The derivative of the cosine is similar to 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 trigonometric reduction formulas for the cosine and sine of the angles. Express one function through another - cosine through sine, and differentiate sine with a complex argument.
Consider the first example of the derivation of the formula (Cos (x)) '
We give a negligible increment Δx to the argument x of the function y = Cos (x). With the new value of the argument x + Δx, we obtain the new value of the function Cos (x + Δx). Then the increment of the function Δy will be equal to Cos (x + Δx) -Cos (x).
The ratio of the increment of the function to Δx will be as follows: (Cos (x + Δx) -Cos (x)) / Δx. We carry out the identical transformations in the numerator of the resulting fraction. Recall the formula for the difference in the cosines of the angles, the result is the product -2Sin (Δx / 2) multiplied by 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 remarkable) limit lim (Sin (Δ / 2) / (Δ / 2)) is equal to 1, and the limit -Sin ( + Δ / 2) is equal to -Sin (x) as Δx tends to to zero.
We write the result: the derivative (Cos (x)) 'is equal to - Sin (x).
Some people like the second way to output the same formula.
From the trigonometry course it is known: 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 get the product Cos (0.5 · ∏-x) · (0.5 · ∏-x) ', because the derivative of the sine of x is equal to the cosine of x. We turn to the second formula Sin (x) = Cos (0.5 · ∏-x) replacing the cosine by the sine, we take into account that (0.5 · ∏-x) '= -1. Now we get -Sin (x).
So, we found the derivative of the cosine, y '= -Sin (x) for the function y = Cos (x).
The derivative of the cosine squared
A frequently used example where a cosine derivative is used. The function y = Cos 2 (x) is complex. First we 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 (2 · x), the sine of the double angle, we obtain the final simplified
answer y '= -Sin (2x)
Hyperbolic functions
They are used in the study of many technical disciplines: in mathematics, for example, facilitate the calculation of integrals, the solution of differential equations. They are expressed in terms of trigonometric functions with an imaginary argument, for example, 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 = (e
x + e
-x ) / 2, this is the hyperbolic cosine ch (x). We use the rule of finding the derivative of the sum of two expressions, the rule of removing the constant factor (Const) by 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 · 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 -1, times e
- x . The difference turned out, and this is the hyperbolic sine sh (x).
Conclusion: (ch (x)) '= sh (x).
Consider, for example, how to calculate the derivative of the function y = ch (x
3 +1).
By the
rule of differentiation of a 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 tabular
When solving examples, there is no need to differentiate them each time according to the proposed scheme; it is enough to use the conclusion.
Example. Differentiate the function y = Cos (x) + Cos 2 (-x) -Ch (5 · x).
It is easy to calculate (we use tabular data), y '= -Sin (x) + Sin (2 · x) -5 · Sh (5 · x).