Textbook Question
Evaluate the integrals in Exercises 33–54.
∫ (e^(√r) / √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 7.7.29
In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
29. y = (1 - t)coth⁻¹(√t)
Verified step by step guidance1
Identify the function given: \(y = (1 - t) \, \text{coth}^{-1}(\sqrt{t})\). We need to find \(\frac{dy}{dt}\), the derivative of \(y\) with respect to \(t\).
Recognize that \(y\) is a product of two functions of \(t\): \(u = (1 - t)\) and \(v = \text{coth}^{-1}(\sqrt{t})\). Use the product rule for differentiation: \(\frac{dy}{dt} = u'v + uv'\).
Compute \(u' = \frac{d}{dt}(1 - t) = -1\).
Find \(v' = \frac{d}{dt} \left( \text{coth}^{-1}(\sqrt{t}) \right)\). Use the chain rule: first find the derivative of \(\text{coth}^{-1}(x)\) with respect to \(x\), then multiply by the derivative of \(\sqrt{t}\) with respect to \(t\).
Recall that \(\frac{d}{dx} \text{coth}^{-1}(x) = -\frac{1}{1 - x^2}\) for \(|x| > 1\). Then, \(\frac{d}{dt} \sqrt{t} = \frac{1}{2\sqrt{t}}\). Combine these to find \(v' = -\frac{1}{1 - (\sqrt{t})^2} \cdot \frac{1}{2\sqrt{t}} = -\frac{1}{1 - t} \cdot \frac{1}{2\sqrt{t}}\).
Substitute \(u\), \(u'\), \(v\), and \(v'\) back into the product rule formula to express \(\frac{dy}{dt} = (-1) \cdot \text{coth}^{-1}(\sqrt{t}) + (1 - t) \cdot \left(-\frac{1}{1 - t} \cdot \frac{1}{2\sqrt{t}}\right)\). Simplify the expression as much as possible.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Inverse Hyperbolic Cotangent Function (coth⁻¹)
The inverse hyperbolic cotangent function, coth⁻¹(x), is the inverse of the hyperbolic cotangent. It can be expressed in terms of logarithms and is defined for |x| > 1. Understanding its domain and derivative formula is essential for differentiating expressions involving coth⁻¹.
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Derivatives of Other Inverse Trigonometric Functions
Chain Rule
The chain rule is a fundamental differentiation technique used when dealing with composite functions. 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. This is crucial when differentiating functions like coth⁻¹(√t).
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05:02Intro to the Chain Rule
Product Rule
The product rule is used to differentiate products of two functions. It states that the derivative of f(t)g(t) is f'(t)g(t) + f(t)g'(t). Since y is given as a product of (1 - t) and coth⁻¹(√t), applying the product rule correctly is necessary to find dy/dt.
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05:18The Product Rule
Related Practice
Textbook Question
Show that the graph of the inverse of f(x)=mx+b, where m and b are constants and m≠0, is a line with slope 1/m and y-intercept -b/m.
Textbook Question
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = e^(5-7x)
Textbook Question
Evaluate the integrals in Exercises 33–54.
∫(from ln3 to ln2) (e^x) dx
Textbook Question
Evaluate the integrals in Exercises 97–110.
105. ∫₀² (log₂(x + 2) / (x + 2)) dx
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Textbook Question
In Exercises 59–86, find the derivative of y with respect to the given independent variable.
77. y = log₃(((x + 1)/(x − 1))^(ln 3))