Upcoming SlideShare
×

# Lesson 16: Inverse Trigonometric Functions (Section 021 slides)

236 views

Published on

We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.

1 Like
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

Views
Total views
236
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
8
0
Likes
1
Embeds 0
No embeds

No notes for slide

### Lesson 16: Inverse Trigonometric Functions (Section 021 slides)

1. 1. Section 3.5 Inverse Trigonometric Functions V63.0121.021, Calculus I New York University November 2, 2010Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations . . . . . .
2. 2. Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 2 / 40
3. 3. Evaluations: The good “Exceptional competence and effectively articulate. (Do not fire him)” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
4. 4. Evaluations: The good “Exceptional competence and effectively articulate. (Do not fire him)” “Good guy” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
5. 5. Evaluations: The good “Exceptional competence and effectively articulate. (Do not fire him)” “Good guy” “He’s the clear man” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
6. 6. Evaluations: The good “Exceptional competence and effectively articulate. (Do not fire him)” “Good guy” “He’s the clear man” “Love the juices” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 3 / 40
7. 7. Evaluations: The bad Too fast, not enough examples . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
8. 8. Evaluations: The bad Too fast, not enough examples Not enough time to do everything Lecture is not the only learning time (recitation and independent study) I try to balance concept and procedure . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
9. 9. Evaluations: The bad Too fast, not enough examples Not enough time to do everything Lecture is not the only learning time (recitation and independent study) I try to balance concept and procedure Too many proofs . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
10. 10. Evaluations: The bad Too fast, not enough examples Not enough time to do everything Lecture is not the only learning time (recitation and independent study) I try to balance concept and procedure Too many proofs In this course we care about concepts There will be conceptual problems on the exam Concepts are the keys to overcoming templated problems . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 4 / 40
11. 11. Evaluations: technological comments Smart board issues laser pointer visibility slides sometimes move fast . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 5 / 40
12. 12. Evaluations: The ugly “If class was even remotely interesting this class would be awesome.” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
13. 13. Evaluations: The ugly “If class was even remotely interesting this class would be awesome.” “Sometimes condescending/rude.” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
14. 14. Evaluations: The ugly “If class was even remotely interesting this class would be awesome.” “Sometimes condescending/rude.” “Can’t pick his nose without checking his notes, and he still gets it wrong the first time.” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
15. 15. Evaluations: The ugly “If class was even remotely interesting this class would be awesome.” “Sometimes condescending/rude.” “Can’t pick his nose without checking his notes, and he still gets it wrong the first time.” “If I were chained to a desk and forced to see this guy teach, I would chew my arm off in order to get free.” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 6 / 40
16. 16. A slide on slides Pro “Powerpoints explain topics carefully step-by-step” “Powerpoint and lesson flow smoothly” “Can visualize material well” “I like that you post slides beforehand” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 7 / 40
17. 17. A slide on slides Pro “Powerpoints explain topics carefully step-by-step” “Powerpoint and lesson flow smoothly” “Can visualize material well” “I like that you post slides beforehand” Con “I would like to have him use the chalkboard more.” “It’s so unnatural to learn math via powerpoint.” “I hate powerpoint.” . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 7 / 40
18. 18. A slide on slides Pro “Powerpoints explain topics carefully step-by-step” “Powerpoint and lesson flow smoothly” “Can visualize material well” “I like that you post slides beforehand” Con “I would like to have him use the chalkboard more.” “It’s so unnatural to learn math via powerpoint.” “I hate powerpoint.” Why I like them Board handwriting not an issue Easy to put online; notetaking is more than transcription . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 7 / 40
19. 19. My handwriting . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 8 / 40
