider two values ​​x 1 and x 2 from this interval (let x 2 > x 1). By Lagrand's theorem on [x 1, x 2] there is a point c at which
. Hence f(x 2) –f(x 1) =f`(c)(x 2 –x 1). Then for f`(c) > 0 the left side of the inequality is positive, i.e. f(x 2) >f(x 1), and the function is increasing. Whenf`(c)< 0 левая часть неравенства отрицательна, т.е.f(х 2)

The theorem has been proven.

Geometric interpretation of the condition for the monotonicity of a function: if the tangents to the curve in a certain interval are directed at acute angles to the abscissa axis, then the function increases, and if at obtuse angles, then it decreases (see Figure 3.6).

Note: the necessary condition for monotonicity is weaker. If a function increases (decreases) over a certain interval, then the derivative is non-negative (non-positive) on this interval (that is, at individual points the derivative of a monotonic function can be equal to zero).

Prove formulas 3 and 5 yourself.

BASIC RULES OF DIFFERENTIATION

Using the general method of finding the derivative using the limit, one can obtain the simplest differentiation formulas. Let u=u(x),v=v(x)– two differentiable functions of a variable x.

Prove formulas 1 and 2 yourself.

Proof of Formula 3.

Let y = u(x) + v(x). For argument value x+Δ x we have y(x+Δ x)=u(x+Δ x) + v(x+Δ x).

Δ y=y(x+Δ x) – y(x) = u(x+Δ x) + v(x+Δ x) – u(x) – v(x) = Δ u +Δ v.

Hence,

Proof of formula 4.

Let y=u(x)·v(x). Then y(x+Δ x)=u(x+Δ x)· v(x+Δ x), That's why

Δ y=u(x+Δ x)· v(x+Δ x) – u(x)· v(x).

Note that since each of the functions u And v differentiable at the point x, then they are continuous at this point, which means u(x+Δ x)→u(x), v(x+Δ x)→v(x), at Δ x→0.

Therefore we can write

Based on this property, one can obtain a rule for differentiating the product of any number of functions.

Let, for example, y=u·v·w. Then,

y " = u "·( v w) + u·( v·w) " = u "· v·w + u·( v"·w+ v·w ") = u "· v·w + u· v"·w+ u·v·w ".

Proof of formula 5.

Let . Then

In the proof we used the fact that v(x+Δ x)→v(x) at Δ x→0.

Examples.

THEOREM ON THE DERIVATIVE OF COMPLEX FUNCTION

Let y = f(u), A u= u(x). We get the function y depending on the argument x: y = f(u(x)). The last function is called a function of a function or complex function.

Function definition domain y = f(u(x)) is either the entire domain of definition of the function u=u(x) or that part in which the values ​​are determined u, not leaving the domain of definition of the function y= f(u).

The function-from-function operation can be performed not just once, but any number of times.

Let us establish a rule for differentiating a complex function.

Theorem. If the function u= u(x) has at some point x 0 derivative and takes the value at this point u 0 = u(x 0), and the function y=f(u) has at the point u 0 derivative y" u = f "(u 0), then a complex function y = f(u(x)) at the specified point x 0 also has a derivative, which is equal to y" x = f "(u 0)· u "(x 0), where instead of u the expression must be substituted u= u(x).

Thus, the derivative of a complex function is equal to the product of the derivative of a given function with respect to the intermediate argument u to the derivative of the intermediate argument with respect to x.

Proof. For a fixed value X 0 we will have u 0 =u(x 0), at 0 =f(u 0 ). For a new argument value x 0+Δ x:

Δ u= u(x 0 + Δ x) – u(x 0), Δ y=f(u 0+Δ u) – f(u 0).

Because u– differentiable at a point x 0, That u– is continuous at this point. Therefore, at Δ x→0 Δ u→0. Similarly for Δ u→0 Δ y→0.

By condition . From this relation, using the definition of the limit, we obtain (at Δ u→0)

where α→0 at Δ u→0, and, consequently, at Δ x→0.

Let us rewrite this equality as:

Δ y=y" uΔ u+α·Δ u.

The resulting equality is also valid for Δ u=0 for arbitrary α, since it turns into the identity 0=0. At Δ u=0 we will assume α=0. Let us divide all terms of the resulting equality by Δ x

.

By condition . Therefore, passing to the limit at Δ x→0, we get y" x = y"u·u" x. The theorem has been proven.

So, to differentiate a complex function y = f(u(x)), you need to take the derivative of the "external" function f, treating its argument simply as a variable, and multiply by the derivative of the "inner" function with respect to the independent variable.

If the function y=f(x) can be represented in the form y=f(u), u=u(v), v=v(x), then finding the derivative y " x is carried out by sequential application of the previous theorem.

According to the proven rule, we have y" x = y" u u"x. Applying the same theorem for u"x we get, i.e.

y" x = y" x u" v v" x = f"u( u)· u" v ( v)· v" x ( x).

Examples.

THE CONCEPT OF AN INVERSE FUNCTION

Let's start with an example. Consider the function y= x 3. We will consider the equality y= x 3 as an equation relative x. This is the equation for each value at defines a single value x: . Geometrically, this means that every straight line parallel to the axis Ox intersects the graph of a function y= x 3 only at one point. Therefore we can consider x as a function of y. A function is called the inverse of a function y= x 3.

Before moving on to the general case, we introduce definitions.

Function y = f(x) called increasing on a certain segment, if the larger value of the argument x from this segment corresponds to a larger value of the function, i.e. If x 2 >x 1, then f(x 2 ) > f(x 1 ).

The function is called similarly decreasing, if a smaller value of the argument corresponds to a larger value of the function, i.e. If X 2 < X 1, then f(x 2 ) > f(x 1 ).

