What is the derivative of y? Rules for calculating derivatives

Derivative calculation- one of the most important operations in differential calculus. Below is a table for finding derivatives of simple functions. For more complex differentiation rules, see other lessons:

Table of derivatives of exponential and logarithmic functions

Use the given formulas as reference values. They will help in solving differential equations and problems. In the picture, in the table of derivatives of simple functions, there is a “cheat sheet” of the main cases of finding a derivative in a form that is understandable for use, next to it are explanations for each case.

Derivatives of simple functions

1. The derivative of a number is zero
с´ = 0
Example:
5´ = 0

Explanation:
The derivative shows the rate at which the value of a function changes when its argument changes. Since the number does not change in any way under any conditions, the rate of its change is always zero.

2. Derivative of a variable equal to one
x´ = 1

Explanation:
With each increment of the argument (x) by one, the value of the function (the result of the calculation) increases by the same amount. Thus, the rate of change in the value of the function y = x is exactly equal to the rate of change in the value of the argument.

3. The derivative of a variable and a factor is equal to this factor
сx´ = с
Example:
(3x)´ = 3
(2x)´ = 2
Explanation:
In this case, every time the function argument changes ( X) its value (y) increases in With once. Thus, the rate of change of the function value in relation to the rate of change of the argument is exactly equal to the value With.

Whence it follows that
(cx + b)" = c
that is, the differential of the linear function y=kx+b is equal to the slope of the line (k).

4. Modulo derivative of a variable equal to the quotient of this variable to its modulus
|x|"= x / |x| provided that x ≠ 0
Explanation:
Since the derivative of a variable (see formula 2) is equal to one, the derivative of the module differs only in that the value of the rate of change of the function changes to the opposite when crossing the point of origin (try drawing a graph of the function y = |x| and see for yourself. This is exactly what value and returns the expression x / |x|. When x< 0 оно равно (-1), а когда x >0 - one. That is, for negative values of the variable x, with each increase in the change in the argument, the value of the function decreases by exactly the same value, and for positive values, on the contrary, it increases, but by exactly the same value.

5. Derivative of a variable to a power equal to the product of a number of this power and a variable to the power reduced by one
(x c)"= cx c-1, provided that x c and cx c-1 are defined and c ≠ 0
Example:
(x 2)" = 2x
(x 3)" = 3x 2
To remember the formula:
Move the degree of the variable down as a factor, and then reduce the degree itself by one. For example, for x 2 - the two was ahead of the x, and then the reduced power (2-1 = 1) simply gave us 2x. The same thing happened for x 3 - we “move down” the triple, reduce it by one and instead of a cube we have a square, that is, 3x 2. A little "unscientific" but very easy to remember.

6.Derivative of a fraction 1/x
(1/x)" = - 1 / x 2
Example:
Since a fraction can be represented as raising to a negative power
(1/x)" = (x -1)", then you can apply the formula from rule 5 of the table of derivatives
(x -1)" = -1x -2 = - 1 / x 2

7. Derivative of a fraction with a variable of arbitrary degree in the denominator
(1 / x c)" = - c / x c+1
Example:
(1 / x 2)" = - 2 / x 3

8. Derivative of the root(derivative of variable under square root)
(√x)" = 1 / (2√x) or 1/2 x -1/2
Example:
(√x)" = (x 1/2)" means you can apply the formula from rule 5
(x 1/2)" = 1/2 x -1/2 = 1 / (2√x)

9. Derivative of a variable under the root of an arbitrary degree
(n √x)" = 1 / (n n √x n-1)

In this lesson we will learn to apply formulas and rules of differentiation.

Examples. Find derivatives of functions.

1. y=x 7 +x 5 -x 4 +x 3 -x 2 +x-9. Applying the rule I, formulas 4, 2 and 1. We get:

y’=7x 6 +5x 4 -4x 3 +3x 2 -2x+1.

2. y=3x 6 -2x+5. We solve similarly, using the same formulas and formula 3.

y’=3∙6x 5 -2=18x 5 -2.

Applying the rule I, formulas 3, 5 And 6 And 1.

Applying the rule IV, formulas 5 And 1 .

