Textbook Question
In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
61. y = √(θ + 3) sin θ
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 7.6.29
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
29. y=arcsec(1/t), 0<t<1
Verified step by step guidance1
Identify the function given: \(y = \arcsec\left(\frac{1}{t}\right)\), where \(0 < t < 1\).
Recall the derivative formula for \(y = \arcsec(u)\) with respect to \(t\):
\(\frac{dy}{dt} = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot \frac{du}{dt}\), where \(u\) is a function of \(t\).
Set \(u = \frac{1}{t}\) and compute its derivative with respect to \(t\):
\(\frac{du}{dt} = \frac{d}{dt} \left(\frac{1}{t}\right) = -\frac{1}{t^2}\).
Substitute \(u\) and \(\frac{du}{dt}\) back into the derivative formula:
\(\frac{dy}{dt} = \frac{1}{\left| \frac{1}{t} \right| \sqrt{\left(\frac{1}{t}\right)^2 - 1}} \cdot \left(-\frac{1}{t^2}\right)\).
Simplify the expression step-by-step, paying attention to the absolute value and the square root, to express \(\frac{dy}{dt}\) in terms of \(t\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative of Inverse Trigonometric Functions
Inverse trigonometric functions like arcsec(x) have specific derivative formulas. For arcsec(x), the derivative is 1 / (|x|√(x² - 1)). Understanding these formulas is essential to differentiate functions involving inverse trig expressions.
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Derivatives of Other Inverse Trigonometric Functions
Chain Rule
The chain rule is used to differentiate composite functions. When y = arcsec(1/t), you first differentiate arcsec(u) with respect to u, then multiply by the derivative of u = 1/t. This rule allows handling nested functions systematically.
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05:02Intro to the Chain Rule
Domain Considerations for arcsec(x)
The domain of arcsec(x) is |x| ≥ 1, but here the argument is 1/t with 0 < t < 1, so 1/t > 1, which fits the domain. Recognizing domain restrictions ensures the function and its derivative are valid in the given interval.
Recommended video:
5:10Finding the Domain and Range of a Graph
Related Practice
Textbook Question
In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
64. y = 1/(t(t+1)(t+2))
Textbook Question
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
51. lim (θ → 0) (θ - sin θ cos θ) / (tan θ - θ)
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Textbook Question
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
27. y=arccsc(x²+1)
Textbook Question
Rewrite the expressions in Exercises 5–10 in terms of exponentials and simplify the results as much as you can.
8. cosh(3x) - sinh(3x)
Textbook Question
Verify the integration formulas in Exercises 37–40.
39. ∫x coth⁻¹(x)dx = ((x²-1)/2)coth⁻¹(x) + x/2 + C
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