Find the derivative of a function given parametrically online with a solution. Derivative of a Parametrically Defined Function
Logarithmic differentiation
Derivatives of elementary functions
Basic rules for differentiation
Differential function
home linear part function increments A D x in the definition of the differentiability of the function
D f = f(x)- f(x 0)= A(x - x 0)+ o(x - x 0), x®x 0
is called the differential of the function f(x) at the point x 0 and denoted
df(x 0)= f ¢(x 0) D x = A D x.
Differential depends on the point x 0 and from the increment D x. On D x at the same time they look at it as an independent variable, so that at each point the differential is a linear function of the increment D x.
If we consider as a function f(x)= x, then we get dx = D x, dy = Adx... This is consistent with Leibniz's notation
Geometric interpretation of the differential as increments of the ordinate of the tangent.
Rice. 4.3
1) f = const , f ¢ = 0, df = 0D x = 0.
2) f = u + v, f ¢ = u ¢ + v ¢, df = du + dv.
3) f = uv, f ¢ = u ¢ v + v ¢ u, df = u dv + v du.
Consequence. (cf(x))¢ = cf ¢(x), (c 1 f 1 (x)+… + C n f n(x))¢ = c 1 f ¢ 1 (x)+… + C n f ¢ n(x)
4) f = u / v, v(x 0) ¹0 and the derivative exists, then f ¢ =(u ¢ v-v¢ u)/v 2 .
For brevity, we will denote u = u(x), u 0 = u(x 0), then
Passing to the limit at D x® 0 we obtain the required equality.
5) Derivative of a complex function.
Theorem. If there are f ¢(x 0), g ¢(x 0)and x 0 = g(t 0), then in some neighborhood of t 0 the complex function f(g(t)), it is differentiable at the point t 0 and
Proof.
f(x)- f(x 0)= f ¢(x 0)(x-x 0)+ a ( x)(x-x 0), xÎ U(x 0).
f(g(t))- f(g(t 0))= f ¢(x 0)(g(t)- g(t 0))+ a ( g(t))(g(t)- g(t 0)).
We divide both sides of this equality by ( t - t 0) and pass to the limit at t®t 0 .
6) Calculation of the derivative of the inverse function.
Theorem. Let f be continuous and strictly monotone on[a, b]... Let at the point x 0 Î( a, b)there is f ¢(x 0) ¹ 0 , then the inverse function x = f -1 (y)has at the point y 0 derivative equal to
Proof... We count f strictly monotonically increasing, then f -1 (y) is continuous, monotonically increasing on [ f(a), f(b)]. We put y 0 = f(x 0), y = f(x), x - x 0 = D x,
y - y 0 = D y... By virtue of the continuity of the inverse function D y®0 Þ D x®0, we have
Passing to the limit, we obtain the required equality.
7) Derivative even function odd, derivative odd function even.
Indeed, if x® - x 0 , then - x® x 0 , therefore
For an even function for an odd function
1) f = const, f ¢(x)=0.
2) f(x)= x, f ¢(x)=1.
3) f(x)= e x, f ¢(x)= e x ,
4) f(x)= a x,(a x)¢ = a x ln a.
5) ln a.
6) f(x) = ln x,
Consequence. (the derivative of an even function is odd)
7) (x m )¢= m x m -1 , x>0, x m = e m ln x .
8) (sin x)¢= cos x,
9) (cos x)¢=- sin x,(cos x)¢= (sin ( x + p / 2)) ¢= cos ( x + p / 2) = - sin x.
10) (tg x)¢= 1 / cos 2 x.
11) (ctg x)¢= -1 / sin 2 x.
16) sh x, ch x.
f (x),, whence it follows that f ¢(x)= f(x) (ln f(x))¢ .
The same formula can be obtained differently f(x)= e ln f(x) , f ¢ = e ln f(x) (ln f(x))¢.
Example. Calculate the derivative of a function f = x x.
= x x = x x = x x = x x(ln x + 1).
Locus of points on a plane
will be called the graph of the function, given parametrically... They also talk about the parametric definition of a function.
