Textbook Question
Indeterminate Powers and Products
Find the limits in Exercises 53–68.
53. lim (x → 1⁺) x^(1/(1 - x))
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 7.6.45
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
45. y=cos(x-arccos(x))
Verified step by step guidance1
Identify the function given: \(y = \cos\left(x - \arccos(x)\right)\). We need to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).
Apply the chain rule to differentiate \(y = \cos(u)\) where \(u = x - \arccos(x)\). The derivative of \(\cos(u)\) with respect to \(x\) is \(-\sin(u) \cdot \frac{du}{dx}\).
Find \(\frac{du}{dx}\) where \(u = x - \arccos(x)\). Differentiate each term separately: the derivative of \(x\) with respect to \(x\) is 1, and the derivative of \(\arccos(x)\) with respect to \(x\) is \(-\frac{1}{\sqrt{1 - x^2}}\).
Combine the derivatives to get \(\frac{du}{dx} = 1 - \left(-\frac{1}{\sqrt{1 - x^2}}\right) = 1 + \frac{1}{\sqrt{1 - x^2}}\).
Substitute back into the chain rule formula: \(\frac{dy}{dx} = -\sin\left(x - \arccos(x)\right) \cdot \left(1 + \frac{1}{\sqrt{1 - x^2}}\right)\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative of Trigonometric Functions
Understanding how to differentiate trigonometric functions like cosine is essential. The derivative of cos(u) with respect to x is -sin(u) times the derivative of u, applying the chain rule to handle composite functions.
Recommended video:
06:35
Derivatives of Other Inverse Trigonometric Functions
Chain Rule
The chain rule is used to differentiate composite functions. It states that the derivative of f(g(x)) is f'(g(x)) multiplied by g'(x). This is crucial when differentiating expressions like cos(x - arccos(x)).
Recommended video:
05:02Intro to the Chain Rule
Derivative of Inverse Trigonometric Functions
Knowing the derivative of inverse trig functions, such as arccos(x), is important. The derivative of arccos(x) is -1 divided by the square root of (1 - x^2), which is needed when differentiating terms involving arccos(x).
Recommended video:
06:35Derivatives of Other Inverse Trigonometric Functions
Related Practice
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Textbook Question
Use the results of Exercise 55 to show that the functions in Exercises 56–60 have inverses over their domains. Find a formula for df⁻¹/dx using Theorem 1.
f(x) = 27x³