Mathematics
Derivatives of Sec, Csc and Cot
The derivatives of secant, cosecant, and cotangent functions can be found using the chain rule and the fundamental trigonometric identities. The derivative of secant is secant multiplied by tangent, the derivative of cosecant is negative cosecant multiplied by cotangent, and the derivative of cotangent is negative cosecant squared. These derivatives are important in calculus and are used in various applications.
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7 Key excerpts on "Derivatives of Sec, Csc and Cot"
eBook - PDFCalculus
Resequenced for Students in STEM
- David Dwyer, Mark Gruenwald(Authors)
- 2017(Publication Date)
- Wiley(Publisher)
Solution d dx [sec x] = d dx 1 cos x = cos x · d dx [1] - 1 · d dx [cos x] cos 2 x = cos x · 0 - 1(- sin x) cos 2 x = sin x cos 2 x = 1 cos x sin x cos x = sec x tan x For convenience, we summarize the differentiation formulas for all six trigono- metric functions below. Proofs of the differentiation formulas for csc x and cot x are left for Exercises 21 and 22. Derivatives of Trigonometric Functions d dx [sin x] = cos x d dx [cos x] = - sin x d dx [tan x] = sec 2 x d dx [cot x] = - csc 2 x d dx [sec x] = sec x tan x d dx [csc x] = - csc x cot x It is helpful to memorize these formulas. Fortunately, there is a pattern that cuts our work in half. To express this relationship, we need to define cofunctions. Figure 3.29 shows a right triangle with an acute angle θ. Because the sum of the measures of the angles in a triangle is 180 ◦ , or π, the other acute angle must be the complement to θ; that is, it must be π 2 - θ. Since the side adjacent to θ is 160 CHAPTER 3. THE DERIVATIVE the side opposite π 2 - θ and vice versa, the following relationships hold: sin π 2 - θ = cos θ cos π 2 - θ = sin θ tan π 2 - θ = cot θ cot π 2 - θ = tan θ sec π 2 - θ = csc θ csc π 2 - θ = sec θ It is for this reason that, for example, sine and cosine are said to be cofunctions of each other. (The “co-” stands for complementary.) Thus, tangent and cotangent are cofunctions, as are secant and cosecant. More generally, if g(x) = f ( π 2 - x ) , then we’ll say that f and g are cofunctions. Now we can express the pattern in the a b c θ π 2 -θ Figure 3.29 trigonometric differentiation formulas as follows: Rule of Thumb The derivative of a cofunction is the negative cofunction of the derivative. Thus, for example, if we can remember that the derivative of tan x is sec 2 x, then the formula for the derivative of cot x comes for free—we take the negative of the cofunction of sec 2 x to obtain - csc 2 x.
eBook - PDF- Michael A. Calter, Paul A. Calter, Paul Wraight, Sarah White(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
799 Section 33–2 ◆ Derivatives of the Tangent, Cotangent, Secant, and Cosecant Functions and, since u u 1 cos sec , 2 2 = dy dx u du dx sec 2 = Thus, Derivative of the Tangent d u dx u du dx (tan ) sec 2 = 352 The derivative of the tangent of some function is the secant squared of that function, multiplied by the derivative of that function. ◆◆◆ Example 10: (a) If y = 3 tan x 2 , then y x x x x 3(sec )(2 ) 6 sec 2 2 2 2 ′ = = (b) If y = 2 sin 3x tan 3x, then, by the product rule, y x x x x x x x 2[sin 3 (sec 3 )(3) tan 3 (cos 3 )(3)] 6 sin 3 sec 3 6 sin 3 2 2 ′ = + = + Derivatives of cot u, sec u, and csc u Each of these derivatives can be obtained using the rules for the sine, cosine, and tangent already derived and using the following identities: u u u u u u u cot cos sin sec 1 cos csc 1 sin = = = We list those derivatives here, together with those already found. Derivatives of the Trigonometric Functions d u dx u du dx (sin ) cos = 350 d u dx u du dx (cos ) sin = - 351 d u dx u du dx (tan ) sec 2 = 352 d u dx u du dx (cot ) csc 2 = - 353 d u dx u u du dx (sec ) sec tan = 354 d u dx u u du dx (csc ) csc cot = - 355 ◆◆◆ You should memorize at least the first three of these. Notice how the signs alternate. The derivative of each cofunction is negative. 800 Chapter 33 ◆ Derivatives of Trigonometric, Logarithmic, and Exponential Functions ◆◆◆ Example 11: (a) If y = sec(x 3 - 2x), then y x x x x x x x x x x [sec( 2 ) tan( 2 )](3 2) (3 2) sec( 2 ) tan( 2 ) 3 3 2 2 3 3 ′ = - - - = - - - (b) If y = cot 3 5x, then, by the power rule and the chain rule, y x x x x 3(cot 5 ) ( csc 5 )(5) 15 cot 5 csc 5 2 2 2 2 ′ = - = - ◆◆◆ Example 12: In Fig. 33-7, link L is pivoted at B but slides along the fixed pin P. As slider C moves in the slot at a constant rate of 4.26 cm/s, the angle θ changes. Find the rate at which θ is changing when S = 6.00 cm. Estimate: When S = 6.00 cm, θ is equal to arctan ( ) 8.63 6.00 = 55.2°.
