5128_Ch03_pp098-184

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Section 3.8 Derivatives of Inverse Trigonometric Functions 171 Standardized Test Questions You may use a graphing calculator to solve the following problems. 35. True or False The domain of y sin 1 x is 1 x 1. Justify your answer. True. By definition of the function. 36. True or False The domain of y tan 1 x is 1 x 1. Justify your answer. False. The domain is all real numbers. d 37. Multiple Choice Which of the following is sin 1 d x x 2 ? (A) 4 2 x 2 (B) 1 4 x 2 (C) 2 4 x 2 (D) 4 2 x 2 (E) 1 4 x 2 d 38. Multiple Choice Which of the following is tan 1 (3x)? d x (A) 1 39x 2 (B) 1 19x 2 (C) 1 19x 2 E D Extending the Ideas 47. Identities Confirm the following identities for x 0. (a) cos 1 x sin 1 x p2 (b) tan 1 x cot 1 x p2 (c) sec 1 x csc 1 x p2 48. Proof Without Words The figure gives a proof without words that tan 1 1 tan 1 2 tan 1 3 p. Explain what is going on. 3 (D) 1 39x 2 (E) 1 9x 2 d 39. Multiple Choice Which of the following is sec 1 (x 2 )? d x 2 (A) xx 4 1 (B) 2 xx 2 1 (C) 2 x1 x 4 (D) x1 2 x 2 (E) 2x 1 x 4 40. Multiple Choice Which of the following is the slope of the tangent line to y tan 1 (2x) at x 1? C (A) 25 (B) 15 (C) 25 (D) 52 (E) 5 Explorations In Exercises 41–46, find (a) the right end behavior model, (b) the left end behavior model, and (c) any horizontal tangents for the function if they exist. 41. y tan 1 x 42. y cot 1 x 43. y sec 1 x 44. y csc 1 x 45. y sin 1 x 46. y cos 1 x A 49. (Continuation of Exercise 48) Here is a way to construct tan 1 1, tan 1 2, and tan 1 3 by folding a square of paper. Try it and explain what is going on. Fold 2 Fold 1 Fold 3 tan –1 3 tan –1 2 tan –1 1 1 13. . ⏐2s 1⏐s 2 s 2 15. (x 2 1)x 2 2 1 17. 1 t 2 1 19. 2t t 1 d 32. cos 1 d (x) d x d x 2 sin1 x d 0 sin 1 x d x 1 1 x 2 d 33. cot 1 d x d x d x 2 tan1 x d 0 tan 1 x d x 1 1 x 2 d 34. csc 1 d (x) d x d x d 0 sec 1 x d x 1 ⏐x⏐x 2 1 2 sec1 x 41. (a) y 2 (b) y 2 (c) None 42. (a) y 0 (b) y (c) None 43. (a) y 2 (b) y 2 (c) None 44. (a) y 0 (b) y 0 (c) None 45. (a) None (b) None (c) None 46. (a) None (b) None (c) None
172 Chapter 3 Derivatives 3.9 What you’ll learn about • Derivative of e x • Derivative of a x • Derivative of lnx • Derivative of log a x • Power Rule for Arbitrary Real Powers . . . and why The relationship between exponential and logarithmic functions provides a powerful differentiation tool called logarithmic differentiation. X –.03 –.02 –.01 0 .01 .02 .03 X=0 [–4.9, 4.9] by [–2.9, 2.9] (a) Y1 .98515 .99007 .99502 ERROR 1.005 1.0101 1.0152 (b) Figure 3.56 (a) The graph and (b) the table support the conclusion that 1 lim h→0 eh 1. h Derivatives of Exponential and Logarithmic Functions Derivative of e x At the end of the brief review of exponential functions in Section 1.3, we mentioned that the function y e x was a particularly important function for modeling exponential growth. The number e was defined in that section to be the limit of 1 1x x as x→. This intriguing number shows up in other interesting limits as well, but the one with the most interesting implications for the calculus of exponential functions is this one: 1 lim h→0 eh 1. h (The graph and the table in Figure 3.56 provide strong support for this limit being 1. A formal algebraic proof that begins with our limit definition of e would require some rather subtle limit arguments, so we will not include one here.) The fact that the limit is 1 creates a remarkable relationship between the function e x and its derivative, as we will now see. d e d x x lim e xh e x h→0 h h e x lim e x • h→0 eh lim h→0 ( e x • eh 1 h e x • lim h→0 ( e x • 1 e x ) ) 1 eh h In other words, the derivative of this particular function is itself! d e d x x e x If u is a differentiable function of x, then we have d e d x u e u d u . dx We will make extensive use of this formula when we study exponential growth and decay in Chapter 6. EXAMPLE 1 Using the Formula Find dydx if y e (xx2) . SOLUTION Let u x x 2 then y e u . Then dy e u du du , and 1 2x. dx dx dx dy Thus, e u du e (xx2) (1 2x). dx dx Now try Exercise 9.
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5128_Ch03_pp098-184

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