In the fifth example, according to the rule I the derivative of the sum is equal to the sum of the derivatives, and we just found the derivative of the 1st term (example 4 ), therefore, we will find derivatives 2nd And 3rd terms, and for 1st summand we can immediately write the result.

Let's differentiate 2nd And 3rd terms according to the formula 4 . To do this, we transform the roots of the third and fourth powers in the denominators to powers with negative exponents, and then, according to 4 formula, we find derivatives of powers.

Look at this example and the result. Did you catch the pattern? Fine. This means we have a new formula and can add it to our derivatives table.

Let's solve the sixth example and derive another formula.

Let's use the rule IV and formula 4 . Let's reduce the resulting fractions.

Let's look at this function and its derivative. You, of course, understand the pattern and are ready to name the formula:

Learning new formulas!

Examples.

1. Find the increment of the argument and the increment of the function y= x 2, if the initial value of the argument was equal to 4 , and new - 4,01 .

Solution.

New argument value x=x 0 +Δx. Let's substitute the data: 4.01=4+Δх, hence the increment of the argument Δх=4.01-4=0.01. The increment of a function, by definition, is equal to the difference between the new and previous values of the function, i.e. Δy=f (x 0 +Δx) - f (x 0). Since we have a function y=x2, That Δу=(x 0 +Δx) 2 - (x 0) 2 =(x 0) 2 +2x 0 · Δx+(Δx) 2 - (x 0) 2 =2x 0 · Δx+(Δx) 2 =

2 · 4 · 0,01+(0,01) 2 =0,08+0,0001=0,0801.

Answer: argument increment Δх=0.01; function increment Δу=0,0801.

The function increment could be found differently: Δy=y (x 0 +Δx) -y (x 0)=y(4.01) -y(4)=4.01 2 -4 2 =16.0801-16=0.0801.

2. Find the angle of inclination of the tangent to the graph of the function y=f(x) at the point x 0, If f "(x 0) = 1.

Solution.

The value of the derivative at the point of tangency x 0 and is the value of the tangent of the tangent angle (the geometric meaning of the derivative). We have: f "(x 0) = tanα = 1 → α = 45°, because tg45°=1.

Answer: the tangent to the graph of this function forms an angle with the positive direction of the Ox axis equal to 45°.

3. Derive the formula for the derivative of the function y=xn.

Differentiation is the action of finding the derivative of a function.

When finding derivatives, use formulas that were derived based on the definition of a derivative, in the same way as we derived the formula for the derivative degree: (x n)" = nx n-1.

These are the formulas.

Table of derivatives It will be easier to memorize by pronouncing verbal formulations:

1. The derivative of a constant quantity is zero.

2. X prime is equal to one.

3. The constant factor can be taken out of the sign of the derivative.

4. The derivative of a degree is equal to the product of the exponent of this degree by a degree with the same base, but the exponent is one less.

5. The derivative of a root is equal to one divided by two equal roots.

6. The derivative of one divided by x is equal to minus one divided by x squared.

7. The derivative of the sine is equal to the cosine.

8. The derivative of the cosine is equal to minus sine.

9. The derivative of the tangent is equal to one divided by the square of the cosine.

10. The derivative of the cotangent is equal to minus one divided by the square of the sine.

We teach differentiation rules.

1. The derivative of an algebraic sum is equal to the algebraic sum of the derivatives of the terms.

2. The derivative of a product is equal to the product of the derivative of the first factor and the second plus the product of the first factor and the derivative of the second.

3. The derivative of “y” divided by “ve” is equal to a fraction in which the numerator is “y prime multiplied by “ve” minus “y multiplied by ve prime”, and the denominator is “ve squared”.

4. A special case of the formula 3.

Let's learn together!

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

Geometric and physical meaning of derivative

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

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

Otherwise it can be written like this:

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

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

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

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

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

Rule one: set a constant

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

Example. Let's calculate the derivative:

Rule two: derivative of the sum of functions

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

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

Find the derivative of the function:

Rule three: derivative of the product of functions

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

Example: find the derivative of a function:

Solution:

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

In the above example we come across the expression:

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

Rule four: derivative of the quotient of two functions

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

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

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

When deriving the very first formula of the table, we will proceed from the definition of the derivative function at a point. Let's take where x– any real number, that is, x– any number from the domain of definition of the function. Let us write down the limit of the ratio of the increment of the function to the increment of the argument at :

It should be noted that under the limit sign the expression is obtained, which is not the uncertainty of zero divided by zero, since the numerator does not contain an infinitesimal value, but precisely zero. In other words, the increment of a constant function is always zero.

