Textbook Question
Use Theorem 3.10 to evaluate the following limits.
lim x🠂0 (tan 5x) / x
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.6.5
Suppose w(t) is the weight (in pounds) of a golden retriever puppy t weeks after it is born. Interpret the meaning of w'(15) = 1.75.
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Understand that w(t) represents the weight of the puppy as a function of time, where t is measured in weeks.
Recognize that w'(t) is the derivative of w(t) with respect to t, which represents the rate of change of the puppy's weight with respect to time.
The expression w'(15) = 1.75 indicates that at t = 15 weeks, the rate of change of the puppy's weight is 1.75 pounds per week.
Interpret this to mean that when the puppy is 15 weeks old, its weight is increasing at a rate of 1.75 pounds each week.
This rate of change is an instantaneous rate, meaning it describes how fast the weight is increasing exactly at 15 weeks, not over an interval of time.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative Interpretation
The derivative of a function at a given point represents the rate of change of that function at that point. In this context, w'(15) indicates how the weight of the puppy is changing at 15 weeks. Specifically, it tells us that the weight is increasing at a rate of 1.75 pounds per week at that time.
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Function Notation
Function notation, such as w(t), is used to express the relationship between an independent variable (t, in this case, time in weeks) and a dependent variable (w, the weight of the puppy). Understanding this notation is crucial for interpreting the behavior of the function over time and how changes in t affect w.
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Contextual Interpretation
Interpreting the meaning of mathematical results in context is essential. Here, w'(15) = 1.75 means that at 15 weeks, the puppy's weight is not just increasing, but it is doing so at a specific rate, which can inform decisions about its health and growth patterns. This contextual understanding helps relate mathematical concepts to real-world scenarios.
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Related Practice
Textbook Question
Evaluate the derivative of the following functions.
f(x) = sec-1 (ln x)
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Textbook Question
15–48. Derivatives Find the derivative of the following functions.
y = 4^-x sin x
Textbook Question
Given that f(1) = 5, f′(1) = 4, g(1) = 2, and g′(1) = 3 , find d/dx (f(x)g(x))∣ ∣x=1 and d/dx (f(x) / g(x)) ∣ x=1.
Textbook Question
How are the derivatives of sin^−1 x and cos^−1 x related?
Textbook Question
51–56. Second derivatives Find d²y/dx².
x⁴+y⁴ = 64