20. 20. A slide on slides Pro “Powerpoints explain topics carefully step-by-step” “Powerpoint and lesson flow smoothly” “Can visualize material well” “I like that you post slides beforehand” Con “I would like to have him use the chalkboard more.” “It’s so unnatural to learn math via powerpoint.” “I hate powerpoint.” Why I like them Board handwriting not an issue Easy to put online; notetaking is more than transcription What we can do if you have suggestions for details to put in, I’m listening Feel free to ask me to fill in something on the board . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 9 / 40
21. 21. Objectives Know the definitions, domains, ranges, and other properties of the inverse trignometric functions: arcsin, arccos, arctan, arcsec, arccsc, arccot. Know the derivatives of the inverse trignometric functions. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 10 / 40
22. 22. OutlineInverse Trigonometric FunctionsDerivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent ArcsecantApplications . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 11 / 40
23. 23. What is an inverse function?DefinitionLet f be a function with domain D and range E. The inverse of f is thefunction f−1 defined by: f−1 (b) = a,where a is chosen so that f(a) = b. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 12 / 40
24. 24. What is an inverse function?DefinitionLet f be a function with domain D and range E. The inverse of f is thefunction f−1 defined by: f−1 (b) = a,where a is chosen so that f(a) = b.So f−1 (f(x)) = x, f(f−1 (x)) = x . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 12 / 40
25. 25. What functions are invertible?In order for f−1 to be a function, there must be only one a in Dcorresponding to each b in E. Such a function is called one-to-one The graph of such a function passes the horizontal line test: any horizontal line intersects the graph in exactly one point if at all. If f is continuous, then f−1 is continuous. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 13 / 40
26. 26. Graphing the inverse function y . If b = f(a), then f−1 (b) = a. . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
27. 27. Graphing the inverse function y . If b = f(a), then f−1 (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . . b, a) ( . . . a, b) ( . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
28. 28. Graphing the inverse function y . y . =x If b = f(a), then f−1 (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . . b, a) ( On the xy-plane, the point . (b, a) is the reflection of (a, b) in the line y = x. . . a, b) ( . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
29. 29. Graphing the inverse function y . y . =x If b = f(a), then f−1 (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . . b, a) ( On the xy-plane, the point . (b, a) is the reflection of (a, b) in the line y = x. . . a, b) ( . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
30. 30. Graphing the inverse function y . y . =x If b = f(a), then f−1 (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . . b, a) ( On the xy-plane, the point . (b, a) is the reflection of (a, b) in the line y = x. . . a, b) ( . x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
31. 31. Graphing the inverse function y . y . =x If b = f(a), then f−1 (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . . b, a) ( On the xy-plane, the point . (b, a) is the reflection of (a, b) in the line y = x. . . a, b) ( Therefore: . x .FactThe graph of f−1 is the reflection of the graph of f in the line y = x. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
32. 32. Graphing the inverse function y . y . =x If b = f(a), then f−1 (b) = a. So if (a, b) is on the graph of f, then (b, a) is on the graph of f−1 . . b, a) ( On the xy-plane, the point . (b, a) is the reflection of (a, b) in the line y = x. . . a, b) ( Therefore: . x .FactThe graph of f−1 is the reflection of the graph of f in the line y = x. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 14 / 40
33. 33. arcsinArcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . x . π π s . in − . . 2 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
34. 34. arcsinArcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . . x . π π s . in − . . . 2 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
35. 35. arcsinArcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . y . =x . . . . x . π π s . in − . . . 2 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
36. 36. arcsinArcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . rcsin a . . . . x . π π s . in − . . . 2 2 . The domain of arcsin is [−1, 1] [ π π] The range of arcsin is − , 2 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 15 / 40
37. 37. arccosArccos is the inverse of the cosine function after restriction to [0, π] y . c . os . . x . 0 . . π . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
38. 38. arccosArccos is the inverse of the cosine function after restriction to [0, π] y . . c . os . . x . 0 . . π . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
39. 39. arccosArccos is the inverse of the cosine function after restriction to [0, π] y . y . =x . c . os . . x . 0 . . π . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
40. 40. arccosArccos is the inverse of the cosine function after restriction to [0, π] . . rccos a y . . c . os . . . x . 0 . . π . The domain of arccos is [−1, 1] The range of arccos is [0, π] . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 16 / 40