Thus, derivative of a constant functionis equal to zero throughout the entire domain of definition.

Derivative of a power function.

The formula for the derivative of a power function has the form , where the exponent p– any real number.

Let us first prove the formula for the natural exponent, that is, for p = 1, 2, 3, …

We will use the definition of derivative. Let us write down the limit of the ratio of the increment of a power function to the increment of the argument:

To simplify the expression in the numerator, we turn to the Newton binomial formula:

Hence,

This proves the formula for the derivative of a power function for a natural exponent.

Derivative of an exponential function.

We present the derivation of the derivative formula based on the definition:

We have arrived at uncertainty. To expand it, we introduce a new variable, and at . Then . In the last transition, we used the formula for transitioning to a new logarithmic base.

Let's substitute into the original limit:

If we recall the second remarkable limit, we arrive at the formula for the derivative of the exponential function:

Derivative of a logarithmic function.

Let us prove the formula for the derivative of a logarithmic function for all x from the domain of definition and all valid values of the base a logarithm By definition of derivative we have:

As you noticed, during the proof the transformations were carried out using the properties of the logarithm. Equality is true due to the second remarkable limit.

Derivatives of trigonometric functions.

To derive formulas for derivatives of trigonometric functions, we will have to recall some trigonometry formulas, as well as the first remarkable limit.

By definition of the derivative for the sine function we have .

Let's use the difference of sines formula:

It remains to turn to the first remarkable limit:

Thus, the derivative of the function sin x There is cos x.

The formula for the derivative of the cosine is proved in exactly the same way.

Therefore, the derivative of the function cos x There is –sin x.

We will derive formulas for the table of derivatives for tangent and cotangent using proven rules of differentiation (derivative of a fraction).

Derivatives of hyperbolic functions.

The rules of differentiation and the formula for the derivative of the exponential function from the table of derivatives allow us to derive formulas for the derivatives of the hyperbolic sine, cosine, tangent and cotangent.

Derivative of the inverse function.

To avoid confusion during presentation, let's denote in subscript the argument of the function by which differentiation is performed, that is, it is the derivative of the function f(x) By x.

Now let's formulate rule for finding the derivative of an inverse function.

Let the functions y = f(x) And x = g(y) mutually inverse, defined on the intervals and respectively. If at a point there is a finite non-zero derivative of the function f(x), then at the point there is a finite derivative of the inverse function g(y), and . In another post .

This rule can be reformulated for any x from the interval , then we get .

Let's check the validity of these formulas.

Let's find the inverse function for the natural logarithm (Here y is a function, and x- argument). Having resolved this equation for x, we get (here x is a function, and y– her argument). That is, and mutually inverse functions.

From the table of derivatives we see that And .

Let’s make sure that the formulas for finding the derivatives of the inverse function lead us to the same results:

Calculation of the derivative is often found in Unified State Examination tasks. This page contains a list of formulas for finding derivatives.

Rules of differentiation

(k⋅ f(x))′=k⋅ f ′(x).
(f(x)+g(x))′=f′(x)+g′(x).
(f(x)⋅ g(x))′=f′(x)⋅ g(x)+f(x)⋅ g′(x).
Derivative of a complex function. If y=F(u), and u=u(x), then the function y=f(x)=F(u(x)) is called a complex function of x. Equal to y′(x)=Fu′⋅ ux′.
Derivative of an implicit function. The function y=f(x) is called an implicit function defined by the relation F(x,y)=0 if F(x,f(x))≡0.
Derivative of the inverse function. If g(f(x))=x, then the function g(x) is called the inverse function of the function y=f(x).
Derivative of a parametrically defined function. Let x and y be specified as functions of the variable t: x=x(t), y=y(t). They say that y=y(x) is a parametrically defined function on the interval x∈ (a;b), if on this interval the equation x=x(t) can be expressed as t=t(x) and the function y=y( t(x))=y(x).
Derivative of a power-exponential function. Found by taking logarithms to the base of the natural logarithm.

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