Textbook Question
Find the first and second derivatives of the functions in Exercises 33–38.
w = ((1 + 3z) / 3z) (3 − z)
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Ch. 3 - Derivatives
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 3, Problem 3.7.20
Second Derivatives
In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.
x²/³ + y²/³ = 1
Verified step by step guidance1
Start by differentiating both sides of the equation x^(2/3) + y^(2/3) = 1 with respect to x. Use implicit differentiation for the y term.
For the term x^(2/3), apply the power rule: differentiate to get (2/3)x^(-1/3).
For the term y^(2/3), apply the chain rule: differentiate to get (2/3)y^(-1/3) * (dy/dx).
Set the derivative of the left side equal to the derivative of the right side (which is 0), and solve for dy/dx.
To find the second derivative d²y/dx², differentiate the expression for dy/dx with respect to x, applying implicit differentiation again, and express the result in terms of x and y.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. It involves differentiating both sides of an equation with respect to a variable, typically x, while treating other variables as implicit functions of x. This method is essential for finding dy/dx when y is not isolated.
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05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This rule is crucial when applying implicit differentiation, as it allows for the differentiation of terms involving y with respect to x.
Recommended video:
05:02Intro to the Chain Rule
Second Derivative
The second derivative, denoted as d²y/dx², represents the derivative of the first derivative, providing information about the curvature or concavity of a function. In the context of implicit differentiation, finding the second derivative involves differentiating the expression for dy/dx again with respect to x, often requiring the use of the chain rule and implicit differentiation techniques.
Recommended video:
06:02The Second Derivative Test: Finding Local Extrema
Related Practice
Textbook Question
A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?
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Textbook Question
Second Derivatives
In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.
y² – 2x = 1 – 2y
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Textbook Question
In Exercises 53 and 54, find dr/ds.
2rs - r - s + s² = -3
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Textbook Question
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Find the linearization of ƒ(x) = 2/ (1 - x) + √1 + x - 3.1 at x = 0.
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Textbook Question
In Exercises 51 and 52, find dp/dq.
p³ + 4pq - 3q² = 2
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