Textbook Question
Evaluate the integrals in Exercises 31–78.
33. ∫e^x sec²(e^x - 7)dx
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.17
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
17. y = ln(arccos(x))
Verified step by step guidance1
Identify the function given: \(y = \ln(\arccos(x))\). We need to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).
Recall the chain rule for derivatives: if \(y = \ln(u)\), then \(\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}\). Here, \(u = \arccos(x)\).
Find the derivative of the inner function \(u = \arccos(x)\). The derivative is \(\frac{du}{dx} = -\frac{1}{\sqrt{1 - x^2}}\).
Apply the chain rule by substituting \(u\) and \(\frac{du}{dx}\) into the formula: \(\frac{dy}{dx} = \frac{1}{\arccos(x)} \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right)\).
Simplify the expression to write the derivative as \(\frac{dy}{dx} = -\frac{1}{\arccos(x) \sqrt{1 - x^2}}\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Chain Rule
The chain rule is a fundamental differentiation technique used to find the derivative of composite functions. It states that the derivative of f(g(x)) is f'(g(x)) multiplied by g'(x). In this problem, since y = ln(arccos(x)) is a composition of ln(u) and u = arccos(x), the chain rule is essential.
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Intro to the Chain Rule
Derivative of the Natural Logarithm Function
The derivative of ln(u), where u is a differentiable function of x, is 1/u times the derivative of u. This rule helps differentiate logarithmic functions by simplifying the process to the derivative of the inner function divided by the original function.
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05:18Derivative of the Natural Logarithmic Function
Derivative of the Arccosine Function
The derivative of arccos(x) with respect to x is -1 divided by the square root of (1 - x^2). This formula is crucial for differentiating inverse trigonometric functions and is needed to find the derivative of the inner function in the given problem.
Recommended video:
06:30Derivatives of Other Trig Functions
Related Practice
Textbook Question
Use l’Hôpital’s Rule to find the limits in Exercises 85–108.
85. lim(x→1) (x² + 3x - 4)/(x - 1)
Textbook Question
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
1. y = 10e^(-x/5)
Textbook Question
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
3. y = (1/4)xe^(4x) - (1/16)e^(4x)
Textbook Question
113. The function f(x) = e^x + x, being differentiable and one-to-one, has a differentiable inverse f^(-1)(x). Find the value of df^(-1)/dx at the point f (ln 2).
Textbook Question
Evaluate the integrals in Exercises 31–78.
37. ∫(from -1 to 1)dx/(3x-4)
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