The basis of the proof is the determination of the limit of the function. You can use another method using trigonometric reduction formulas for cosine and sine angles. Express one function through another - cosine through sine, and differentiate the sine with a complex argument.
Let's consider the first example of derivation of the formula (Cos(x))"
We give a negligibly small increment Δx to the argument x of the function y = Cos(x). With a new value of the argument x+Δx, we obtain a new value of the function Cos(x+Δx). Then the increment of the function Δу will be equal to Cos(x+Δx)-Cos(x).
The ratio of the function increment to Δx will be as follows: (Cos(x+Δx)-Cos(x))/Δx. Let us carry out identical transformations in the numerator of the resulting fraction. Let us recall the formula for the difference between the cosines of angles, the result will be the product -2Sin(Δx/2) multiplied by Sin(x+Δx/2). We find the limit of the partial lim of this product on Δx as Δx tends to zero. It is known that the first (it is called remarkable) limit lim(Sin(Δx/2)/(Δx/2)) is equal to 1, and the limit -Sin(x+Δx/2) is equal to -Sin(x) with Δx tending to zero.
Let's write down the result: the derivative (Cos(x))" is equal to - Sin(x).
Some people prefer the second way to derive the same formula
From the trigonometry course we know: Cos(x) is equal to Sin(0.5·∏-x), similarly Sin(x) is equal to Cos(0.5·∏-x). Then we differentiate the complex function - the sine of the additional angle (instead of the cosine x).
We obtain the product Cos(0.5·∏-х)·(0.5·∏-х)", because the derivative of sine x is equal to cosine x. We turn to the second formula Sin(x) = Cos(0.5·∏- x) replacing cosine with sine, we take into account that (0.5·∏-x)" = -1. Now we get -Sin(x).
So, the derivative of the cosine has been found, y" = -Sin(x) for the function y = Cos(x).
A frequently used example where the derivative of the cosine is used. The function y = Cos 2 (x) is complex. We first find the differential of a power function with exponent 2, it will be 2 Cos(x), then multiply it by the derivative (Cos(x))", which is equal to -Sin(x). We get y" = -2 Cos(x) Sin(x). When we apply the formula Sin(2 x), the sine of the double angle, we obtain the final simplified
answer y" = -Sin(2 x)
Hyperbolic functions
They are used in the study of many technical disciplines: in mathematics, for example, they facilitate the calculation of integrals, solutions. They are expressed through trigonometric functions with an imaginary argument, for example, hyperbolic cosine ch(x) = Cos(i x), where i is the imaginary unit, hyperbolic sine sh(x) = Sin(i x).
The derivative of the hyperbolic cosine is calculated quite simply.
Consider the function y = (e x +e -x)/2, this is the hyperbolic cosine ch(x). We use the rule for finding the derivative of the sum of two expressions, the rule for moving the constant factor (Const) beyond the sign of the derivative. The second term 0.5 e -x is a complex function (its derivative is equal to -0.5 e -x), 0.5 e x is the first term. (ch(x)) "=((e x +e - x)/2)" can be written differently: (0.5 e x +0.5 e - x)" = 0.5 e x -0.5 e - x, because the derivative (e - x)" is equal to -1 multiplied by e - x. The result is a difference, and this is the hyperbolic sine sh(x).
Conclusion: (ch(x))" = sh(x).
Let's look at an example of how to calculate the derivative of the function y = ch(x 3 +1).
According to the hyperbolic cosine with a complex argument, y" = sh(x 3 +1)·(x 3 +1)", where (x 3 +1)" = 3·x 2 +0.
Answer: the derivative of this function is 3 x 2 sh(x 3 +1).
The derivatives of the considered functions y = ch(x) and y = Cos(x) are tabulated
When solving examples, there is no need to differentiate them each time according to the proposed scheme; it is enough to use the derivation.
Example. Differentiate the function y = Cos(x)+Cos 2 (-x)-Ch(5 x).
