Increment function for dummies. Open Library - an open library of educational information. Derivative of a complex function

Let X– argument (independent variable); y=y(x)– function.

Let's take a fixed argument value x=x 0 and calculate the value of the function y 0 =y(x 0 ) . Now let's arbitrarily set increment (change) of the argument and denote it  X ( X can be of any sign).

Increment argument is a dot X 0 +  X. Let's say it also contains a function value y=y(x 0 +  X)(see picture).

Thus, with an arbitrary change in the value of the argument, a change in the function is obtained, which is called increment function values:

and is not arbitrary, but depends on the type of function and value
.

Argument and function increments can be final, i.e. expressed as constant numbers, in which case they are sometimes called finite differences.

In economics, finite increments are considered quite often. For example, the table shows data on the length of the railway network of a certain state. Obviously, the increment in network length is calculated by subtracting the previous value from the subsequent one.

We will consider the length of the railway network as a function, the argument of which will be time (years).

Railway length as of December 31, thousand km.	Increment	Average annual growth

In itself, the increment of a function (in this case, the length of the railway network) does not characterize the change in function well. In our example, from the fact that 2,5>0,9 it cannot be concluded that the network grew faster in 2000-2003 years than in 2004 g., because the increment 2,5 refers to a three-year period, and 0,9 - in just one year. Therefore, it is quite natural that an increment in a function leads to a unit change in the argument. The increment of the argument here is periods: 1996-1993=3; 2000-1996=4; 2003-2000=3; 2004-2003=1 .

We get what is called in economic literature average annual growth.

You can avoid the operation of reducing the increment to the unit of argument change if you take the function values for argument values that differ by one, which is not always possible.

In mathematical analysis, in particular in differential calculus, infinitesimal (IM) increments of argument and function are considered.

Differentiation of a function of one variable (derivative and differential) Derivative of a function

Increments of argument and function at a point X 0 can be considered as comparable infinitesimal quantities (see topic 4, comparison of BM), i.e. BM is of the same order.

Then their ratio will have a finite limit, which is defined as the derivative of the function in t X 0 .

Limit of the ratio of the increment of a function to the BM increment of the argument at a point x=x 0 called derivative functions at a given point.

The symbolic designation of a derivative by a stroke (or rather, by the Roman numeral I) was introduced by Newton. You can also use a subscript, which shows which variable the derivative is calculated with, for example, . Another notation proposed by the founder of the calculus of derivatives, the German mathematician Leibniz, is also widely used:
. You will learn more about the origin of this designation in the section Function differential and argument differential.

This number estimates speed changes in the function passing through a point
.

Let's install geometric meaning derivative of a function at a point. For this purpose, we will plot the function y=y(x) and mark on it the points that determine the change y(x) in the interim

Tangent to the graph of a function at a point M 0
we will consider the limiting position of the secant M 0 M given that
(dot M slides along the graph of a function to a point M 0 ).

Let's consider
. Obviously,
.

If the point M direct along the graph of the function towards the point M 0 , then the value
will tend to a certain limit, which we denote
. Wherein.

Limit angle  coincides with the angle of inclination of the tangent drawn to the graph of the function incl. M 0 , so the derivative
numerically equal tangent slope at the specified point.

geometric meaning of the derivative of a function at a point.

Thus, we can write the tangent and normal equations ( normal - this is a straight line perpendicular to the tangent) to the graph of the function at some point X 0 :

Tangent - .

Normal -
.

Of interest are cases when these lines are located horizontally or vertically (see Topic 3, special cases of the position of a line on a plane). Then,

If
;

If
.

The definition of derivative is called differentiation functions.

 If the function is at a point X 0 has a finite derivative, then it is called differentiable at this point. A function that is differentiable at all points of a certain interval is called differentiable on this interval.

 Theorem . If the function y=y(x) differentiable incl. X 0 , then it is continuous at this point.

Thus, continuity– a necessary (but not sufficient) condition for the differentiability of a function.

Definition 1

If for each pair $(x,y)$ of values of two independent variables from some domain a certain value $z$ is associated, then $z$ is said to be a function of two variables $(x,y)$. Notation: $z=f(x,y)$.

In relation to the function $z=f(x,y)$, let's consider the concepts of general (total) and partial increments of a function.

Let a function $z=f(x,y)$ be given of two independent variables $(x,y)$.

Note 1

Since the variables $(x,y)$ are independent, one of them can change, while the other remains constant.

Let's give the variable $x$ an increment of $\Delta x$, while keeping the value of the variable $y$ unchanged.

