derivative of cosine is similar to the derivative of the sine, the basis of the evidence - definition of the limit function.You can use the other method using trigonometric formulas for bringing the sine and cosine of angles.To express one function through another - through a sine cosine and sine differentiate with a complex argument.
Consider the first example of the derivation of (Cos (x)) '
Give a negligible increment △ x x argument of the function y = Cos (x).With the new value of the argument x + △ x we obtain a new value of the function Cos (x + △ x).Then increment Δu will still function Cos (x + Δx) -Cos (x).
same ratio to the increment of the function will be the △ x: (Cos (x + Δx) -Cos (x)) / △ x.We carry out identity transformations in the numerator of the resulting fractions.Recall the formula for the difference of cosines, the result is the product of -2Sin (△ x / 2) multiplied by Sin (x + △ x / 2).We find the limit of the private lim this work on when △ x △ x approaches zero.It is known that the first (called remarkable) limit lim (Sin (△ x / 2) / (△ x / 2)) is 1 and the limit -Sin (x + △ x / 2) is -Sin (x) during Δx, tends tozero.
record the results: the derivative (Cos (x)) 'is - Sin (x).
Some prefer the second method of deriving the same formula
Of course we know trigonometry: Cos (x) is Sin (0,5 · Π-x), similar to Sin (x) is equal to Cos (0,5 · Π-x).Then differentiable complex function - the sine of an additional angle (instead of the cosine of X).
obtain a product of Cos (0,5 · Π-x) · (0,5 · Π-x) ', because the derivative of the sine of x is equal to cosine of x.We appeal to the second formula Sin (x) = Cos (0,5 · Π-x) replace the sine cosine, take into account that (0,5 · Π-x) = -1.Now we get -Sin (x).
So, we find the derivative of the cosine, y '= -Sin (x) for the function y = Cos (x).
derivative of cosine squared
often used an example where the derivative of the cosine is used.The function y = Cos2 (x) complex.We find the first differential power function with an exponent 2, it will be 2 · Cos (x), then multiply it by the derivative (Cos (x)) ', which is equal to -Sin (x).Obtain y '= -2 · Cos (x) · Sin (x).When we apply the formula Sin (2 * x) sine of double angle, we get the final answer simple
y '= -Sin (2 * x)
Hyperbolic functions
applied in the study of many technical disciplines in mathematics, for example, make it easier to calculate integralssolution of differential equations.They are expressed in terms of trigonometric functions with imaginary argument, so the hyperbolic cosine ch (x) = Cos (i · x), where i - imaginary unit, the hyperbolic sine sh (x) = Sin (i · x).
hyperbolic cosine is calculated simply.
Consider the function y = (ex + ex) / 2, this is the hyperbolic cosine ch (x).Use the rule for finding the derivative of the sum of two expressions, the right to make a constant factor (Const) for the sign of the derivative.The second term is 0.5 x e s - a complex function of (its derivative is equal to 0.5 · e-x), 0.5 x Ex first term.(ch (x)) = ((EX + ex) / 2) 'can be written differently: (0.5 + 0.5 · EX · e-x) = 0.5 · 0.5 · EX-e-x, because the derivative of the (ex) 'is -1, umnnozhennaya on ex.The result was the difference, and this is the hyperbolic sine sh (x).
Conclusion: (ch (x)) '= sh (x).
Rassmitrim an example of how to calculate the derivative of the function y = ch (x3 + 1).
the rule for differentiating hyperbolic cosine with a complex argument of the '= sh (x3 + 1) · (x3 + 1)' where (x3 + 1) = 3 · x2 + 0.
Answer: the derivative of this function is 3 · x2 · sh (x3 + 1).
derivatives discussed functions in = ch (x) and y = Cos (x) table
In solving examples of each time there is no need to differentiate them on the proposed scheme, it is sufficient to use the output.
example.Differentiate the function y = Cos (x) + Cos2 (-x) -Ch (5 · x).
easy to compute (use tabular data), from '= -Sin (x) + Sin (2 * x) -5 · Sh (5 · x).
