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.
x²/³ + y²/³ = 1
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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 3.54
In Exercises 53 and 54, find dr/ds.
2rs - r - s + s² = -3
Verified step by step guidance1
First, identify the equation given: 2rs - r - s + s² = -3. We need to find the derivative dr/ds.
To find dr/ds, we will use implicit differentiation. Differentiate both sides of the equation with respect to s.
When differentiating 2rs with respect to s, apply the product rule: d/ds(2rs) = 2r(ds/ds) + 2s(dr/ds).
Differentiate the remaining terms: d/ds(-r) = -dr/ds, d/ds(-s) = -1, and d/ds(s²) = 2s.
Set the derivative of the right side of the equation to zero, as d/ds(-3) = 0. Combine all differentiated terms to form an equation involving dr/ds, and solve for dr/ds.
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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 differentiate equations where the dependent and independent variables are not explicitly separated. In this case, we differentiate both sides of the equation with respect to one variable while treating the other variable as a function of the first. This allows us to find derivatives like dr/ds even when r and s are intertwined.
Recommended video:
05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When applying the chain rule, we multiply the derivative of the outer function by the derivative of the inner function. In the context of finding dr/ds, it helps us relate the rates of change of r with respect to s, especially when r is expressed as a function of s.
Recommended video:
05:02Intro to the Chain Rule
Solving for Derivatives
Solving for derivatives involves isolating the desired derivative after applying differentiation techniques. In this problem, after differentiating the given equation implicitly, we will rearrange the resulting equation to express dr/ds explicitly. This step is crucial for obtaining the final answer and understanding the relationship between the variables.
Recommended video:
5:02Solving Logarithmic Equations
Related Practice
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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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
Find the derivatives of the functions in Exercises 1–42.
𝔂 = 3 .
(5x² + sin 2x)³/²
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Textbook Question
Find the derivatives of the functions in Exercises 1–42.
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s = √ t .
1 + √ t
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Textbook Question
In Exercises 51 and 52, find dp/dq.
p³ + 4pq - 3q² = 2
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