41. 41. arctanArctan is the inverse of the tangent function after restriction to[−π/2, π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 t .an . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
42. 42. arctanArctan is the inverse of the tangent function after restriction to[−π/2, π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 t .an . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
43. 43. arctan y . =xArctan is the inverse of the tangent function after restriction to[−π/2, π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 t .an . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
44. 44. arctanArctan is the inverse of the tangent function after restriction to[−π/2, π/2]. y . π . a . rctan 2 . x . π − . 2 The domain of arctan is (−∞, ∞) ( π π) The range of arctan is − , 2 2 π π lim arctan x = , lim arctan x = − x→∞ 2 x→−∞ 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 17 / 40
45. 45. arcsecArcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π, 3π/2]. y . . x . 3π π π 3π − . − . . . 2 2 2 2 s . ec . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
46. 46. arcsecArcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π, 3π/2]. y . . . x . 3π π π 3π − . − . . . . 2 2 2 2 s . ec . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
47. 47. arcsecArcsecant is the inverse of secant after restriction to . = x y[0, π/2) ∪ (π, 3π/2]. y . . . x . 3π π π 3π − . − . . . . 2 2 2 2 s . ec . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
48. 48. arcsec 3π . 2Arcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π, 3π/2]. . . y π . 2 . . . x . . The domain of arcsec is (−∞, −1] ∪ [1, ∞) [ π ) (π ] The range of arcsec is 0, ∪ ,π 2 2 π 3π lim arcsec x = , lim arcsec x = x→∞ 2 x→−∞ 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 18 / 40
49. 49. Values of Trigonometric Functions π π π π x 0 6 4 3 2 √ √ 1 2 3 sin x 0 1 2 2 2 √ √ 3 2 1 cos x 1 0 2 2 2 1 √ tan x 0 √ 1 3 undef 3 √ 1 cot x undef 3 1 √ 0 3 2 2 sec x 1 √ √ 2 undef 3 2 2 2 csc x undef 2 √ √ 1 2 3 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 19 / 40
50. 50. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 20 / 40
51. 51. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2Solution π 6 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 20 / 40
52. 52. What is arctan(−1)? . 3 . π/4 . . . . − . π/4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
53. 53. What is arctan(−1)? . ( ) 3 . π/4 3π . Yes, tan = −1 4 √ 2 s . in(3π/4) = 2 . √ . 2 . os(3π/4) = − c 2 . − . π/4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
54. 54. What is arctan(−1)? . ) ( 3 . π/4 3π . Yes, tan = −1 4 √ But, the) ( π π range of arctan is 2 s . in(3π/4) = − , 2 2 2 . √ . 2 . os(3π/4) = − c 2 . − . π/4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
55. 55. What is arctan(−1)? . ( ) 3 . π/4 3π . Yes, tan = −1 4 But, the) ( π π range of arctan is √ − , 2 2 2 c . os(π/4) = . 2 Another angle whose . π tangent is −1 is − , and √ 4 2 this is in the right range. . in(π/4) = − s 2 . − . π/4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
56. 56. What is arctan(−1)? . ( ) 3 . π/4 3π . Yes, tan = −1 4 But, the) ( π π range of arctan is √ − , 2 2 2 c . os(π/4) = . 2 Another angle whose . π tangent is −1 is − , and √ 4 2 this is in the right range. . in(π/4) = − s π 2 So arctan(−1) = − 4 . − . π/4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 21 / 40
57. 57. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2Solution π 6 π − 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 22 / 40
58. 58. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2Solution π 6 π − 4 3π 4 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 22 / 40
59. 59. Caution: Notational ambiguity . in2 x =.(sin x)2 s . in−1 x = (sin x)−1 s sinn x means the nth power of sin x, except when n = −1! The book uses sin−1 x for the inverse of sin x, and never for (sin x)−1 . 1 I use csc x for and arcsin x for the inverse of sin x. sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 23 / 40
60. 60. OutlineInverse Trigonometric FunctionsDerivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent ArcsecantApplications . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 24 / 40
61. 61. The Inverse Function TheoremTheorem (The Inverse Function Theorem)Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in anopen interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b))In Leibniz notation we have dx 1 = dy dy/dx . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 25 / 40
62. 62. The Inverse Function TheoremTheorem (The Inverse Function Theorem)Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in anopen interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b))In Leibniz notation we have dx 1 = dy dy/dxUpshot: Many times the derivative of f−1 (x) can be found by implicitdifferentiation and the derivative of f: . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 25 / 40
