How to derive the cosine derivative

The cosine derivative is analogous to the derivative of the sine, the basis of the proof is the definition of the limit of the function. You can use a different method, using trigonometric formulas for casting cosine and sine angles. To express one function through another is a cosine through a sine, and to differentiate the sine with a complex argument.

Consider the first example of the derivation of the formula (Cos (x)) '

We give an infinitesimal increment Δx to the argument x of the function y = Cos (x). With the new value of the argument x + Δx, we get a new value of the function Cos (x + Δx). Then the increment of the function Δy will be 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 perform the identical transformations in the numerator of the resulting fraction. Recall the formula for the difference of 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 partial lim of this product on Δx for Δx tending to zero. It is known that the first (it is called remarkable) limit lim (Sin (Δx / 2) / (Δx / 2)) is 1, and the limit -Sin (x + Δx / 2) is -Sin (x) for Δx tending to Zero.
Write the result: the derivative (Cos (x)) 'is - Sin (x).

Some people like the second way of deriving the same formula

From the course of trigonometry is known: Cos (x) is equal to Sin (0.5 · Π-x), similarly to Sin (x) is Cos (0,5 · Π-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 · Π-x) · (0.5 · Π-x) ', because the derivative of the sine x is equal to the cosine of x. We turn to the second formula Sin (x) = Cos (0.5 · Π-x) of the cosine-to-sine change; we take into account that (0.5 · Π-x) '= -1. Now we get -Sin (x).
Thus, we have found the cosine derivative, y '= -Sin (x) for the function y = Cos (x).

Square cosine derivative

Often 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, this will be 2 · Cos (x), then multiply it by the derivative (Cos (x)) ', which is -Sin (x). We obtain y '= -2 · Cos (x) · Sin (x). When we apply the formula Sin (2 · x), the sine of the double angle, we get the final simplified
The answer y '= -Sin (2 · x)

Hyperbolic functions

Applied in the study of many technical disciplines: in mathematics, for example, facilitate the calculation of integrals, the solution of differential equations. They are expressed through trigonometric functions with an imaginary argument, so the hyperbolic cosine ch (x) = Cos (i · x), where i is the imaginary unit, the hyperbolic sine sh (x) = Sin (i · x).
The hyperbolic cosine derivative is easily calculated.
Consider the function y = (e ^x + e ^-x ) / 2, this is the hyperbolic cosine of ch (x). We use the rule of finding the derivative of the sum of two expressions, the rule for carrying out a constant factor (Const) behind 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 summand. (X (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, multiplied by e ^{- x} . The result is a difference, and this is the hyperbolic sine of sh (x).
Conclusion: (ch (x)) '= sh (x).
Let's consider the example of how to calculate the derivative of the function y = ch (x ³ +1).
By the rule for differentiating 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 functions y = ch (x) and y = Cos (x) are tabular

When solving examples it is not necessary to differentiate them each time according to the proposed scheme, it is sufficient to use the derivation.
Example. Differentiate the function y = Cos (x) + Cos ² (-x) -Ch (5 · x).
It is easy to calculate (use the tabular data), y '= -Sin (x) + Sin (2 · x) -5 · Sh (5 · x).