eBook - PDF- James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Table of Derivatives of Trigonometric Functions d dx ssin xd - cos x d dx scsc xd - 2csc x cot x d dx scos xd - 2sin x d dx ssec xd - sec x tan x d dx stan xd - sec 2 x d dx scot xd - 2csc 2 x When you memorize this table, it is helpful to notice that the minus signs go with the derivatives of the “cofunctions,” that is, cosine, cosecant, and cotangent. EXAMPLE 2 Differentiate f s xd - sec x 1 1 tan x . For what values of x does the graph of f have a horizontal tangent? SOLUTION The Quotient Rule gives f 9 s xd - s1 1 tan xd d dx ssec xd 2 sec x d dx s1 1 tan xd s1 1 tan xd 2 - s1 1 tan xd sec x tan x 2 sec x sec 2 x s1 1 tan xd 2 - sec x stan x 1 tan 2 x 2 sec 2 xd s1 1 tan xd 2 - sec x stan x 2 1d s1 1 tan xd 2 ssec 2 x - tan 2 x 1 1d The graph of f has horizontal tangents when f 9 s xd - 0. Because sec x is never 0, we see that f 9 s xd - 0 when tan x - 1, and this occurs when x - y4 1 n, where n is an integer (see Figure 3). n Trigonometric functions are often used in modeling real-world phenomena. In par- ticular, vibrations, waves, elastic motions, and other quantities that vary in a periodic manner can be described using trigonometric functions. In the following example we discuss an instance of simple harmonic motion. EXAMPLE 3 An object fastened to the end of a vertical spring is stretched 4 cm beyond its rest position and released at time t - 0. (See Figure 4 and note that the downward direction is positive.) Its position at time t is s - f std - 4 cos t Find the velocity and acceleration at time t and use them to analyze the motion of the object. 3 _3 _π 2π π FIGURE 3 The horizontal tangents in Example 2 s 0 4 _4 FIGURE 4 Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
No longer available |Learn moreFoundations of Mathematics
Algebra, Geometry, Trigonometry and Calculus
- Philip Brown(Author)
- 2016(Publication Date)
- Mercury Learning and Information(Publisher)
= = - d dx x x d dx x x sin( ) cos( ), cos( ) sin( ) (8.22) The derivatives of the remaining four trigonometric ratios can be obtained by means of the reciprocal trigonometric identities and the quotient and product rules. For example, d dx x d dx x x x x x x x x x x x x tan( ) sin( ) cos( ) cos( ) cos( ) sin( ) ( sin( )) cos ( ) cos ( ) sin ( ) cos ( ) 1 cos ( ) sec ( ). 2 2 2 2 2 2 = = ⋅ - ⋅ - = + = = The proofs of the remaining trigonometric ratios are left as exercises. It is well worthwhile memorizing their derivatives given in table 8.1. TABLE 8.1. Derivatives of trigonometric ratios d dx x x sin( ) cos( ) = d dx x x cos( ) sin( ) = - d dx x x tan( ) sec 2 = d dx x x cot( ) csc 2 = - d dx x x x sec( ) sec( ) tan( ) = d dx x x x csc( ) csc( ) cot( ) = - The following examples demonstrate the product rule combined with the derivatives of trigo- nometric functions. EXAMPLE 8.6.1. (i) d dx x x x x x x x x sin( ) sin( ) cos( ) sin( ) cos( ) = + = + (ii) d dx x x x x x x x sec( ) 2 sec( ) sec( ) tan( ) 2 2 = + 8.7 SOME BASIC APPLICATIONS OF CALCULUS By means of the examples below, we demonstrate how the techniques of calculus can be used to solve many kinds of practical problems. In the first example, we introduce the important notion of instantaneous rate of change and explain how it relates to the derivative of a time–displacement function of an object in motion. The general study or science of motion is called dynamics, and, because of this relationship of the Differential Calculus • 263 derivative to instantaneous change, calculus is an essential tool in dynamics. A more general and sophisticated expression for a moving body than a simple time–displacement function is a differen- tial equation. Students often ask why they need to learn calculus. The best answer to this question is that calculus is the foundation for the mathematics of differential equations.
