Textbook Question
In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
13. y = (x+2)^(x+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.P.29
In Exercises 25–30, use logarithmic differentiation to find the derivative of y with respect to the appropriate variable.
29. y = (sin θ)^√θ
Verified step by step guidance1
Start by taking the natural logarithm of both sides of the equation to simplify the differentiation process: \(\ln y = \ln \left( (\sin \theta)^{\sqrt{\theta}} \right)\).
Use the logarithm power rule to bring the exponent down: \(\ln y = \sqrt{\theta} \cdot \ln (\sin \theta)\).
Differentiate both sides with respect to \(\theta\). Remember to use implicit differentiation on the left side and the product rule on the right side: \(\frac{1}{y} \frac{dy}{d\theta} = \frac{d}{d\theta} \left( \sqrt{\theta} \cdot \ln (\sin \theta) \right)\).
Apply the product rule to the right side: \(\frac{d}{d\theta} \left( \sqrt{\theta} \right) \cdot \ln (\sin \theta) + \sqrt{\theta} \cdot \frac{d}{d\theta} \left( \ln (\sin \theta) \right)\).
Calculate each derivative separately: \(\frac{d}{d\theta} \left( \sqrt{\theta} \right) = \frac{1}{2 \sqrt{\theta}}\) and \(\frac{d}{d\theta} \left( \ln (\sin \theta) \right) = \frac{\cos \theta}{\sin \theta} = \cot \theta\). Then substitute back into the expression and multiply both sides by \(y\) to solve for \(\frac{dy}{d\theta}\).
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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. By taking the natural logarithm of both sides, the expression simplifies, allowing the use of implicit differentiation and the properties of logarithms to find the derivative more easily.
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Logarithmic Differentiation
Chain Rule
The chain rule is a fundamental differentiation rule used when differentiating composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This is essential when differentiating expressions like sin(θ) raised to a function of θ.
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05:02Intro to the Chain Rule
Properties of Logarithms
Properties of logarithms, such as log(a^b) = b log(a), allow us to simplify complex expressions involving exponents. Applying these properties helps transform the original function into a form that is easier to differentiate, especially when the exponent itself is a function of the variable.
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05:36Change of Base Property
Related Practice
Textbook Question
118. A particle is traveling upward and to the right along the curve y=ln(x). Its x-coordinate is increasing at the rate (dx/dt)=√x m/sec. At what rate is the y-coordinate changing at the point (e², 2)?
Textbook Question
111. True, or false? Give reasons for your answers.
c. x = o(x + ln(x))
Textbook Question
7. What integrals lead to logarithms? Give examples. What are the integrals of tan x, cot x, sec x, and csc x?
Textbook Question
111. True, or false? Give reasons for your answers.
e. arctan x = O(1)
Textbook Question
19. Center of mass Find the center of mass of a thin plate of constant density covering the region in the first and fourth quadrants enclosed by the curves y=1/(1+x²) and y=-1/(1+x²) and by the lines x=0 and x=1.