Remark 1. If x, y continuous on [a, b] and x(t) strictly monotone on the segment (for example, it increases strictly monotonically), then on [ a, b], a = x(a) , b = x(b) function defined f(x)= y(t(x))where t(x) – function inverse to x (t). The graph of this function is the same as the graph of the function
If the scope a parametrically given function can be divided into a finite number of segments , k = 1,2,..., n, on each of which the function x(t) is strictly monotone, then the parametrically defined function splits into a finite number of ordinary functions f k(x)= y(t -1 (x)) with scopes [ x(a k), x(b k)] for ascending sections x(t) and with scopes [ x(b k), x(a k)] for segments of decreasing function x(t). The functions obtained in this way are called single-valued branches of a parametrically given function.
The figure shows a graph of a parametrically specified function
With the selected parameterization, the domain is divided into five sections of strict monotonicity of the function sin (2 t), exactly: tÎ tÎ ,tÎ ,tÎ , and, accordingly, the graph splits into five unambiguous branches corresponding to these sections.
Rice. 4.4
Rice. 4.5
You can choose another parameterization of the same locus of points
In this case, there will be only four such branches. They will correspond to areas of strict monotony. tÎ ,tÎ , tÎ ,tÎ function sin (2 t).
Rice. 4.6
The four sections of monotonicity of the sin (2 t) on a long segment.
Rice. 4.7
The display of both graphs in one figure allows you to approximately display the graph of a parametrically given function using the monotonic regions of both functions.
For example, consider the first branch corresponding to the segment tÎ . At the ends of this section, the function x = sin (2 t) takes values -1 and 1 , so this branch will be defined at [-1,1]. After that, you need to look at the sections of monotony of the second function y = cos ( t), on her two areas of monotony . This allows us to say that the first branch has two sections of monotony. Having found the end points of the graph, you can connect them with straight lines in order to indicate the nature of the monotonicity of the graph. Having done this with each branch, we will obtain areas of monotony of single-valued branches of the graph (in the figure they are highlighted in red)
Rice. 4.8
First unambiguous branch f 1 (x)= y(t(x)) corresponding to the site will be defined for xÎ [-1,1] . First unambiguous branch tÎ , xÎ [-1,1].
All the other three branches will also have a domain of definition [-1,1] .
Rice. 4.9
Second branch tÎ xÎ [-1,1].
Rice. 4.10
Third branch tÎ xÎ [-1,1]
Rice. 4.11
Fourth branch tÎ xÎ [-1,1]
Rice. 4.12
Comment 2. The same function can have different parametric assignments. Differences may relate to how the functions themselves x(t), y(t) , and areas of definition these functions.
An example of various parametric assignments of the same function
and tÎ [-1, 1] .
Remark 3. If x, y are continuous on , x(t) - strictly monotone on the segment and there are derivatives y ¢(t 0),x ¢(t 0) ¹0, then there exists f ¢(x 0)= .
Really, .
The last statement also applies to single-valued branches of a parametrically given function.
4.2 Higher order derivatives and differentials
Senior derivatives and differentials. Differentiation of functions specified parametrically. Leibniz's formula.
Let the function be given parametrically:
(1)
where is some variable called a parameter. And let the functions and have derivatives at some value of the variable. Moreover, the function also has an inverse function in some neighborhood of the point. Then function (1) has a derivative at the point, which, at parametric form, is determined by the formulas:
(2)
Here and are derivatives of functions and with respect to a variable (parameter). They are often written as follows:
;
.
Then system (2) can be written as follows:
Proof
By condition, the function has an inverse function. Let's denote it as
.
Then the original function can be represented as a complex function:
.
Let us find its derivative using the rules for differentiating complex and inverse functions:
.
The rule is proven.
Proof in the second way
Let us find the derivative in the second way, based on the definition of the derivative of the function at the point:
.
Let's introduce the notation:
.
Then the previous formula takes the form:
.
We will use the fact that the function has an inverse function in the vicinity of the point.
Let us introduce the notation:
;
;
;
.
Divide the numerator and denominator of the fraction by:
.
At , . Then
.
The rule is proven.
Higher order derivatives
To find the derivatives of higher orders, it is necessary to carry out the differentiation several times. Suppose we need to find the second-order derivative of a parametrically defined function of the following form:
(1)
Using formula (2), we find the first derivative, which is also determined parametrically:
(2)
Let's denote the first derivative by a variable:
.