Then the function $z=f(x,y)$ will receive an increment, which will be called the partial increment of the function $z=f(x,y)$ with respect to the variable $x$. Designation:

Similarly, we will give the variable $y$ an increment of $\Delta y$, while keeping the value of the variable $x$ unchanged.

Then the function $z=f(x,y)$ will receive an increment, which will be called the partial increment of the function $z=f(x,y)$ with respect to the variable $y$. Designation:

If the argument $x$ is given an increment of $\Delta x$, and the argument $y$ is given an increment of $\Delta y$, then the total increment is obtained given function$z=f(x,y)$. Designation:

Thus we have:

$\Delta _(x) z=f(x+\Delta x,y)-f(x,y)$ - partial increment of the function $z=f(x,y)$ over $x$;

$\Delta _(y) z=f(x,y+\Delta y)-f(x,y)$ - partial increment of the function $z=f(x,y)$ by $y$;

$\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)$ - total increment of the function $z=f(x,y)$.

Example 1

Solution:

$\Delta _(x) z=x+\Delta x+y$ - partial increment of the function $z=f(x,y)$ over $x$;

$\Delta _(y) z=x+y+\Delta y$ - partial increment of the function $z=f(x,y)$ with respect to $y$.

$\Delta z=x+\Delta x+y+\Delta y$ - total increment of the function $z=f(x,y)$.

Example 2

Calculate the partial and total increment of the function $z=xy$ at the point $(1;2)$ for $\Delta x=0.1;\, \, \Delta y=0.1$.

Solution:

By definition of partial increment we find:

$\Delta _(x) z=(x+\Delta x)\cdot y$ - partial increment of the function $z=f(x,y)$ over $x$

$\Delta _(y) z=x\cdot (y+\Delta y)$ - partial increment of the function $z=f(x,y)$ by $y$;

By definition of total increment we find:

$\Delta z=(x+\Delta x)\cdot (y+\Delta y)$ - total increment of the function $z=f(x,y)$.

Hence,

\[\Delta _(x) z=(1+0.1)\cdot 2=2.2\] \[\Delta _(y) z=1\cdot (2+0.1)=2.1 \] \[\Delta z=(1+0.1)\cdot (2+0.1)=1.1\cdot 2.1=2.31.\]

Note 2

The total increment of a given function $z=f(x,y)$ is not equal to the sum of its partial increments $\Delta _(x) z$ and $\Delta _(y) z$. Mathematical notation: $\Delta z\ne \Delta _(x) z+\Delta _(y) z$.

Example 3

Check assertion remarks for function

Solution:

$\Delta _(x) z=x+\Delta x+y$; $\Delta _(y) z=x+y+\Delta y$; $\Delta z=x+\Delta x+y+\Delta y$ (obtained in example 1)

Let's find the sum of partial increments of a given function $z=f(x,y)$

\[\Delta _(x) z+\Delta _(y) z=x+\Delta x+y+(x+y+\Delta y)=2\cdot (x+y)+\Delta x+\Delta y.\]

\[\Delta _(x) z+\Delta _(y) z\ne \Delta z.\]

Definition 2

If for each triple $(x,y,z)$ of values of three independent variables from some domain a certain value $w$ is associated, then $w$ is said to be a function of three variables $(x,y,z)$ in this area.

Notation: $w=f(x,y,z)$.

Definition 3

If for each set $(x,y,z,...,t)$ of values of independent variables from some domain a certain value $w$ is associated, then $w$ is said to be a function of the variables $(x,y, z,...,t)$ in this area.

Notation: $w=f(x,y,z,...,t)$.

For a function of three or more variables, in the same way as for a function of two variables, partial increments are determined for each of the variables:

$\Delta _(z) w=f(x,y,z+\Delta z)-f(x,y,z)$ - partial increment of the function $w=f(x,y,z,...,t )$ by $z$;

$\Delta _(t) w=f(x,y,z,...,t+\Delta t)-f(x,y,z,...,t)$ - partial increment of the function $w=f (x,y,z,...,t)$ by $t$.

Example 4

Write partial and total increment functions

Solution:

By definition of partial increment we find:

$\Delta _(x) w=((x+\Delta x)+y)\cdot z$ - partial increment of the function $w=f(x,y,z)$ over $x$

$\Delta _(y) w=(x+(y+\Delta y))\cdot z$ - partial increment of the function $w=f(x,y,z)$ over $y$;

$\Delta _(z) w=(x+y)\cdot (z+\Delta z)$ - partial increment of the function $w=f(x,y,z)$ over $z$;

By definition of total increment we find:

$\Delta w=((x+\Delta x)+(y+\Delta y))\cdot (z+\Delta z)$ - total increment of the function $w=f(x,y,z)$.