So, let's be given an increasing or decreasing function y=f(x), defined on some interval [ a; b]. For definiteness, we will consider an increasing function (for a decreasing one everything is similar).

Consider two different values X 1 and X 2. Let y 1 =f(x 1 ), y 2 =f(x 2 ). From the definition of an increasing function it follows that if x 1 <x 2, then at 1 <at 2. Therefore, two different values X 1 and X 2 corresponds to two different function values at 1 and at 2. The opposite is also true, i.e. If at 1 <at 2, then from the definition of an increasing function it follows that x 1 <x 2. Those. again two different values at 1 and at 2 corresponds to two different values x 1 and x 2. Thus, between the values x and their corresponding values y a one-to-one correspondence is established, i.e. equation y=f(x) for everyone y(taken from the range of the function y=f(x)) defines a single value x, and we can say that x there is some argument function y: x= g(y).

This function is called reverse for function y=f(x). Obviously, the function y=f(x) is the inverse of the function x=g(y).

Note that the inverse function x=g(y) found by solving the equation y=f(x) relatively X.

Example. Let the function be given y= e x . This function increases at –∞< x <+∞. Она имеет обратную функцию x= log y. Domain of inverse function 0< y < + ∞.

Let's make a few comments.

Note 1. If an increasing (or decreasing) function y=f(x) is continuous on the interval [ a; b], and f(a)=c, f(b)=d, then the inverse function is defined and continuous on the interval [ c; d].

Note 2. If the function y=f(x) is neither increasing nor decreasing on a certain interval, then it can have several inverse functions.

Example. Function y=x2 defined at –∞<x<+∞. Она не является ни возрастающей, ни убывающей и не имеет обратной функции. Однако, если мы рассмотриминтервал 0≤x<+∞, то здесь функция является возрастающей и обратной для нее будет . На интервале – ∞ <x≤ 0 function – decreases and its inverse.

Note 3. If the functions y=f(x) And x=g(y) are mutually inverse, then they express the same relationship between variables x And y. Therefore, the graph of both is the same curve. But if we denote the argument of the inverse function again by x, and the function through y and plot them in the same coordinate system, we will get two different graphs. It is easy to notice that the graphs will be symmetrical with respect to the bisector of the 1st coordinate angle.

THEOREM ON THE DERIVATIVE INVERSE FUNCTION

Let us prove a theorem that allows us to find the derivative of the function y=f(x), knowing the derivative of the inverse function.

Theorem. If for the function y=f(x) there is an inverse function x=g(y), which at some point at 0 has a derivative g "(v 0), nonzero, then at the corresponding point x 0=g(x 0) function y=f(x) has a derivative f "(x 0), equal to , i.e. the formula is correct.

Proof. Because x=g(y) differentiable at the point y 0, That x=g(y) is continuous at this point, so the function y=f(x) continuous at a point x 0=g(y 0). Therefore, at Δ x→0 Δ y→0.

Let's show that .

Let . Then, by the property of the limit . Let us pass in this equality to the limit at Δ y→0. Then Δ x→0 and α(Δx)→0, i.e. .

Hence,

,

Q.E.D.

This formula can be written in the form .

Let's look at the application of this theorem using examples.

Solving physical problems or examples in mathematics is completely impossible without knowledge of the derivative and methods for calculating it. The derivative is one of the most important concepts in mathematical analysis. We decided to devote today’s article to this fundamental topic. What is a derivative, what is its physical and geometric meaning, how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?

Geometric and physical meaning of derivative

Let there be a function f(x) , specified in a certain interval (a, b) . Points x and x0 belong to this interval. When x changes, the function itself changes. Changing the argument - the difference in its values x-x0 . This difference is written as delta x and is called argument increment. A change or increment of a function is the difference between the values ​​of a function at two points. Definition of derivative:

The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.

Otherwise it can be written like this:

What's the point of finding such a limit? And here's what it is:

the derivative of a function at a point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at a given point.

Physical meaning of the derivative: the derivative of the path with respect to time is equal to the speed of rectilinear motion.

Indeed, since school days everyone knows that speed is a particular path x=f(t) and time t . Average speed over a certain period of time:

To find out the speed of movement at a moment in time t0 you need to calculate the limit:

Rule one: set a constant

The constant can be taken out of the derivative sign. Moreover, this must be done. When solving examples in mathematics, take it as a rule - If you can simplify an expression, be sure to simplify it .

Example. Let's calculate the derivative:

Rule two: derivative of the sum of functions

The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.

We will not give a proof of this theorem, but rather consider a practical example.

Find the derivative of the function:

Rule three: derivative of the product of functions

The derivative of the product of two differentiable functions is calculated by the formula:

Example: find the derivative of a function:

Solution:

It is important to talk about calculating derivatives of complex functions here. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable.

In the above example we come across the expression:

In this case, the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first calculate the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate argument itself with respect to the independent variable.

Rule four: derivative of the quotient of two functions

Formula for determining the derivative of the quotient of two functions:

We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it seems, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.

With any questions on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult test and understand the tasks, even if you have never done derivative calculations before.

The operation of finding the derivative is called differentiation.

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

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

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

Example 1. Find the derivative of a function

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

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

Example 2. Find the derivative of a function

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

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

Table of derivatives of simple functions

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

Rules of differentiation

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

Rule 1.If the functions

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

and

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

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

Rule 2.If the functions

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

and

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

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

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

For example, for three multipliers:

Rule 3.If the functions

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

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

Where to look for things on other pages

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

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

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

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

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

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

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

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

Example 3. Find the derivative of a function

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

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

We substitute the found derivatives into the sum of products and obtain the derivative of the entire function required by the condition of the problem:

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

Example 4. Find the derivative of a function

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

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

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

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

Example 5. Find the derivative of a function

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

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

Example 6. Find the derivative of a function

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

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