Derivatives of Inverse Trigonometric Functions

4 Key excerpts on "Derivatives of Inverse Trigonometric Functions"

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  eBook - ePub

  Foundations of Mathematics

  Algebra, Geometry, Trigonometry and Calculus

  • Philip Brown(Author)
  • 2016(Publication Date)
  • Mercury Learning and Information

    (Publisher)

  XAMPLE 8.5.1. Using the sum, product, and quotient rules, respectively, (i)

  8.6DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

    We will use formulas (8.3) and (8.6) to compute the derivatives of trigonometric functions.

  If f(x) = sin(x) and g(x) = cos(x), then

  (These limits are given in formula (7.5).) Furthermore, we can compute f′(x) and g′(x) for any x by making use of the trigonometric identities for sin(A + B) and cos(A + B), the rules for limits and the two limits computed above:

  We have thus obtained the following derivative formulas. 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,

  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

  The following examples demonstrate the product rule combined with the derivatives of trigonometric functions.

  EXAMPLE 8.6.1.

  (i)
  (ii)

  8.7SOME 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 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 differential 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.

  EXAMPLE 8.7.1. A car approaches an intersection with time–displacement function s(t) = t2 −4, where s is measured in meters and t is measured in seconds. (The letter s is the variable that physicists typically use to represent displacement or position, and t is the letter used to represent time, so s(t) is a function that gives the position of the car at any given time.) The value s

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  No longer available |Learn more

  Trigonometry For Dummies

  • Mary Jane Sterling(Author)
  • 2014(Publication Date)
  • For Dummies

    (Publisher)

  Part IV

  Equations and Applications

  Find the geometric properties that are most used when studying trigonometry in a free article at www.dummies.com/extras/trigonometry .

  In this part…

  • Become acquainted with inverse trig functions.
  • Identify the domains and ranges of the inverse trig functions.
  • Recognize the pairings of the quadrants used by each inverse function.
  • Solve trig equations using identities and inverse functions.
  • Write expressions to include infinitely many answers.
  • Find the areas of triangles using trig functions in the formulas.

  Passage contains an image

  Chapter 15

  Investigating Inverse Trig Functions

  In This Chapter

  Acquainting yourself with inverse notation

  Setting limits on inverse trig functions

  Determining domain and range of inverse trig functions

  A s thrilling and fulfilling as the original six trig functions are, they just aren't complete without their inverses. An inverse trig function behaves like the inverse of any other type of function — it undoes what the original function did. In mathematics, functions can have inverses if they're one-to-one, meaning each output value occurs only once. This whole inverse idea is going to take some fast talking when it comes to trig functions, because they keep repeating values over and over as angles are formed with every full rotation of the circle — so you're going to wonder how these functions and inverses can be one-to-one. If you need a refresher on basic inverse functions, flip on back to Chapter 3 for the lowdown on them and how you determine one.

  Writing It Right

  You use inverse trig functions to solve equations such as

  or sec x = –2, or tan 2x = 1. In typical algebra equations, you can solve for the value of x by dividing each side of the equation by the coefficient or by adding the same thing to each side, and so on. But you can't do that with the function .

  Would it make sense to divide each side by sine? “Out, out thou sine!” Here's what you'd get:

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  eBook - ePub

  Elementary Calculus

  An Infinitesimal Approach

  • H. Jerome Keisler(Author)
  • 2013(Publication Date)
  • Dover Publications

    (Publisher)

  π/2, we have

  Thus

  Figure 7.3.6

  We shall next study the derivatives of the inverse trigonometric functions. Here is a general theorem which tells us when the derivative of the inverse function exists and gives a rule for computing its value.

  INVERSE FUNCTION THEOREM

  Suppose a real function f is differentiable on an open interval I and f has an inverse function g. Let x be a point in I where f′(x) ≠ 0 and let y = f(x). Then

  (i)g′(y) exists,

  (ii)

  We omit the proof that g′(y) exists. Intuitively, the curve y = f(x) has a nonhorizontal tangent line, so the curve x = g(y) should have a nonvertical tangent line and thus g′(y) should exist.

  The Inverse Function Rule from Chapter 2 says that (ii) is true if we assume (i). The proof of (ii) from (i) is an application of the Chain Rule:

  The Inverse Function Theorem shows that all the inverse trigonometric functions have derivatives. We now evaluate these derivatives.

  THEOREM 2

  PROOF We prove the first part of (i) and (iii). Since the derivatives exist we may use implicit differentiation.

  (i) Let y = arcsin x. Then

  From sin2 y + cos2 y = 1 we get

  Since –π/2 ≤ y ≤ π/2, cos y ≥ 0. Then

  Substituting,

  (iii) Let y = arcsec x.

  Then

  From tan2 y + 1 = sec2 y we get .

  Since 0 ≤ y ≤ π, tan y and have the same sign.

  Therefore and When we turn these formulas for derivatives around we get some surprising new integration formulas.

  THEOREM 3

  From part (i), arcsin x and –arccos x must differ only by a constant. We already knew this from Example 5,

  Before now we were not able to find the area of the regions under the curves It is a remarkable and quite unexpected fact that these areas are given by inverse trigonometric functions.

  EXAMPLE 6 (a) Find the area of the region under the curve

  for –1 ≤ x ≤ 1.

  (b)Find the area of the region under the same curve for –∞ < x < ∞. The regions are shown in Figure 7.3.7

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  eBook - ePub

  The Calculus

  A Genetic Approach

  • Otto Toeplitz(Author)
  • 2018(Publication Date)
  • University of Chicago Press

    (Publisher)

  30. INVERSE TRIGONOMETRIC FUNCTIONS Purely formally, we obtain:

  FIG. 100

  FIG. 101

  FIG. 102

  Now for exact definitions. Figure 100 shows the graph of sin x , x being given in arc measurement. Each monotonic part of this function gives rise to an inverse function. Even when aiming at a greatest possible connected part, we

  FIG. 103

  FIG. 104

  FIG. 105

  would still have the choice between infinitely many inverse functions. Mathematicians have agreed to use the interval − π/2 ≤ x < π/2.

  In the adjacent group of figures (Figs, 101–6 ) the so-called principal branch of each of the inverse trigonometric functions is shown. The other “branches” for arc tan x are obtained from the principal one by merely adding integer multiples of π: . . . − 2π + arc tan x , − π + arc tan x ,

  arc tan x ,

  π + arc tan x , 2π + arc tan x ,. . . .

  FIG. 106

  31. FUNCTIONS OF SEVERAL FUNCTIONS If we have to differentiate a function such as

  it would be convenient to place cos x = u , sin x = v , such that

  where u and v are each functions of x . How do we now find f ′(x )? Placing u(x) = u1 and we have

  Now, as x 1 approaches

  x, (u1

  − u )/x 1 − x ) approaches u ′(x ) and (v 1 − v )/x 1 − x ) approaches v ′(x ). Therefore

  approaches the derivative of F (u, v ) with respect to v , in which u plays the role of a constant. We designate this derivative as. More difficult is the question as to what becomes of

  For, as x 1 approaches

  x, v 1

  does not remain constant as did u but approaches v . Now it is very plausible that the quotient in question does approach the derivative of F (u, v ) with respect to u , with v being kept constant, which we designate by F ′1 (u , v ); but this is a point which we shall have to consider more closely later on (so far the third point postponed for later discussion). Accepting this result, we then have

  The second expression gives a new way of writing the derivatives F ′1 and F ′2 , where ∂ indicates that only the one of the variables u and v varies while the other stays constant. (As in Section 24, we should really devote an additional discussion to the case of vanishing denominators u 1 − u and v 1 − v

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