Then, to find the second derivative of a function with respect to a variable, you need to find the first derivative of a function with respect to a variable. The dependence of a variable on a variable is also specified parametrically:
(3)
Comparing (3) with formulas (1) and (2), we find:
Now let's express the result in terms of functions and. To do this, we substitute and apply the formula for the derivative of the fraction:
.
Then
.
From here we get the second derivative of the function with respect to the variable:
It is also parametric. Note that the first line can also be written like this:
.
Continuing the process, you can get the derivatives of the function of the variable of the third and higher orders.
Note that it is possible to omit the notation for the derivative. It can be written like this:
;
.
Example 1
Find the derivative of a function defined parametrically:
Solution
Find derivatives and with respect to.
From the table of derivatives we find:
;
.
We apply:
.
Here .
.
Here .
The desired derivative:
.
Answer
Example 2
Find the derivative of the function expressed in terms of a parameter:
Solution
Let's expand the brackets using formulas for power functions and roots:
.
Find the derivative:
.
Find the derivative. To do this, we introduce a variable and apply the formula for the derivative of a complex function.
.
Find the desired derivative:
.
Answer
Example 3
Find the derivatives of the second and third orders of the function given parametrically in example 1:
Solution
In Example 1, we found the first-order derivative:
Let us introduce the notation. Then the function is derivative with respect to. It is defined parametrically:
To find the second derivative with respect to, we need to find the first derivative with respect to.
Differentiate by.
.
We found the derivative with respect to example 1:
.
The second-order derivative with respect to is equal to the first-order derivative with respect to:
.
So, we have found the second-order derivative in parametric form:
Now we find the third-order derivative. Let us introduce the notation. Then we need to find the first-order derivative of the function, which is given parametrically:
Find the derivative with respect to. To do this, we rewrite it in an equivalent form:
.
From
.
The third-order derivative with respect to is equal to the first-order derivative with respect to:
.
Comment
You can omit the variables and, which are derived from and, respectively. Then you can write it like this:
;
;
;
;
;
;
;
;
.
Answer
In parametric representation, the second-order derivative has the following form:
Derivative of the third order.
Do not strain, in this paragraph everything is also quite simple. You can write the general formula for a parametrically defined function, but, in order to make it clear, I will immediately write specific example... In parametric form, the function is given by two equations:. Often, equations are written not under curly braces, but sequentially:,.
The variable is called a parameter and can take values from "minus infinity" to "plus infinity". Consider, for example, a value and substitute it into both equations: 
... Or humanly: "if x is equal to four, then y is equal to one." A point can be marked on the coordinate plane, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the "te" parameter. As for the "normal" function, for the American Indians of a parametrically defined function, all rights are also respected: you can plot a graph, find derivatives, and so on. By the way, if there is a need to plot a graph of a parametrically given function, download my geometric program on the page Mathematical formulas and tables.
In the simplest cases, it is possible to represent the function explicitly. Let us express the parameter from the first equation: 
- and substitute it into the second equation: 
... The result is an ordinary cubic function.
In more "severe" cases, this trick does not work. But this does not matter, because to find the derivative of a parametric function, there is a formula:
Find the derivative of the "game with respect to the te variable":
All the rules of differentiation and the table of derivatives are, of course, also valid for the letter, thus, there is no novelty in the process of finding derivatives... Just mentally replace all the "x" in the table with the letter "te".
Find the derivative of "x with respect to the te variable": 
Now it only remains to substitute the found derivatives into our formula: 
Ready. The derivative, like the function itself, also depends on the parameter.
As for the designations, in the formula, instead of writing, it could simply be written without a subscript, since this is the "usual" derivative "by x". But in the literature there is always a variant, so I will not deviate from the standard.
Example 6
We use the formula
In this case:
Thus: 
A feature of finding the derivative of a parametric function is the fact that at each step, it is beneficial to simplify the result as much as possible... So, in the considered example, when I found it, I expanded the parentheses under the root (although I could not do this). Chances are great that when substituted and into the formula, many things will be reduced well. Although, of course, there are examples with clumsy answers.
