Textbook Question
Evaluate the integrals in Exercises 31–78.
47. ∫(1/r)csc²(1+ln(r))dr
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.PE.23
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
23. y = arccsc(secθ), 0<θ<π/2
Verified step by step guidance1
Recognize that the function is given as \(y = \arccsc(\sec \theta)\), where \(0 < \theta < \frac{\pi}{2}\). Our goal is to find \(\frac{dy}{d\theta}\).
Recall the derivative formula for the inverse cosecant function: if \(y = \arccsc(u)\), then \(\frac{dy}{dx} = -\frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dx}\), where \(u\) is a function of \(x\).
Identify the inner function \(u = \sec \theta\). Next, find the derivative of \(u\) with respect to \(\theta\): \(\frac{du}{d\theta} = \sec \theta \tan \theta\).
Substitute \(u = \sec \theta\) and \(\frac{du}{d\theta} = \sec \theta \tan \theta\) into the derivative formula for \(\arccsc(u)\):
\[\frac{dy}{d\theta} = -\frac{1}{|\sec \theta| \sqrt{(\sec \theta)^2 - 1}} \cdot \sec \theta \tan \theta.\]
Simplify the expression inside the square root using the identity \(\sec^2 \theta - 1 = \tan^2 \theta\), and simplify the entire expression accordingly to express \(\frac{dy}{d\theta}\) in terms of \(\theta\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Trigonometric Functions
Inverse trigonometric functions, like arccsc and sec, reverse the effect of their corresponding trigonometric functions. Understanding their domains and ranges is essential, as well as knowing how to express and differentiate them. For example, arccsc(x) is the inverse of csc(x), defined for |x| ≥ 1.
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06:35
Derivatives of Other Inverse Trigonometric Functions
Chain Rule
The chain rule is a differentiation technique used when a function is composed of another function, such as y = arccsc(secθ). It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Recommended video:
05:02Intro to the Chain Rule
Derivatives of Inverse Trigonometric Functions
Each inverse trig function has a specific derivative formula, for example, d/dx[arccsc(x)] = -1 / (|x|√(x² - 1)). Knowing these formulas allows you to differentiate expressions involving inverse trig functions accurately, especially when combined with the chain rule.
Recommended video:
06:35Derivatives of Other Inverse Trigonometric Functions
Related Practice
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Textbook Question
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
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