Textbook Question
Evaluate the integrals in Exercises 53–76.
65. ∫3dr/√(1-4(r-1)²)
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 7.3.115
In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.
115. y = (x + 1)ˣ
Verified step by step guidance1
Recognize that the function is of the form \(y = (x + 1)^x\), where both the base and the exponent depend on \(x\). This makes logarithmic differentiation a suitable method.
Take the natural logarithm of both sides to simplify the expression: \(\ln y = \ln \left( (x + 1)^x \right)\).
Use the logarithm power rule to bring the exponent down: \(\ln y = x \cdot \ln (x + 1)\).
Differentiate both sides with respect to \(x\). Remember to use implicit differentiation on the left side and the product rule on the right side: \(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( x \ln (x + 1) \right)\).
Apply the product rule to the right side: \(\frac{d}{dx} \left( x \ln (x + 1) \right) = \ln (x + 1) + x \cdot \frac{1}{x + 1}\), then multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Logarithmic Differentiation
Logarithmic differentiation is a technique used to differentiate functions where the variable appears both in the base and the exponent, such as y = (x + 1)^x. By taking the natural logarithm of both sides, the expression simplifies, allowing the use of implicit differentiation to find dy/dx more easily.
Recommended video:
06:30
Logarithmic Differentiation
Implicit Differentiation
Implicit differentiation involves differentiating both sides of an equation with respect to x when y is defined implicitly. After taking the logarithm, y is expressed in terms of ln(y), so differentiating requires applying the chain rule to ln(y), treating y as a function of x.
Recommended video:
05:14Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental differentiation rule used when differentiating composite functions. In logarithmic differentiation, it is applied to differentiate ln(y), resulting in (1/y) * dy/dx, which helps isolate dy/dx and find the derivative of the original function.
Recommended video:
05:02Intro to the Chain Rule
Related Practice
Textbook Question
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
47. lim (t → ∞) (e^t + t²) / (e^t - t)
1
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Textbook Question
In Exercises 5–8, show that each function is a solution of the given initial value problem.
7. Differential Equation: xy' + y = -sin(x), x>0
Initial condition: y(π/2) = 0
Solution candidate: y = cos(x)/x
Textbook Question
Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
75. lim (x → ∞) e^(x²) / (x e^x)
1
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Textbook Question
In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
13. y = 6sinh(x/3)
Textbook Question
Each of Exercises 1–4 gives a value of sinh x or cosh x. Use the definitions and the identity cosh²x - sinh²x = 1 to find the values of the remaining five hyperbolic functions.
2. sinh x = 4/3