eBook - PDF- Geoffrey Berresford, Andrew Rockett(Authors)
- 2015(Publication Date)
- Cengage Learning EMEA(Publisher)
(Review Exercises 48–49.) ● ● Find the derivative of one of these other functions. (Review Exercises 50–55.) d dt tan t 5 sec 2 t d dt cot t 5 2 csc 2 t d dt sec t 5 sec t tan t d dt csc t 5 2 csc t cot t y x t r ( x , y ) y x t r ( x , y ) Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 583 Review Exercises and Chapter Test 8 Review Exercises and Chapter Test indicate a Chapter Test exercise. ● Find the tangent line to a curve at a given point. (Review Exercise 56.) ● Find the indefinite integral of one of the other trigonometric functions. (Review Exercises 57–62.) ● Find the definite integral of one of the other trigonometric functions. (Review Exercises 63–64.) ● Find the area under a trigonometric curve. (Review Exercises 65– 66.) Sections 8.3, 8.4, and 8.5 ● Solve a more involved graphing, differentiation, integration, or numerical integration problem in-volving a trigonometric function. (Review Exercises 67–86.) Hints and Suggestions ● We measure angles in radians because it simplifies the differentiation and integration formulas. All equiv-alent measures can be found by using the equation 180° 5 p radians. ● Think of p as a natural unit for radian measure, just as it occurs naturally in the formulas for the circumfer-ence and area of a circle. ● The “triangle” definitions such as sin t 5 opp hyp hold only for acute angles, while the “coordinate” definitions such as sin t 5 y r hold for all angles. ● We can find sin t and cos t exactly for some angles, such as p 6 , p 4 , and p 3 , using 30° 2 60° 2 90° and 45° 2 45° 2 90° triangles.
eBook - PDFCalculus
Early Transcendental Single Variable
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
79 2 One of the crowning achievements of calculus is its ability to capture continuous motion mathematically, allowing that motion to be analyzed instant by instant. THE DERIVATIVE Many real-world phenomena involve changing quantities—the speed of a rocket, the inflation of currency, the number of bacteria in a culture, the shock intensity of an earthquake, the voltage of an electrical signal, and so forth. In this chapter we will develop the concept of a “derivative,” which is the mathematical tool for studying the rate at which one quantity changes relative to another. The study of rates of change is closely related to the geometric concept of a tangent line to a curve, so we will also be discussing the general definition of a tangent line and methods for finding its slope and equation. 2.1 TANGENT LINES AND RATES OF CHANGE In this section we will discuss three ideas: tangent lines to curves, the velocity of an object moving along a line, and the rate at which one variable changes relative to another. Our goal is to show how these seemingly unrelated ideas are, in actuality, closely linked. TANGENT LINES In Example 1 of Section 1.1 we used an informal argument to find the equation of a tangent line to a curve. However, at that stage in the text we did not have a precise definition of a tangent line. Now that limits have been defined precisely we can give a mathematical definition of the tangent line to a curve y = f (x) at a point P(x 0 , f (x 0 )) on the curve. As illustrated in Figure 2.1.1, the slope m PQ of the secant line through P and a second point Q(x, f (x)) on the graph of f is m PQ = f (x) − f (x 0 ) x − x 0 If we let x approach x 0 , then the point Q will move along the curve and approach the point P. Suppose the slope m PQ of the secant line through P and Q approaches a limit as x → x 0 . In that case we can take the value of the limit to be the slope m tan of the tangent line at P. Thus, we make the following definition. Figure 2.1.1
eBook - PDFCalculus
Late Transcendentals
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
59 2 One of the crowning achievements of calculus is its ability to capture continuous motion mathematically, allowing that motion to be analyzed instant by instant. THE DERIVATIVE Many real-world phenomena involve changing quantities—the speed of a rocket, the inflation of currency, the number of bacteria in a culture, the shock intensity of an earthquake, the voltage of an electrical signal, and so forth. In this chapter we will develop the concept of a “derivative,” which is the mathematical tool for studying the rate at which one quantity changes relative to another. The study of rates of change is closely related to the geometric concept of a tangent line to a curve, so we will also be discussing the general definition of a tangent line and methods for finding its slope and equation. 2.1 TANGENT LINES AND RATES OF CHANGE In this section we will discuss three ideas: tangent lines to curves, the velocity of an object moving along a line, and the rate at which one variable changes relative to another. Our goal is to show how these seemingly unrelated ideas are, in actuality, closely linked. TANGENT LINES In Example 1 of Section 1.1 we used an informal argument to find the equation of a tangent line to a curve. However, at that stage in the text we did not have a precise definition of a tangent line. Now that limits have been defined precisely we can give a mathematical definition of the tangent line to a curve y = f (x) at a point P(x 0 , f (x 0 )) on the curve. As illustrated in Figure 2.1.1, the slope m PQ of the secant line through P and a second point Q(x, f (x)) on the graph of f is m PQ = f (x) − f (x 0 ) x − x 0 If we let x approach x 0 , then the point Q will move along the curve and approach the point P. Suppose the slope m PQ of the secant line through P and Q approaches a limit as x → x 0 . In that case we can take the value of the limit to be the slope m tan of the tangent line at P. Thus, we make the following definition. Figure 2.1.1
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