Example 7
Find the derivative of a parametrically defined function 
This is an example for a do-it-yourself solution.
The article The simplest typical tasks with derivative we considered examples in which it was required to find the second derivative of a function. For a parametrically given function, you can also find the second derivative, and it is found by the following formula:. It is quite obvious that in order to find the second derivative, one must first find the first derivative.
Example 8
Find the first and second derivatives of a function given parametrically
First, let's find the first derivative.
We use the formula
In this case: 
Substitutes found derivatives into the formula. For the sake of simplification, we use the trigonometric formula: 
I noticed that in the problem of finding the derivative of a parametric function, quite often, for simplification purposes, one has to use trigonometric formulas ... Remember them or keep them close at hand, and do not miss the opportunity to simplify each intermediate result and answers. What for? Now we have to take the derivative of, and this is clearly better than finding the derivative of.
Let's find the second derivative.
We use the formula:.
Let's take a look at our formula. The denominator has already been found in the previous step. It remains to find the numerator - the derivative of the first derivative with respect to the variable "te":
It remains to use the formula: 
To consolidate the material, I propose a couple more examples for an independent solution.
Example 9
Example 10
Find and for a function defined parametrically 
Wish you success!
I hope this lesson was useful, and now you can easily find derivatives from functions that are specified implicitly and from parametric functions.
Solutions and Answers:
Example 3: Solution:
Thus:
Consider the definition of a line on a plane, in which the variables x, y are functions of the third variable t (called a parameter):
For each value t from a certain interval correspond to certain values x and y, and, therefore, a certain point M (x, y) of the plane. When t runs through all values from the specified interval, then the point M (x, y) describes some line L... Equations (2.2) are called parametric equations of the line L.
If the function x = φ (t) has an inverse t = Ф (x), then substituting this expression into the equation y = g (t), we obtain y = g (Ф (x)), which specifies y as a function of x... In this case, we say that equations (2.2) define the function y parametrically.
Example 1. Let be M (x, y)- an arbitrary point of a circle of radius R and centered at the origin. Let be t- the angle between the axis Ox and radius OM(see fig. 2.3). Then x, y expressed through t:
Equations (2.3) are parametric equations of the circle. We exclude the parameter t from equations (2.3). To do this, we square each of the equations and add, we get: x 2 + y 2 = R 2 (cos 2 t + sin 2 t) or x 2 + y 2 = R 2 - the equation of a circle in a Cartesian coordinate system. It defines two functions: Each of these functions is given by parametric equations (2.3), but for the first function, and for the second.
Example 2... Parametric Equations
define an ellipse with semiaxes a, b(fig. 2.4). Eliminating the parameter from the equations t, we get canonical equation ellipse:
Example 3... A cycloid is a line described by a point lying on a circle, if this circle rolls without sliding along a straight line (Fig. 2.5). Let us introduce the parametric equations of the cycloid. Let the radius of the rolling circle be a, point M describing the cycloid, at the beginning of the movement coincided with the origin of coordinates. 
Determine the coordinates x, y points M after the circle has turned through an angle t
(fig. 2.5), t = РMCB... Arc length MB equal to the length of the segment OB, since the circle rolls without sliding, therefore
OB = at, AB = MD = asint, CD = acost, x = OB - AB = at - asint = a (t - sint),
y = AM = CB - CD = a - acost = a (1 - cost).
So, the parametric equations of the cycloid are obtained:
When changing a parameter t from 0 to 2π the circle rotates one revolution, while the point M describes one arc of the cycloid. Equations (2.5) define y as a function of x... Although the function x = a (t - sint) has an inverse function, but it is not expressed in terms of elementary functions, therefore the function y = f (x) not expressed in terms of elementary functions.
Consider the differentiation of a function defined parametrically by equations (2.2). The function x = φ (t) on some interval of t has the inverse function t = Ф (x), then y = g (Ф (x))... Let be x = φ (t), y = g (t) have derivatives, and x "t ≠ 0... According to the rule of differentiation of a complex function y "x = y" t × t "x. Based on the rule of differentiation of the inverse function, therefore:
The resulting formula (2.6) allows one to find the derivative for a function given parametrically.