Example 5

Calculate the partial and total increment of the function $w=xyz$ at the point $(1;2;1)$ for $\Delta x=0,1;\, \, \Delta y=0,1;\, \, \Delta z=0.1$.

Solution:

By definition of partial increment we find:

$\Delta _(x) w=(x+\Delta x)\cdot y\cdot z$ - partial increment of the function $w=f(x,y,z)$ over $x$

$\Delta _(y) w=x\cdot (y+\Delta y)\cdot z$ - partial increment of the function $w=f(x,y,z)$ by $y$;

$\Delta _(z) w=x\cdot y\cdot (z+\Delta z)$ - partial increment of the function $w=f(x,y,z)$ over $z$;

By definition of total increment we find:

$\Delta w=(x+\Delta x)\cdot (y+\Delta y)\cdot (z+\Delta z)$ - total increment of the function $w=f(x,y,z)$.

Hence,

\[\Delta _(x) w=(1+0.1)\cdot 2\cdot 1=2.2\] \[\Delta _(y) w=1\cdot (2+0.1)\ cdot 1=2.1\] \[\Delta _(y) w=1\cdot 2\cdot (1+0.1)=2.2\] \[\Delta z=(1+0.1) \cdot (2+0.1)\cdot (1+0.1)=1.1\cdot 2.1\cdot 1.1=2.541.\]

WITH geometric point In terms of view, the total increment of the function $z=f(x,y)$ (by definition $\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)$) is equal to the increment of the applicate of the graph of the function $z =f(x,y)$ when moving from point $M(x,y)$ to point $M_(1) (x+\Delta x,y+\Delta y)$ (Fig. 1).

Picture 1.

Let x be an arbitrary point in some neighborhood of a fixed point x 0 . the difference x – x 0 is usually called the increment of the independent variable (or argument increment) at the point x 0 and is denoted Δx. Thus,

Δx = x –x 0 ,

whence it follows that

Function increment – the difference between two function values.

Let the function be given at = f(x), defined with the value of the argument equal to X 0 . Let's give the argument an increment D X, ᴛ.ᴇ. consider the value of the argument equal to x 0+D X. Let's assume that this argument value is also within the scope of this function. Then the difference D y = f(x 0+D X) – f(x 0) It is commonly called the increment of a function. Function increment f(x) at point x- a function usually denoted by Δ x f from the new variable Δ x defined as

Δ x f(Δ x) = f(x + Δ x) − f(x).

Find the increment of the argument and the increment of the function at the point x 0 if

Example 2. Find the increment of the function f(x) = x 2 if x = 1, ∆x = 0.1

Solution: f(x) = x 2, f(x+∆x) = (x+∆x) 2

Let's find the increment of the function ∆f = f(x+∆x) - f(x) = (x+∆x) 2 - x 2 = x 2 +2x*∆x+∆x 2 - x 2 = 2x*∆x + ∆x 2 /

Substitute the values x=1 and ∆x= 0.1, we get ∆f = 2*1*0.1 + (0.1) 2 = 0.2+0.01 = 0.21

Find the increment of the argument and the increment of the function at the point x 0

2.f(x) = 2x 3. x 0 =3 x=2.4

3. f(x) = 2x 2 +2 x 0 =1 x=0.8

4. f(x) = 3x+4 x 0 =4 x=3.8

Definition: Derivative of a function at a point, it is customary to call the limit (if it exists and is finite) of the ratio of the increment of the function to the increment of the argument, provided that the latter tends to zero.

The most commonly used derivative notations are:

Thus,

Finding the derivative is usually called differentiation . Introduced definition of a differentiable function: A function f that has a derivative at each point of a certain interval is usually called differentiable on this interval.

Let a function be defined in a certain neighborhood of a point. The derivative of a function is usually called a number such that the function in the neighborhood U(x 0) can be represented as

f(x 0 + h) = f(x 0) + Ah + o(h)

if exists.

Determining the derivative of a function at a point.

Let the function f(x) defined on the interval (a; b), and are the points of this interval.

Definition. Derivative of a function f(x) at a point it is customary to call the limit of the ratio of the increment of a function to the increment of the argument at . Denoted by .

When the last limit takes on a specific final value, we speak of the existence finite derivative at the point. If the limit is infinite, then we say that derivative is infinite at a given point. If the limit does not exist, then the derivative of the function at this point does not exist.

