The derivative of the cosine is found by analogy with the derivative of the sine, the basis of the proof is the definition of the limit of the function. You can use another method, using the trigonometric reduction formulas for the cosine and sine of angles. Express one function in terms of another - cosine in terms of sine, and differentiate the sine with a complex argument.
Consider the first example of deriving the formula (Cos(x))'
Give a negligibly small increment Δx to the argument x of the function y=Cos(x). With a new value of the argument х+Δх, we obtain a new value of the function Cos(х+Δх). Then the function increment Δy will be equal to Cos(х+Δx)-Cos(x).
The ratio of the function increment to Δх will be: (Cos(х+Δx)-Cos(x))/Δх. Let's carry out identical transformations in the numerator of the resulting fraction. Recall the formula for the difference of the cosines of the angles, the result will be the product -2Sin (Δx / 2) times Sin (x + Δx / 2). We find the limit of the quotient lim of this product on Δx as Δx tends to zero. It is known that the first(it is called wonderful) the limit lim(Sin(Δx/2)/(Δx/2)) is equal to 1, and the limit -Sin(x+Δx/2) is equal to -Sin(x) as Δx tends to zero. Write down the result: the derivative of (Cos(x))' is equal to - Sin(x).
Some people prefer the second way of deriving the same formula
It is known from the course of trigonometry: Cos(x) is equal to Sin(0, 5 ∏-x), similarly Sin(x) is equal to Cos(0, 5 ∏-x). Then we differentiate a complex function - the sine of the additional angle (instead of the cosine x).
We get the product Cos(0, 5 ∏-x) (0, 5 ∏-x)', because the derivative of the sine x is equal to the cosine X. We turn to the second formula Sin(x)=Cos(0.5 ∏-x) of replacing cosine with sine, taking into account that (0.5 ∏-x)'=-1. Now we get -Sin(x). So, the derivative of the cosine is found, y'=-Sin(x) for the function y=Cos(x).
Squared cosine derivative
A commonly used example where the cosine derivative is used. The function y=Cos2(x) is hard. We first find the differential of the power function with exponent 2, it will be 2·Cos(x), then we 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(2x), the sine of a double angle, we get the final simplifiedanswer y'=-Sin(2x)
Hyperbolic functions
They are used in the study of many technical disciplines: in mathematics, for example, they facilitate the calculation of integrals, the solution of differential equations. They are expressed in terms of trigonometric functions with imaginaryargument, so the hyperbolic cosine ch(x)=Cos(i x), where i is the imaginary unit, the hyperbolic sine sh(x)=Sin(i x).
The derivative of the hyperbolic cosine is calculated quite simply.
Consider the function y=(ex+e-x)/2, this and is the hyperbolic cosine ch(x). We use the rule for finding the derivative of the sum of two expressions, the rule for taking the constant factor (Const) out of the sign of the derivative. The second term 0.5 e-x is a complex function (its derivative is -0.5 e-x), 0.5 eх― the first term. (ch(x)) '=((ex+e-x)/2)' can be written in another way: (0, 5 ex+0, 5 e-x)'=0, 5 ex-0, 5 e-x, because the derivative (e -x)' equals -1 times e-x. The result is a difference, and this is the hyperbolic sine sh(x).Output: (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 differentiation rule with complex argument y'=sh(x
3+1) (x 3+1)', where (x3+1)'=3 x2+0. Answer: the derivative of this function is 3 x
2sh(x3+1).
Derivatives of the considered functions y=ch(x) and y=Cos(x) tabular
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)+Cos2(-x)-Ch(5 x). Easy to calculate (use tabular data), y'=-Sin(x) +Sin(2 x)-5 Sh(5 x).
