Textbook Question
Water flows into a conical tank at a rate of 2 ft³/min. If the radius of the top of the tank is 4 ft and the height is 6 ft, determine how quickly the water level is rising when the water is 2 ft deep in the tank.
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 120a
Let f(x) = x².
a. Show that f(x)−f(y) / x−y = f′(x+y²), for all x≠y.
Verified step by step guidance1
Start by understanding the problem: We need to show that the difference quotient (f(x) - f(y)) / (x - y) is equal to the derivative of f evaluated at (x + y²).
First, calculate f(x) and f(y) using the given function f(x) = x². This gives us f(x) = x² and f(y) = y².
Substitute these into the difference quotient: (f(x) - f(y)) / (x - y) = (x² - y²) / (x - y).
Recognize that x² - y² is a difference of squares, which can be factored as (x - y)(x + y). Substitute this into the difference quotient to simplify: ((x - y)(x + y)) / (x - y).
Cancel the (x - y) terms in the numerator and denominator, assuming x ≠ y, to get x + y. Now, find the derivative f'(x) = 2x, and evaluate it at x + y²: f'(x + y²) = 2(x + y²). Compare this with the simplified expression x + y to verify the equality.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Difference Quotient
The difference quotient is a fundamental concept in calculus that represents the average rate of change of a function over an interval. It is defined as (f(x) - f(y)) / (x - y) for x ≠ y. This expression is crucial for understanding the derivative, as it approaches the instantaneous rate of change as y approaches x.
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Derivative
The derivative of a function at a point measures how the function's output changes as its input changes. It is denoted as f'(x) and can be interpreted as the slope of the tangent line to the function's graph at that point. In this question, we need to show that the difference quotient equals the derivative of f at a specific point, which involves applying the definition of the derivative.
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Function Composition
Function composition involves combining two functions to create a new function, where the output of one function becomes the input of another. In this context, we need to evaluate f' at (x + y²), which requires understanding how to apply the derivative to a function that is itself a composition of variables. This concept is essential for manipulating and simplifying expressions involving derivatives.
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Related Practice
Textbook Question
A general proof of the Chain Rule Let f and g be differentiable functions with h(x)=f(g(x)). For a given constant a, let u=g(a) and v=g(x), and define H (v) = <1x1 matrix>
c. Show that h′(a) = lim x→a ((H(g(x))+f′(g(a)))⋅g(x)−g(a)/x−a).
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Textbook Question
Derivatives of even and odd functions Recall that f is even if f(−x) = f(x), for all x in the domain of f, and f is odd if f(−x) = −f(x) for all x in the domain of f.
b. If f is a differentiable, odd function on its domain, determine whether f' is even, odd, or neither.
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