Function f(x) is said to be differentiable at a point when it has a finite derivative at it.

In case the function f(x) differentiable at each point of some interval (a; b), then the function is called differentiable on this interval. Τᴀᴋᴎᴍ ᴏϬᴩᴀᴈᴏᴍ, any point x from between (a; b) we can match the value of the derivative of the function at this point, that is, we have the opportunity to define a new function, which is called the derivative of the function f(x) on the interval (a; b).

The operation of finding the derivative is usually called differentiation.

In life we are not always interested in the exact values of any quantities. Sometimes it is interesting to know the change in this quantity, for example, the average speed of the bus, the ratio of the amount of movement to the period of time, etc. To compare the value of a function at a certain point with the values of the same function at other points, it is convenient to use concepts such as “function increment” and “argument increment.”

The concepts of "function increment" and "argument increment"

Let's say x is some arbitrary point that lies in some neighborhood of the point x0. The increment of the argument at the point x0 is the difference x-x0. The increment is designated as follows: ∆х.

∆x=x-x0.

Sometimes this quantity is also called the increment of the independent variable at point x0. From the formula it follows: x = x0+∆x. In such cases, they say that the initial value of the independent variable x0 received an increment ∆x.

If we change the argument, then the value of the function will also change.

f(x) - f(x0) = f(x0 + ∆x) - f(x0).

Increment of the function f at the point x0, the corresponding increment ∆х is the difference f(x0 + ∆х) - f(x0). The increment of a function is denoted as follows: ∆f. Thus we get, by definition:

∆f= f(x0 +∆x) - f(x0).

Sometimes, ∆f is also called the increment of the dependent variable and ∆у is used for this designation if the function was, for example, y=f(x).

Geometric meaning of increment

Look at the following picture.

As you can see, the increment shows the change in the ordinate and abscissa of a point. And the ratio of the increment of the function to the increment of the argument determines the angle of inclination of the secant passing through the initial and final position of the point.

Let's look at examples of incrementing a function and argument

Example 1. Find the increment of the argument ∆x and the increment of the function ∆f at the point x0, if f(x) = x 2, x0=2 a) x=1.9 b) x =2.1

Let's use the formulas given above:

a) ∆х=х-х0 = 1.9 - 2 = -0.1;

∆f=f(1.9) - f(2) = 1.9 2 - 2 2 = -0.39;

b) ∆x=x-x0=2.1-2=0.1;

∆f=f(2.1) - f(2) = 2.1 2 - 2 2 = 0.41.

Example 2. Calculate the increment ∆f for the function f(x) = 1/x at point x0 if the increment of the argument is equal to ∆x.

Again, we will use the formulas obtained above.

∆f = f(x0 + ∆x) - f(x0) =1/(x0-∆x) - 1/x0 = (x0 - (x0+∆x))/(x0*(x0+∆x)) = - ∆x/((x0*(x0+∆x)).

Definition 1

If for each pair $(x,y)$ of values of two independent variables from some domain a certain value $z$ is associated, then $z$ is said to be a function of two variables $(x,y)$. Notation: $z=f(x,y)$.

In relation to the function $z=f(x,y)$, let's consider the concepts of general (total) and partial increments of a function.

Let a function $z=f(x,y)$ be given of two independent variables $(x,y)$.

Note 1

Since the variables $(x,y)$ are independent, one of them can change, while the other remains constant.

Let's give the variable $x$ an increment of $\Delta x$, while keeping the value of the variable $y$ unchanged.

Then the function $z=f(x,y)$ will receive an increment, which will be called the partial increment of the function $z=f(x,y)$ with respect to the variable $x$. Designation:

Similarly, we will give the variable $y$ an increment of $\Delta y$, while keeping the value of the variable $x$ unchanged.

Then the function $z=f(x,y)$ will receive an increment, which will be called the partial increment of the function $z=f(x,y)$ with respect to the variable $y$. Designation:

If the argument $x$ is given an increment $\Delta x$, and the argument $y$ is given an increment $\Delta y$, then the full increment of the given function $z=f(x,y)$ is obtained. Designation:

Thus we have:

$\Delta _(x) z=f(x+\Delta x,y)-f(x,y)$ - partial increment of the function $z=f(x,y)$ over $x$;

$\Delta _(y) z=f(x,y+\Delta y)-f(x,y)$ - partial increment of the function $z=f(x,y)$ by $y$;

$\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)$ - total increment of the function $z=f(x,y)$.