63. 63. Illustrating the Inverse Function Theorem.ExampleUse the inverse function theorem to find the derivative of the square rootfunction.. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 26 / 40
64. 64. Illustrating the Inverse Function Theorem.ExampleUse the inverse function theorem to find the derivative of the square rootfunction.Solution (Newtonian notation) √Let f(x) = x2 so that f−1 (y) = y. Then f′ (u) = 2u so for any b > 0 we have 1 (f−1 )′ (b) = √ 2 b. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 26 / 40
65. 65. Illustrating the Inverse Function Theorem.ExampleUse the inverse function theorem to find the derivative of the square rootfunction.Solution (Newtonian notation) √Let f(x) = x2 so that f−1 (y) = y. Then f′ (u) = 2u so for any b > 0 we have 1 (f−1 )′ (b) = √ 2 bSolution (Leibniz notation)If the original function is y = x2 , then the inverse function is defined by x = y2 .Differentiate implicitly: dy dy 1 1 = 2y =⇒ = √ dx dx 2 x. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 26 / 40
66. 66. Derivation: The derivative of arcsinLet y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
67. 67. Derivation: The derivative of arcsinLet y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a righttriangle: . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
68. 68. Derivation: The derivative of arcsinLet y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a righttriangle: . . y . = arcsin x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
69. 69. Derivation: The derivative of arcsinLet y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a righttriangle: 1 .. .. x y . = arcsin x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
70. 70. Derivation: The derivative of arcsinLet y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a righttriangle: 1 .. .. x y . = arcsin x . √ . 1 − x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
71. 71. Derivation: The derivative of arcsinLet y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a righttriangle: √ cos(arcsin x) = 1 − x2 1 .. .. x y . = arcsin x . √ . 1 − x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
72. 72. Derivation: The derivative of arcsinLet y = arcsin x, so x = sin y. Then dy dy 1 1 cos y = 1 =⇒ = = dx dx cos y cos(arcsin x)To simplify, look at a righttriangle: √ cos(arcsin x) = 1 − x2 So 1 .. .. xFact y . = arcsin x d 1 . √ arcsin(x) = √ . 1 − x2 dx 1 − x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 27 / 40
73. 73. Graphing arcsin and its derivative 1 .√ 1 − x2 The domain of f is [−1, 1], but the domain of f′ is . . rcsin a (−1, 1) lim f′ (x) = +∞ x→1− lim f′ (x) = +∞ . | . . | x→−1+ − . 1 1 . . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 28 / 40
74. 74. Composing with arcsinExampleLet f(x) = arcsin(x3 + 1). Find f′ (x). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 29 / 40
75. 75. Composing with arcsinExampleLet f(x) = arcsin(x3 + 1). Find f′ (x).SolutionWe have d 1 d 3 arcsin(x3 + 1) = √ (x + 1) dx 1 − (x3 + 1)2 dx 3x2 =√ −x6 − 2x3 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 29 / 40
76. 76. Derivation: The derivative of arccosLet y = arccos x, so x = cos y. Then dy dy 1 1 − sin y = 1 =⇒ = = dx dx − sin y − sin(arccos x) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 30 / 40
77. 77. Derivation: The derivative of arccosLet y = arccos x, so x = cos y. Then dy dy 1 1 − sin y = 1 =⇒ = = dx dx − sin y − sin(arccos x)To simplify, look at a righttriangle: √ sin(arccos x) = 1 − x2So 1 . √ . 1 − x2Fact y . = arccos x d 1 . arccos(x) = − √ x . dx 1 − x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 30 / 40
78. 78. Graphing arcsin and arccos . . rccos a . . rcsin a . | . |. . − . 1 1 . . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 31 / 40
79. 79. Graphing arcsin and arccos . . rccos a Note (π ) cos θ = sin −θ . . rcsin a 2 π =⇒ arccos x = − arcsin x 2 . | . |. . So it’s not a surprise that their − . 1 1 . derivatives are opposites. . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 31 / 40
80. 80. Derivation: The derivative of arctanLet y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 y . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
81. 81. Derivation: The derivative of arctanLet y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 yTo simplify, look at a righttriangle: . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
82. 82. Derivation: The derivative of arctanLet y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 yTo simplify, look at a righttriangle: . y . = arctan x . . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
83. 83. Derivation: The derivative of arctanLet y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 yTo simplify, look at a righttriangle: .. x y . = arctan x . . 1 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
84. 84. Derivation: The derivative of arctanLet y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 yTo simplify, look at a righttriangle: √ . 1 + x2 .. x y . = arctan x . . 1 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
