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College Calculus: Level I Inverse Trigonometric Functions

Section 3: Derivatives, part 1: Lecture 8 | 27:05 min

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Inverse Trigonometric Functions

Check with your instructor as to whether or not you should memorize these formulas!
One type of problem here simply incorporates inverse trigonometric functions into differentiation problems involving, for example, the Chain Rule.
You can use implicit differentiation to derive these formulas.
Check with your instructor as to whether or not you can use the “triangle approach” to derive these formulas.

Inverse Trigonometric Functions

Find f′(x) if f(x) = e^arcsinx

f′(x) = [d/dx] e^arcsinx
= e^arcsinx [d/dx] arcsinx
= e^arcsinx [1/(√{1 − x²})] =

[(e^arcsinx)/(√{1 − x²})]

Find f′(x) if f(x) = cos(arcsinx)

f′(x) = −sin(arcsinx) [d/dx] arcsinx
= [(−sin(arcsinx))/(√{1− x²})] =

[(−x)/(√{1 − x²})]

Find f′(x) if f(x) = [(e^x)/arccosx]

f′(x) = [(arccosx [d/dx] e^x − e^x [d/dx] arccosx)/((arccosx)²)]
= [(e^x arccosx − e^x [(−1)/(√{1 − x²})])/((arccosx)²)]

[(e^x arccosx + [(e^x)/(√{1 − x²})])/((arccosx)²)]

Find f′(x) if f(x) = lnx (arctanx)

f′(x) = lnx [d/dx] arctanx + arctanx [d/dx] lnx

lnx ([1/(1 + x²)]) + arctanx ([1/x])

Find f′(x) if f(x) = √{x² + cot⁻¹ x}

u = x² + cot⁻¹ x , u′ = 2x − [1/(1 + x²)]
f′(x) = [1/2] u^−[1/2] u′
= [1/(2 √u)] u′

[1/(2 √{x² + cot⁻¹ x})] (2x − [1/(1 + x²)])

Find f′(x) if f(x) = sec⁻¹ (x³ − x)

u = x³ − x , u′ = 3x² − 1
f′(x) = [1/(|u| √{u² − 1 })] u′

[(3x² − 1)/(|x³ − x| √{(x³ − x)² − 1})]

Find dy given y = csc⁻¹ e^x

u = e^x , u′ = e^x
dy = [(−1)/(|u| √{u² − 1})] u′
= [(−u′)/(|u| √{u² − 1})]
= [(−e^x)/(|e^x| √{(e^x)² − 1})]
= [(−e^x)/(e^x √{e^2x − 1})]

[(−1)/(√{e^2x − 1})]

Find f′(x) if f(x) = ln(arcsin(x)arccos(x))

f(x) = lnarcsinx + lnarccosx
f′(x) = [1/arcsinx] [d/dx] arcsinx + [1/arccosx] [d/dx] arccosx
= [1/(√{1 − x²} arcsinx)] − [1/(√{1 − x²} arccosx)]

[1/(√{1 − x²})]([1/arcsinx] − [1/arccosx])

Find f′(x) if f(x) = [(cos⁻¹ x )/cosx]

f′(x) = [(cosx [d/dx] cos⁻¹ x + cos⁻¹ x [d/dx] cosx)/(cos² x)]
= [(cosx [(−1)/(√{1 − x²})] + cos⁻¹ x (−sinx))/(cos² x)]

[([(−cosx)/(√{1 − x²})] − sinx cos⁻¹ x)/(cos² x)]

Prove dy = [1/(√{1 − x²})] given y = sin⁻¹ x

y = sin⁻¹ x
siny = sin(sin⁻¹ x)
siny = x
[d/dx] siny = [d/dx] x
cosy dy = 1
dy = [1/cosy]
Using the following identity
cos² y + sin² y = 1
cos² y = 1 − sin² y
cosy = ±√{1 − sin² y}
We have two potential solutions, but we know something about the range and domain of these functions. For y = sin⁻¹ x, the range is [−[(π)/2], [(π)/2]]. Using what we know about y, we can determine acceptable values for cosy. With those inputs, cosy ≥ 0
cosy = √{1 − sin² y} , x = siny
dy = [1/cosy]
dy = [1/(√{1 − sin² y})]

[1/(√{1 − x²})]

Given f([(π)/6]) = [1/2] with f(x) = sinx and g(x) = sin⁻¹ x, find g([1/2])

An inverse function will "reverse" the action done by the original function. Kind of like working backwards.
g(x) is the inverse of f(x)

g([1/2]) = [(π)/6]

Find the domain of f(x) = sin⁻¹x

The only valid inputs for x are given by the range of its inverse, sinx, [-1, 1]

Domain [-1, 1]

Find sin⁻¹([(√2)/2])

sin([(π)/4]) = [(√2)/2]

sin⁻¹([(√2)/2]) = [(π)/4]
There are technically many inputs for sin(x) that give [(√2)/2], but remember it must remain in the range of sin⁻¹ x, which is [−[(π)/2], [(π)/2]].

