Textbook Question
Evaluate the integrals in Exercises 31–78.
39. ∫(from 0 to π)tan(x/3)dx
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
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.21
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
21. y = z arcsec(z) - √(z² - 1), z>1
Verified step by step guidance1
Identify the function to differentiate: \(y = z \cdot \operatorname{arcsec}(z) - \sqrt{z^{2} - 1}\), where \(z > 1\).
Recall the derivative of \(\operatorname{arcsec}(z)\) with respect to \(z\): \(\frac{d}{dz} \operatorname{arcsec}(z) = \frac{1}{|z| \sqrt{z^{2} - 1}}\). Since \(z > 1\), \(|z| = z\).
Apply the product rule to the first term \(z \cdot \operatorname{arcsec}(z)\): \(\frac{d}{dz} [z \cdot \operatorname{arcsec}(z)] = \operatorname{arcsec}(z) \cdot \frac{d}{dz} z + z \cdot \frac{d}{dz} \operatorname{arcsec}(z)\).
Differentiate the second term \(- \sqrt{z^{2} - 1}\) using the chain rule: rewrite as \(-(z^{2} - 1)^{1/2}\) and find its derivative.
Combine the derivatives from the product rule and the chain rule to write the full expression for \(\frac{dy}{dz}\).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4mWas this helpful?
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)) when |x| > 1. Understanding this formula is essential to differentiate expressions involving arcsec.
Recommended video:
06:35
Derivatives of Other Inverse Trigonometric Functions
Product Rule for Differentiation
The product rule is used to differentiate products of two functions. If y = u(x)v(x), then y' = u'v + uv'. Applying this rule correctly is crucial when differentiating terms like z * arcsec(z).
Recommended video:
05:18The Product Rule
Chain Rule and Differentiation of Composite Functions
The chain rule helps differentiate composite functions, such as √(z² - 1). It states that the derivative of f(g(x)) is f'(g(x)) * g'(x). This rule is necessary to handle the square root of an expression involving z.
Recommended video:
03:58The Chain Rule for 3+ Functions
Related Practice
Textbook Question
Evaluate the integrals in Exercises 31–78.
77. ∫dt/((t+1)√(t²+2t-8))
Textbook Question
109. Does f grow faster, slower, or at the same rate as g as x→∞? Give reasons for your answers.
b. f(x)=x, g(x)=x + 1/x
Textbook Question
Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
99. lim(x→0) (2^sin(x) - 1)/(e^x - 1)
Textbook Question
Evaluate the integrals in Exercises 31–78.
67. ∫(from -2 to 2)3dt/(4+3t²)
Textbook Question
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
7. y = log₂(x²/2)