Textbook Question
Evaluate the integrals in Exercises 31–78.
45. ∫(ln x)^(-3)/x dx
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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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- 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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Video duration:
2mKey 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.
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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.
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06:35Derivatives of Other Inverse Trigonometric Functions
Related Practice
Textbook Question
Evaluate the integrals in Exercises 31–78.
55. ∫(from -2 to -1)e^(-(x+1)) dx
Textbook Question
Evaluate the integrals in Exercises 31–78.
47. ∫(1/r)csc²(1+ln(r))dr
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Evaluate the integrals in Exercises 31–78.
43. ∫tan(ln v)/v dv
Textbook Question
Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
104. lim(x→4) (sin²(πx))/(e^(x-4) + 3 - x)
Textbook Question
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
15. y = sin⁻¹√(1-u²), 0<u<1