Example 4. Let the function y depending on x, is given parametrically:
Solution. 
.
Example 5. Find Slope k tangent to the cycloid at the point M 0 corresponding to the value of the parameter.
Solution. From the equations of the cycloid: y "t = asint, x" t = a (1 - cost), therefore 
Slope tangent at point M 0 equal to the value at t 0 = π / 4:
DIFFERENTIAL FUNCTION
Let the function at the point x 0 has a derivative. A-priory: 
therefore, by the properties of the limit (Section 1.8), where a- infinitely small at Δx → 0... From here
Δy = f "(x0) Δx + α × Δx. (2.7)
As Δx → 0, the second term in equality (2.7) is infinitesimal of higher order, in comparison with 
, therefore, Δy and f "(x 0) × Δx are equivalent, infinitesimal (for f" (x 0) ≠ 0).
Thus, the increment of the function Δy consists of two terms, of which the first f "(x 0) × Δx is main part increment Δy, linear with respect to Δx (for f "(x 0) ≠ 0).
Differential of the function f (x) at the point x 0 is called the main part of the increment of the function and is denoted: dy or df (x 0)... Hence,
df (x0) = f "(x0) × Δx. (2.8)
Example 1. Find the differential of a function dy and the increment of the function Δy for the function y = x 2 at:
1) arbitrary x and Δ x; 2) x 0 = 20, Δx = 0.1.
Solution
1) Δy = (x + Δx) 2 - x 2 = x 2 + 2xΔx + (Δx) 2 - x 2 = 2xΔx + (Δx) 2, dy = 2xΔx.
2) If x 0 = 20, Δx = 0.1, then Δy = 40 × 0.1 + (0.1) 2 = 4.01; dy = 40 × 0.1 = 4.
We write equality (2.7) in the form:
Δy = dy + a × Δx. (2.9)
The increment Δy differs from the differential dy by an infinitesimal higher order, compared with Δx, therefore, in approximate calculations, the approximate equality Δy ≈ dy is used if Δx is small enough.
Taking into account that Δy = f (x 0 + Δx) - f (x 0), we obtain an approximate formula:
f (x 0 + Δx) ≈ f (x 0) + dy. (2.10)
Example 2... Calculate approximately.
Solution. Consider:
Using formula (2.10), we get:
Hence, ≈ 2.025.
Consider the geometric meaning of the differential df (x 0)(fig. 2.6). 
Draw the tangent at the point M 0 (x0, f (x 0)) to the graph of the function y = f (x), let φ be the angle between the tangent KM0 and the Ox axis, then f "(x 0) = tgφ. From ΔM0NP:
PN = tgφ × Δx = f "(x 0) × Δx = df (x 0). But PN is the increment of the ordinate of the tangent as x changes from x 0 to x 0 + Δx.
Therefore, the differential of the function f (x) at the point x 0 is equal to the increment of the ordinate of the tangent.
Find the differential of the function
y = x. Since (x) "= 1, then dx = 1 × Δx = Δx. We assume that the differential of the independent variable x is equal to its increment, ie dx = Δx.
If x is an arbitrary number, then from equality (2.8) we obtain df (x) = f "(x) dx, whence 
.
Thus, the derivative for the function y = f (x) is equal to the ratio of its differential to the differential of the argument.
Consider the properties of the differential of a function.
If u (x), v (x) are differentiable functions, then the following formulas are valid:
To prove these formulas, derivative formulas are used for the sum, product and quotient function. Let us prove, for example, formula (2.12):
d (u × v) = (u × v) "Δx = (u × v" + u "× v) Δx = u × v" Δx + u "Δx × v = u × dv + v × du.
Consider the differential of a composite function: y = f (x), x = φ (t), i.e. y = f (φ (t)).
Then dy = y "t dt, but y" t = y "x × x" t, so dy = y "x x" t dt. Considering,
that x "t = dx, we get dy = y" x dx = f "(x) dx.
Thus, the differential of a composite function y = f (x), where x = φ (t), has the form dy = f "(x) dx, the same as in the case when x is an independent variable. This property is called form invariance, the differential a.