85. 85. Derivation: The derivative of arctanLet y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 yTo simplify, look at a righttriangle: 1 cos(arctan x) = √ 1 + x2 √ . 1 + x2 .. x y . = arctan x . . 1 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
86. 86. Derivation: The derivative of arctanLet y = arctan x, so x = tan y. Then dy dy 1 sec2 y = 1 =⇒ = = cos2 (arctan x) dx dx sec2 yTo simplify, look at a righttriangle: 1 cos(arctan x) = √ 1 + x2 √ . 1 + x2 .. x SoFact y . = arctan x d 1 . . arctan(x) = 1 . dx 1 + x2 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 32 / 40
87. 87. Graphing arctan and its derivative y . . /2 π a . rctan 1 . 1 + x2 . x . − . π/2 The domain of f and f′ are both (−∞, ∞) Because of the horizontal asymptotes, lim f′ (x) = 0 x→±∞ . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 33 / 40
88. 88. Composing with arctanExample √Let f(x) = arctan x. Find f′ (x). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 34 / 40
89. 89. Composing with arctanExample √Let f(x) = arctan x. Find f′ (x).Solution d √ 1 d√ 1 1 arctan x = (√ )2 x= · √ dx 1+ x dx 1+x 2 x 1 = √ √ 2 x + 2x x . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 34 / 40
90. 90. Derivation: The derivative of arcsecTry this first. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
91. 91. Derivation: The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x)) . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
92. 92. Derivation: The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a righttriangle: . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
93. 93. Derivation: The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a righttriangle: . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
94. 94. Derivation: The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a righttriangle: y . = arcsec x . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
95. 95. Derivation: The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a righttriangle: x . y . = arcsec x . 1 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
96. 96. Derivation: The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a righttriangle: √ x2 − 1 tan(arcsec x) = 1 √ x . . x2 − 1 y . = arcsec x . 1 . . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
97. 97. Derivation: The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then dy dy 1 1 sec y tan y = 1 =⇒ = = dx dx sec y tan y x tan(arcsec(x))To simplify, look at a righttriangle: √ x2 − 1 tan(arcsec x) = 1 So √ x . . x2 − 1Fact y . = arcsec x . d 1 arcsec(x) = √ 1 . dx x x2 − 1 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 35 / 40
98. 98. Another ExampleExampleLet f(x) = earcsec 3x . Find f′ (x). . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 36 / 40
99. 99. Another ExampleExampleLet f(x) = earcsec 3x . Find f′ (x).Solution 1 f′ (x) = earcsec 3x · √ ·3 3x (3x)2 − 1 3earcsec 3x = √ 3x 9x2 − 1 . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 36 / 40
100. 100. OutlineInverse Trigonometric FunctionsDerivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent ArcsecantApplications . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 37 / 40
101. 101. ApplicationExampleOne of the guiding principles ofmost sports is to “keep youreye on the ball.” In baseball, abatter stands 2 ft away fromhome plate as a pitch is thrownwith a velocity of 130 ft/sec(about 90 mph). At what ratedoes the batter’s angle of gazeneed to change to follow theball as it crosses home plate? . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 38 / 40
102. 102. ApplicationExampleOne of the guiding principles ofmost sports is to “keep youreye on the ball.” In baseball, abatter stands 2 ft away fromhome plate as a pitch is thrownwith a velocity of 130 ft/sec(about 90 mph). At what ratedoes the batter’s angle of gazeneed to change to follow theball as it crosses home plate?SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0. . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 38 / 40
103. 103. SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0. y . 1 . 30 ft/sec . θ . . 2 . ft . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
104. 104. SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · dt 1 + (y/2)2 2 dt y . 1 . 30 ft/sec . θ . . 2 . ft . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
105. 105. SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · dt 1 + (y/2)2 2 dt When y = 0 and y′ = −130,then y . dθ 1 1 = · (−130) = −65 rad/sec 1 . 30 ft/sec dt y=0 1+0 2 . θ . . 2 . ft . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
106. 106. SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ 1 1 dy = · dt 1 + (y/2)2 2 dt When y = 0 and y′ = −130,then y . dθ 1 1 = · (−130) = −65 rad/sec 1 . 30 ft/sec dt y=0 1+0 2The human eye can only track . θ .at 3 rad/sec! . 2 . ft . . . . . . V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 39 / 40
107. 107. Summary y y′ 1 arcsin x √ 1 − x2 1 arccos x − √ Remarkable that the 1 − x2 derivatives of these 1 arctan x transcendental functions 1 + x2 are algebraic (or even 1 rational!) arccot x − 1 + x2 1 arcsec x √ x x2 − 1 1 arccsc x − √ x x2 − 1 . . . . . .V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 40 / 40