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Lesson 17: Inverse Trigonometric Functions

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Information about Lesson 17: Inverse Trigonometric Functions
Education

Published on March 11, 2009

Author: leingang

Source: slideshare.net

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The inverse trig functions are transcendental, but their derivatives are algebraic!
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Section 3.5 Inverse Trigonometric Functions V63.0121, Calculus I March 11–12, 2009 Announcements Get half of your unearned ALEKS points back by March 22 . . . . . .

What functions are invertible? In order for f−1 to be a function, there must be only one a in D corresponding 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. . . . . . .

Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant . . . . . .

arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . x . s . in π π − − . . 2 2 . . . . . .

arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . . . x . s . in π π − − . . 2 2 . . . . . .

arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . . =x y . . . x . s . in π π − − . . 2 2 . . . . . .

arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. y . a . rcsin . . . x . s . in π π − − . . 2 2 The domain of arcsin is [−1, 1] [ π π] The range of arcsin is − , 22 . . . . . .

arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . c . os . . x . π . 0 . . . . . . .

arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . c . os . . x . π . 0 . . . . . . .

arccos Arccos is the inverse of the cosine function after restriction to [0, π] y . . =x y c . os . . x . π . 0 . . . . . . .

arccos Arccos is the inverse of the cosine function after restriction to [0, π] a . rccos y . c . os . . x . π . 0 . The domain of arccos is [−1, 1] The range of arccos is [0, π] . . . . . .

arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. y . . x . π π 3π 3π − − . . . . 2 2 2 2 t . an . . . . . .

arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. y . . x . π π 3π 3π − − . . . . 2 2 2 2 t . an . . . . . .

arctan Arctan is the inverse of the tangent function after restriction to . =x y [−π/2, π/2]. y . . x . π π 3π 3π − − . . . . 2 2 2 2 t . an . . . . . .

arctan Arctan 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 − , 22 π π lim arctan x = , lim arctan x = − 2 x→−∞ 2 x→∞ . . . . . .

Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant . . . . . .

Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) . . . . . .

Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and 1 (f−1 )′ (b) = ′ −1 f (f (b)) “Proof”. If y = f−1 (x), then f(y) = x, So by implicit differentiation dy dy 1 1 f′ (y) = 1 =⇒ =′ = ′ −1 dx dx f (y) f (f (x)) . . . . . .

The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 = 1 =⇒ = = cos y dx dx cos y cos(arcsin x) . . . . . .

The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 = 1 =⇒ = = cos y dx dx cos y cos(arcsin x) To simplify, look at a right triangle: . . . . . . .

The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 = 1 =⇒ = = cos y dx dx cos y cos(arcsin x) To simplify, look at a right triangle: 1 . x . . . . . . . .

The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 = 1 =⇒ = = cos y dx dx cos y cos(arcsin x) To simplify, look at a right triangle: 1 . x . . = arcsin x y . . . . . . .

The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 = 1 =⇒ = = cos y dx dx cos y cos(arcsin x) To simplify, look at a right triangle: 1 . x . . = arcsin x y .√ . 1 − x2 . . . . . .

The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 = 1 =⇒ = = cos y dx dx cos y cos(arcsin x) To simplify, look at a right triangle: √ cos(arcsin x) = 1 − x2 1 . x . . = arcsin x y .√ . 1 − x2 . . . . . .

The derivative of arcsin Let y = arcsin x, so x = sin y. Then dy dy 1 1 = 1 =⇒ = = cos y dx dx cos y cos(arcsin x) To simplify, look at a right triangle: √ cos(arcsin x) = 1 − x2 1 . x . So d 1 arcsin(x) = √ . = arcsin x y 1 − x2 .√ dx . 1 − x2 . . . . . .

Graphing arcsin and its derivative 1 .√ 1 − x2 a . rcsin . | | . . − .1 1 . . . . . . .

The derivative of arccos Let y = arccos x, so x = cos y. Then dy dy 1 1 − sin y = 1 =⇒ = = − sin y − sin(arccos x) dx dx . . . . . .

The derivative of arccos Let y = arccos x, so x = cos y. Then dy dy 1 1 − sin y = 1 =⇒ = = − sin y − sin(arccos x) dx dx To simplify, look at a right triangle: √ sin(arccos x) = 1 − x2 √ 1 . . 1 − x2 So d 1 . = arccos x y arccos(x) = − √ . 1 − x2 dx x . . . . . . .

Graphing arcsin and arccos a . rccos a . rcsin . | | . . − .1 1 . . . . . . .

Graphing arcsin and arccos a . rccos Note (π ) −θ cos θ = sin 2 a . rcsin π =⇒ arccos x = − arcsin x 2 . So it’s not a surprise that their | | . . − .1 1 . derivatives are opposites. . . . . . .

The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = cos2 (arctan x) = 1 =⇒ = sec2 y dx dx . . . . . .

The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = cos2 (arctan x) = 1 =⇒ = sec2 y dx dx To simplify, look at a right triangle: . . . . . . .

The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = cos2 (arctan x) = 1 =⇒ = sec2 y dx dx To simplify, look at a right triangle: x . . 1 . . . . . . .

The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = cos2 (arctan x) = 1 =⇒ = sec2 y dx dx To simplify, look at a right triangle: x . . = arctan x y . 1 . . . . . . .

The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = cos2 (arctan x) = 1 =⇒ = sec2 y dx dx To simplify, look at a right triangle: √ x . . 1 + x2 . = arctan x y . 1 . . . . . . .

The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = cos2 (arctan x) = 1 =⇒ = sec2 y dx dx To simplify, look at a right triangle: 1 cos(arctan x) = √ √ 1 + x2 x . . 1 + x2 . = arctan x y . 1 . . . . . . .

The derivative of arctan Let y = arctan x, so x = tan y. Then dy dy 1 sec2 y = cos2 (arctan x) = 1 =⇒ = sec2 y dx dx To simplify, look at a right triangle: 1 cos(arctan x) = √ √ 1 + x2 x . . 1 + x2 So d 1 . = arctan x y arctan(x) = . 1 + x2 dx 1 . . . . . . .

Graphing arctan and its derivative y . a . rctan 1 . x . 1 + x2 . . . . . .

Example √ x. Find f′ (x). Let f(x) = arctan . . . . . .

Example √ x. Find f′ (x). Let f(x) = arctan Solution √ d√ d 1 1 1 ·√ (√ )2 arctan x = x= 1+x 2 x dx x dx 1+ 1 =√ √ 2 x + 2x x . . . . . .

Recap y′ y 1 √ arcsin x 1 − x2 1 Remarkable that the arccos x − √ 1 − x2 derivatives of these 1 transcendental functions arctan x 1 + x2 are algebraic (or even 1 − arccot x rational!) 1 + x2 1 √ arcsec x x x2 − 1 1 arccsc x − √ x x2 − 1 . . . . . .

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