Example 1

Solution:

$\Delta _(x) z=x+\Delta x+y$ - partial increment of the function $z=f(x,y)$ over $x$;

$\Delta _(y) z=x+y+\Delta y$ - partial increment of the function $z=f(x,y)$ with respect to $y$.

$\Delta z=x+\Delta x+y+\Delta y$ - total increment of the function $z=f(x,y)$.

Example 2

Calculate the partial and total increment of the function $z=xy$ at the point $(1;2)$ for $\Delta x=0.1;\, \, \Delta y=0.1$.

Solution:

By definition of partial increment we find:

$\Delta _(x) z=(x+\Delta x)\cdot y$ - partial increment of the function $z=f(x,y)$ over $x$

$\Delta _(y) z=x\cdot (y+\Delta y)$ - partial increment of the function $z=f(x,y)$ by $y$;

By definition of total increment we find:

$\Delta z=(x+\Delta x)\cdot (y+\Delta y)$ - total increment of the function $z=f(x,y)$.

Hence,

\[\Delta _(x) z=(1+0.1)\cdot 2=2.2\] \[\Delta _(y) z=1\cdot (2+0.1)=2.1 \] \[\Delta z=(1+0.1)\cdot (2+0.1)=1.1\cdot 2.1=2.31.\]

Note 2

Example 3

Check assertion remarks for function

Solution:

$\Delta _(x) z=x+\Delta x+y$; $\Delta _(y) z=x+y+\Delta y$; $\Delta z=x+\Delta x+y+\Delta y$ (obtained in example 1)

Let's find the sum of partial increments of a given function $z=f(x,y)$

\[\Delta _(x) z+\Delta _(y) z=x+\Delta x+y+(x+y+\Delta y)=2\cdot (x+y)+\Delta x+\Delta y.\]

\[\Delta _(x) z+\Delta _(y) z\ne \Delta z.\]

Definition 2

If for each triple $(x,y,z)$ of values of three independent variables from some domain a certain value $w$ is associated, then $w$ is said to be a function of three variables $(x,y,z)$ in this area.

Notation: $w=f(x,y,z)$.

Definition 3

Notation: $w=f(x,y,z,...,t)$.

For a function of three or more variables, in the same way as for a function of two variables, partial increments are determined for each of the variables:

$\Delta _(z) w=f(x,y,z+\Delta z)-f(x,y,z)$ - partial increment of the function $w=f(x,y,z,...,t )$ by $z$;

$\Delta _(t) w=f(x,y,z,...,t+\Delta t)-f(x,y,z,...,t)$ - partial increment of the function $w=f (x,y,z,...,t)$ by $t$.

Example 4

Write partial and total increment functions

Solution:

By definition of partial increment we find:

$\Delta _(x) w=((x+\Delta x)+y)\cdot z$ - partial increment of the function $w=f(x,y,z)$ over $x$

$\Delta _(y) w=(x+(y+\Delta y))\cdot z$ - partial increment of the function $w=f(x,y,z)$ over $y$;

$\Delta _(z) w=(x+y)\cdot (z+\Delta z)$ - partial increment of the function $w=f(x,y,z)$ over $z$;

By definition of total increment we find:

$\Delta w=((x+\Delta x)+(y+\Delta y))\cdot (z+\Delta z)$ - total increment of the function $w=f(x,y,z)$.

Example 5

Calculate the partial and total increment of the function $w=xyz$ at the point $(1;2;1)$ for $\Delta x=0,1;\, \, \Delta y=0,1;\, \, \Delta z=0.1$.

Solution:

By definition of partial increment we find:

$\Delta _(x) w=(x+\Delta x)\cdot y\cdot z$ - partial increment of the function $w=f(x,y,z)$ over $x$

$\Delta _(y) w=x\cdot (y+\Delta y)\cdot z$ - partial increment of the function $w=f(x,y,z)$ by $y$;

$\Delta _(z) w=x\cdot y\cdot (z+\Delta z)$ - partial increment of the function $w=f(x,y,z)$ over $z$;

By definition of total increment we find:

$\Delta w=(x+\Delta x)\cdot (y+\Delta y)\cdot (z+\Delta z)$ - total increment of the function $w=f(x,y,z)$.

Hence,

From a geometric point of view, the total increment of the function $z=f(x,y)$ (by definition $\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)$) is equal to the increment of the applicate of the graph function $z=f(x,y)$ when moving from point $M(x,y)$ to point $M_(1) (x+\Delta x,y+\Delta y)$ (Fig. 1).

Picture 1.