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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 …

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.

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- 1. Section 3.5 Inverse Trigonometric Functions V63.0121.041, Calculus I New York University November 1, 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. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 2 / 32
- 3. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 3 / 32
- 4. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
- 5. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
- 6. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 5 / 32
- 7. OutlineInverse Trigonometric FunctionsDerivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent ArcsecantApplications . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 6 / 32
- 8. arcsinArcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . x . π π s . in − . . 2 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
- 9. arcsinArcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . . x . π π s . in − . . . 2 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
- 10. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
- 11. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
- 12. arccosArccos is the inverse of the cosine function after restriction to [0, π] y . c . os . . x . 0 . . π . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
- 13. arccosArccos is the inverse of the cosine function after restriction to [0, π] y . . c . os . . x . 0 . . π . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
- 14. arccosArccos is the inverse of the cosine function after restriction to [0, π] y . y . =x . c . os . . x . 0 . . π . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
- 15. 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] . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
- 16. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
- 17. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
- 18. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
- 19. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
- 20. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
- 21. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
- 22. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
- 23. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
- 24. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 11 / 32
- 25. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
- 26. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2Solution π 6 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
- 27. What is arctan(−1)? . 3 . π/4 . . . . − . π/4 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
- 28. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
- 29. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
- 30. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
- 31. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
- 32. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2Solution π 6 π − 4 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
- 33. Check: Values of inverse trigonometric functionsExampleFind arcsin(1/2) arctan(−1) ( √ ) 2 arccos − 2Solution π 6 π − 4 3π 4 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
- 34. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 15 / 32
- 35. OutlineInverse Trigonometric FunctionsDerivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent ArcsecantApplications . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 16 / 32
- 36. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
- 37. 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: . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
- 38. Illustrating the Inverse Function Theorem.ExampleUse the inverse function theorem to find the derivative of the square rootfunction.. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
- 39. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
- 40. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
- 41. 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
- 42. 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: . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
- 43. 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 . . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
- 44. 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
- 45. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
- 46. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
- 47. 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 . So d 1 arcsin(x) = √ y . = arcsin x dx 1 − x2 . √ . 1 − x2 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
- 48. 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 . . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 20 / 32
- 49. Composing with arcsinExampleLet f(x) = arcsin(x3 + 1). Find f′ (x). . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
- 50. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
- 51. 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) . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
- 52. 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 − x2 1 . √ . 1 − x2So d 1 y . = arccos x arccos(x) = − √ . dx 1 − x2 x . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
- 53. Graphing arcsin and arccos . . rccos a . . rcsin a . | . |. . − . 1 1 . . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
- 54. 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. . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
- 55. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
- 56. 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: . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
- 57. 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 . . 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
- 58. 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
- 59. 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
- 60. 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
- 61. 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 . So d 1 y . = arctan x arctan(x) = . dx 1 + x2 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
- 62. 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→±∞ . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 25 / 32
- 63. Composing with arctanExample √Let f(x) = arctan x. Find f′ (x). . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
- 64. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
- 65. Derivation: The derivative of arcsecTry this first. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 66. 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)) . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 67. 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: . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 68. 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: . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 69. 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 . . 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 70. 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 71. 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 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 72. 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 So d 1 arcsec(x) = √ . y . = arcsec x dx x x2 − 1 1 . . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
- 73. Another ExampleExampleLet f(x) = earcsec 3x . Find f′ (x). . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
- 74. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
- 75. OutlineInverse Trigonometric FunctionsDerivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent ArcsecantApplications . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 29 / 32
- 76. 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? . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
- 77. 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
- 78. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
- 79. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
- 80. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
- 81. 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
- 82. 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 . . . . . .V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 32 / 32

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