It’s easy to calculate (we use tabular data), y" = -Sin(x)+Sin(2 x)-5 Sh(5 x).
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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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
Derivatives of simple functions
1. The derivative of a number is zeroс´ = 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 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)
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.
The proof and derivation of the formula for the derivative of the cosine - cos(x) is presented. Examples of calculating derivatives of cos 2x, cos 3x, cos nx, cosine squared, cubed and to the power n. Formula for the derivative of the cosine of the nth order.
ContentSee also: Sine and cosine - properties, graphs, formulas
The derivative with respect to the variable x from the cosine of x is equal to minus the sine of x:
(cos x)′ = - sin x.
Proof
To derive the formula for the derivative of the cosine, we use the definition of derivative:
.
Let's transform this expression to reduce it to known mathematical laws and rules. To do this we need to know four properties.
1)
Trigonometric formulas. We will need the following formula:
(1)
;
2)
Continuity property of the sine function:
(2)
;
3)
The meaning of the first remarkable limit:
(3)
;
4)
Property of the limit of the product of two functions:
If and , then
(4)
.
Let's apply these laws to our limit. First we transform the algebraic expression
.
To do this we apply the formula
(1)
;
In our case
; . Then
;
;
;
.
Let's make a substitution. At , . We use the property of continuity (2):
.
Let's make the same substitution and apply the first remarkable limit (3):
.
Since the limits calculated above exist, we apply property (4):
.
Thus, we obtained the formula for the derivative of the cosine.
Examples
Let's look at simple examples of finding derivatives of functions containing a cosine. Let's find derivatives of the following functions:
y = cos 2x; y = cos 3x; y = cos nx; y = cos 2 x; y = cos 3 x and y = cos n x.
Example 1
Find derivatives of cos 2x, cos 3x And cosnx.
The original functions have a similar form. Therefore we will find the derivative of the function y = cosnx. Then, as a derivative of cosnx, substitute n = 2 and n = 3 . And, thus, we obtain formulas for the derivatives of cos 2x And cos 3x .
So, we find the derivative of the function
y = cosnx
.
Let's imagine this function of the variable x as a complex function consisting of two functions:
1)
2)
Then the original function is a complex (composite) function composed of functions and :
.
Let's find the derivative of the function with respect to the variable x:
.
Let's find the derivative of the function with respect to the variable:
.
We apply.
.
Let's substitute:
(P1) .
Now, in formula (A1) we substitute and:
;
.
;
;
.
Example 2
Find the derivatives of cosine squared, cosine cubed and cosine to the power n:
y = cos 2 x; y = cos 3 x; y = cos n x.
In this example, the functions also have a similar appearance. Therefore, we will find the derivative of the most general function - cosine to the power n:
y = cos n x.
Then we substitute n = 2 and n = 3. And, thus, we obtain formulas for the derivatives of cosine squared and cosine cubed.
So we need to find the derivative of the function
.
Let's rewrite it in a more understandable form:
.
Let's imagine this function as a complex function consisting of two functions:
1)
Functions depending on a variable: ;
2)
Functions depending on a variable: .
Then the original function is a complex function composed of two functions and :
.
Find the derivative of the function with respect to the variable x:
.
Find the derivative of the function with respect to the variable:
.
We apply the rule of differentiation of complex functions.
.
Let's substitute:
(P2) .
Now let's substitute and:
;
.
;
;
.
Higher order derivatives
Note that the derivative of cos x first order can be expressed through cosine as follows:
.
Let's find the second-order derivative using the formula for the derivative of a complex function:
.
Here .
Note that differentiation cos x causes its argument to increase by . Then the nth order derivative has the form:
(5)
.
This formula can be proven more strictly using the method of mathematical induction. The proof for the nth derivative of sine is presented on the page “Derivative of sine”. For the nth derivative of the cosine, the proof is exactly the same. You just need to replace sin with cos in all formulas.
See also: