To solve any mathematical exercise the first thing you should do is read it carefully, then you must identify if it is a single trigonometric function or there is more than one function in order to apply the rule of addition, product, subtraction or the quotient of two functions. Once this is done, each of the trigonometric functions is identified and it is possible to evaluate if it is possible to use some of the trigometric identities that will help you to rationalize or factor the functions. Then, when deriving each function you must use the table of derived formulas for these functions, for this you must also take into account if you have to apply the rule of the chain, this is when we present a composite function. First we must review the table of derivatives of trigonometric functions:
(sen x)' = (cos x) (cos x)' = (-sen x)
(tan x)' = (sec2 x) (sec x)' = (sec
x.tan x)
(cosec)' =
(-cosec x.cotg x) (cotang x)' = (-cosec2 x)
It is also useful to consider that (ex)' = (ex) y (Ln
x)' = (1/x)
Calculate the derivative of the following compound trigonometric functions in which the chain rule should be applied:
A) F(x) = sen x2
B) F (x) = cos4 x Here the most external function is g (x) = x4 and the innermost function is h (x) = cos x, so the rule of the chain and the rule must be applied to derive functions with positive integral rational exponents, then F '(x) = F (goh)' = g (h (x)) '. h (x)' = 4cos3 x.(-sen x)
C) F (x) = etan x F(x) = cos4 x Here the most external function is g (x) = ex and the innermost function is h (x) = tan x, so the rule must be applied of the string and the derivative of the tangent function then F '(x) = F (goh)' = g (h (x)) '. h (x)' = etan x.(tan x)' = etan x.(sec2 x)
D) F (x, y) = cos (x + y) To derive this function we must apply the trigonometric identity for the sum of cosines:
cos (x + y) = cos x.cos y + sin x.sen y Then:
F (x, y) '= (cos x.cos y + sin x.sen y)' Here the rule for sum of function derivatives and the product rule of two derivatives applies
F (x, y) '= (-sen x.cos y + cos x.-sin y) + (cos x.sen y + sin x.cos y)
E) F (x) = sin x .sec x Here the rule of the product of two functions must be applied, whether g (x) = sin x and h (x) = sec x Then:
F '(x) = (cos x.sec x) + (sin x. (Sec x.tan x)) Remember to look in the derivative table for the derivative of the sine and tangent function.
Now let's see some examples to derive trigonometric functions applying the quotient rule, but first let's see this table of trigometric identities that can help to rationalize the functions and thus solve them easily: Sen (a + b) = sin a.sen b + cos a.cos b tan x = sin x / cos x cos (a + b) = cos a.cos b + sin a.sen b sen2 x + cos2 x = 1 cos a.sec a = 1 cotg x = cos x / sin x 1) F (x) = 1 / (tan x + 2x) We have a quotient of a constant between a sum of functions so the rule is applied to derive a constant between a function F'(x)= (-1/( tan x + 2x)2).( tan x + 2x)' = (-1/( tan x + 2x)2).( sec2 x + 2)' 2) F (x) = (sin x.tan x) / e We have the product of two trigonometric functions between a constant, the number "e" which is the base of the Neperian logarithms. The rule must be applied to derive a function between a constant. Then F '(x) = f' (x) / constant = (sin x.tan x) '/ e F '(x) = (sin x.tan x)' / e here we have the product of the sine function by the tangent function so the product rule must be applied ((cos x.tan x) + (sen x.sec2 x)) / 2,71828182 Here the rule was applied to derive a function divided by a constant. 3) F (x) = (2cos x) / (x + 1) here we have the quotient of a trigonometric function and a polynomial function then: F '(x) = ((2cos x)'. (X + 1) + (2cos x). (X + 1) ') / (x+ 1)2 F '(x) = ((-2sen x). (X + 1) + (2cos x). (X)) / (x+ 1)2
4) F (x) = (x3sen x) / (cot x)1/2 In this function we have the quotient of the product of two functions and the function cot x with positive rational exponent. Then:1) : [(3x2.sen x + x3.cos x). (cot x)1/2 - (x3sen x).1/2(cot x)-1/2.(-cosec2 x)] / ((cot x)1/2 )25) F (x) = x / (cos x.sec x) In this function we observe that the denominator can be rationalized using the following trigonometric identity:(cos a.sec a) = 1 Then this facilitates the calculation of the derivative. F (x) = x / 1 and F (x) '= (x)' / 1 = 1/1 Let's see some examples to derive trigometric functions using the technique of implicit derivation, if you have a function where "y" is a function of "x" and you ask to calculate dy / dx all the variables must be derived applying the different derivation techniques, but each time the variable "y" is derived, the component "dy / dx" or the component y 'must be added to this result. Observe that the following equations define "y" as the implicit function of "x" calculate the derivative of "y". a) F (x) = x cos y + y = 2x (1.cos y + x.-sin y. (y ')) + y' = 2 then grouping terms where it appears and 'you have y'. - x.sen y) = 2 / cos and finally: y '= (2 - cos y) / (1 - x.sen y) b) F (x) = tan y6 + xsen y = 2x in this equation the product rule and the rule for positive integer powers must be applied then: (Sec2 y6.tan y6)6y5. y '+ (sin y + x.cos y. y') = 2 grouping terms similar to clear to y' you have y'((sec2 y6.tan y6)6y5 + (sin y + x.cos y)) = 2
y '= 2 / ((sec2 y6.tan y6)6y5 + (sin y + x.cos y)) c) F (x) = sin y.ey + 2xy = cos x. ex+1 in the first part of this equation we have the sum of 2 functions in which the product rule must be applied and in the second part the product rule must also be applied. a) (cos y. y'. ey + sen y.ey. y') + 2.y + 2x. y' = (-sen x. ex+1 + cos x. ex+1 ) y'.(cos y. ey + sen y.ey + 2x) = (-sen x. ex+1 + cos x. ex+1 – 2y) y' = (-sen x. ex+1 + cos x. ex+1 – 2y)/(cos y. ey + sen y.ey + 2x) d) F (x) = 5tang xe + 2y2 = csc 5 in the first part of this equation we have a composite function, the most external function is so x and the innermost one is xn then we have:5sec2 xe.(ex) + 4y. y' = -cos 5.cot 5 y' = [-cos 5.cot x - 5sec2 xe.(ex)] /4y e) F (x) = 9cot x3 + y3 = sen x in the first function of this equation is the compound function cot x with xn -cosec2 x3. (3x2) + 3y2. y' = cos 5 y' = cos 5/ (-cos2 x3.(3x2) + 3y2) In this part of the post I want to explain how to use the logarithmic derivation technique to derive trigometric and inverse trigonometric functions, this technique involves the use of the chain rule and is very useful to derive functions that include products, quotients or very complicated powers. To apply this technique you must know the properties of natural logarithms that will help facilitate calculations. If you have a function defined for each x of your domain by f (x) which is positive for all x then the steps to derive logarithmically are:
1) The function must be written in the form y = f (x) 2) Natural logarithm is applied to the function: Ln y = Ln f (x). and it is solved by applying the properties of the natural logarithm. 3) It is derived implicitly with respect to the independent variable in this case x, thus: (Ln y) '= (Ln f (x))' (1 / y). y '= (Ln f (x))' 4) Multiply by and all the equation resulting: y '= y. (Ln f (x)) ' 5) It is changed and by f (x) resulting: f '(x) = y' = f (x). (Ln f (x)) ' Now let's take a brief look at the properties of natural logarithms: Ln (a / b) = Ln a - Ln b Ln (a)x = x. Ln (a) Ln (a.b .... Z) = Ln a + Ln b + ...... + Ln z Let's see 5 examples to better understand the above: 1) y = sen x. x2 + 2x/ (tan x) Ln y = Ln (sen x. x2 + 2xtan x) Ln y = Ln sen x + 2. Ln x + Ln 2x - Ln tan x (1/y). y' = 1/sen x + 2/x + 2/(2x) – (1/tan x).sec2 x y' = y. [ 1/sen x + 2/x + 2/(2x). – (1/tan x).sec2 x] y' = (sen x. x2 + 2x/ (tan x)).[ 1/sen x + 2/x + 2/(2x). – (1/tan x).sec2 x] 2) y = (cos x/sec x)1/3. xtanx Ln y = Ln ((cos x/sec x)1/3. xtanx) Ln y = Ln cos x – 1/3Ln sec x + tan x. Lnx (1/y). y' = -sen x/cos x - 1/3. (tan x.sec x) /sec x + tan x.(1/x) y' = y. [ -sen x/cos x - 1/3. (tan x.sec x) /(sec x ) + tan x.(1/x)] y' = ((cos x/sec x)1/3. xtan x). [ -sen x/cos x - 1/3. (tan x.sec x)/(sec x) + tan x.(1/x)]
3) y = (cotan x)sen x
Ln y = Ln ((cotan x)sen x) = sen x.Ln cotan x
(1/y).
y' = sen x. (1/cotan x). (cotan x) '
(1/y). y' = sen x. (1/cotan x). (cosec2
x)
y' = y. [ sen x. (1/cotan x). (cosec2
x)]
y' = ((cotan x)sen x). [ sen x.
(1/cotan x). (cosec2 x)]
4) y = (tan x)x - x5
Ln y = Ln (tan x)x - Lnx5)
Ln y = x. Ln (tan x) – 5Lnx
Ln y = x. (1/tan x). (tan x)' – 5/x = x. (1/tan
x). (sec2 x) – 5/x
(1/y). y' = x. (1/tan x). (sec2
x) – 5/x
y' = y. [ x. (1/tan x). (sec2
x) – 5/x]
y' = ((tan x)x - x5).[
x. (1/tan x).(sec2 x) – 5/x]
5) y = (tan x)2/ xe
Ln y = Ln (tan x)2 – Ln xe)
Ln y = 2Ln (tan x) - e. Ln x
(1/y). y' = (2/tan x).sec2 x -
e/x
y' = y. [ (2/tan x).sec2 x - e/x]
y' = ((tan x)2/ xe). [ (2/tan x).sec2 x - e/x] Ln y = Ln (sin x, x2 + 2xtan x) Ln y = Ln sin x + 2. Ln x + Ln 2x - Ln tan x Well I hope that these exercises will help you soon I will complement this post adding exercises on how to derive trigonometric functions given in parametric form and derived from inverse trigonometric functions.I hope you have been useful, any suggestions or comments let me know. If you want you can help me grow this blog by sharing on your social networks or with a small donation: BTC 1Eqtwfvkgq4oDFWF3tSNL9X9xfvaZBqHqA LTC LXUGWJkS3oGLj7ycP2RXUkDkNghJDwSNvG ETH 0x8c0B9E801861d83D0a3d2f3D84cF88237C490968
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