Find cos⁻¹(1)

Range of cos⁻¹ x is [0,π].
cos(0) = 1

cos⁻¹(1) = 0

Find sin(cos⁻¹[(√2)/2])

cos([(π)/4]) = [(√2)/2]
cos⁻¹([(√2)/2]) = [(π)/4]
sin(cos⁻¹[(√2)/2]) = sin([(π)/4])

= [(√2)/2]

Find sin⁻¹[1/2]

sin([(π)/6]) = [1/2]
sin⁻¹([1/2]) = [(π)/6]

Find tan(cos⁻¹[(√2)/2])

cos⁻¹[(√2)/2] = [(π)/4]
tan(cos⁻¹[(√2)/2]) = tan[(π)/4]
= 1

Given f(x) = sin² x and g(x) = sin⁻¹ (sin3x), evaluate f(g(x))

g(x) = sin⁻¹ (sin3x)
= 3x

f(g(x)) = sin²3x
SPECIAL NOTE: By convention sin⁻¹ x is the inverse of sinx. (sinx)⁻¹ is cscx.

Find at least one solution for θ given sin(θ) = 0

Take the inverse sin of both sides
sin(θ) = 0
sin⁻¹(sin(θ)) = sin⁻¹(0)

θ = 0

Graph f(x) = 2 sin⁻¹(3x + 1)

We know the input for inverse sin must be within [-1, 1]. Therefore, −1 ≤ 3x + 1 ≤ 1. So [(−2)/3] ≤ x ≤ 0. Domain [[(−2)/3], 0]. Plugging in some values...
f([(−2)/3]) = 2 sin⁻¹(−1) = 2 [(− π)/2] = −π
f([(−1)/3]) = 2 sin⁻¹(0) = 0
f(0) = 2 sin⁻¹1 = 2 [(π)/2] = π
Increasing function with Range [−π, π]

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Answer

Inverse Trigonometric Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Six Basic Inverse Trigonometric Functions

Lecture Example 1

Lecture Example 2

Lecture Example 3

Additional Example 4

Additional Example 5

Intro 0:00
Six Basic Inverse Trigonometric Functions 0:10
Lecture Example 1 1:11
Lecture Example 2 8:53
Lecture Example 3 12:37
Additional Example 4
Additional Example 5

College Calculus 1 Online Course

Section 1: Overview of Functions
	Review of Functions	26:29
	Compositions of Functions	12:29
Section 2: Limits
	Average and Instantaneous Rates of Change	20:59
	Limit Investigations	22:37
	Algebraic Evaluation of Limits	28:19
	Formal Definition of a Limit	23:39
	Continuity and the Intermediate Value Theorem	19:09
Section 3: Derivatives, part 1
	Limit Definition of the Derivative	22:52
	The Power Rule	26:01
	The Product Rule	14:54
	The Quotient Rule	19:17
	Applications of Rates of Change	17:43
	Trigonometric Derivatives	26:58
	The Chain Rule	23:47
	Inverse Trigonometric Functions	27:05
	Equation of a Tangent Line	15:52
Section 4: Derivatives, part 2
	Implicit Differentiation	30:05
	Higher Derivatives	13:16
	Logarithmic and Exponential Function Derivatives	17:42
	Hyperbolic Trigonometric Function Derivatives	14:30
	Related Rates	29:05
	Linear Approximation	23:52
Section 5: Application of Derivatives
	Absolute Minima and Maxima	18:57
	Mean Value Theorem and Rolle's Theorem	20:00
	First Derivative Test, Second Derivative Test	27:11
	L'Hopital's Rule	23:09
	Curve Sketching	40:16
	Applied Optimization	25:37
	Newton's Method	25:13
Section 6: Integrals
	Approximating Areas and Distances	36:50
	Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus	22:02
	Substitution Method for Integration	23:19
Section 7: Application of Integrals, part 1
	Area Between Curves	19:59
	Volume by Method of Disks and Washers	24:22
	Volume by Method of Cylindrical Shells	30:29
	Average Value of a Function	16:31
Section 8: Extra
	Graphs of f, f', f''	23:58
	Slope Fields for Differential Equations	18:32
	Separable Differential Equations	17:04

Related Books

	Calculus by Larson, 9th Edition Authors: Ron Larson, Bruce H. Edwards ISBN: 0547167024 Publisher: Brooks Cole Year: 2009 The book is filled with examples and detailed explanations. Capstone exercises are also included for each section and synthesize the main concepts into a single example.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 1 Author: Ron Larson ISBN: 547213093 Publisher: Brooks Cole Year: 2008 This solutions manual includes worked out solutions to every odd-numbered exercise in Calculus of a Single Variable, 9th edition.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 2 Authors: Ron Larson, Bruce H. Edwards ISBN: 547213107 Publisher: Brooks Cole Year: 2009 This solutions manual includes worked out solutions to every odd-numbered exercise in Multivariable Calculus, 9th edition.

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Professor Switkes

Inverse Trigonometric Functions

Slide Duration:

Table of Contents

Section 1: Overview of Functions

Review of Functions

26m 29s

Intro