Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 28

Chapter 4, Problem 28

Calculate the first derivatives of ƒ(𝓍) = 𝓍²/ (𝓍² + 1) and g(𝓍) = ―1/ (𝓍² + 1) . What can you conclude about the graphs of these functions?

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To find the first derivative of ƒ(𝓍) = 𝓍² / (𝓍² + 1), apply the quotient rule. The quotient rule states that if you have a function h(𝓍) = u(𝓍) / v(𝓍), then its derivative h'(𝓍) is given by (u'(𝓍)v(𝓍) - u(𝓍)v'(𝓍)) / (v(𝓍)²). Here, u(𝓍) = 𝓍² and v(𝓍) = 𝓍² + 1.

Calculate the derivatives of u(𝓍) and v(𝓍). The derivative of u(𝓍) = 𝓍² is u'(𝓍) = 2𝓍. The derivative of v(𝓍) = 𝓍² + 1 is v'(𝓍) = 2𝓍.

Substitute these derivatives into the quotient rule formula: ƒ'(𝓍) = ((2𝓍)(𝓍² + 1) - (𝓍²)(2𝓍)) / (𝓍² + 1)².

For g(𝓍) = -1 / (𝓍² + 1), use the chain rule. The chain rule states that if you have a composite function y = f(g(𝓍)), then its derivative is y' = f'(g(𝓍)) * g'(𝓍). Here, consider g(𝓍) as -1 times the reciprocal of (𝓍² + 1).

The derivative of g(𝓍) = -1 / (𝓍² + 1) is g'(𝓍) = (0 * (𝓍² + 1) - (-1)(2𝓍)) / (𝓍² + 1)² = 2𝓍 / (𝓍² + 1)². Compare the derivatives of ƒ(𝓍) and g(𝓍) to analyze the behavior of their graphs, noting that they are related by symmetry or transformations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function measures the rate at which the function's value changes as its input changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the graph of the function at any given point. Derivatives can be calculated using rules such as the quotient rule, which is particularly useful for functions defined as a ratio of two other functions.

Quotient Rule

The quotient rule is a formula used to find the derivative of a function that is the quotient of two other functions. If you have a function defined as ƒ(𝓍) = u(𝓍)/v(𝓍), the derivative is given by ƒ'(𝓍) = (u'v - uv')/v², where u' and v' are the derivatives of u and v, respectively. This rule is essential for differentiating functions like ƒ(𝓍) = 𝓍²/(𝓍² + 1) and g(𝓍) = -1/(𝓍² + 1).

Graph Behavior and Critical Points

Analyzing the first derivative of a function helps determine its critical points, where the function's slope is zero or undefined. These points are crucial for understanding the behavior of the graph, such as identifying local maxima, minima, and points of inflection. By evaluating the first derivatives of ƒ(𝓍) and g(𝓍), one can draw conclusions about the increasing or decreasing nature of the functions and their overall shape.

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Critical Points

Related Practice

Textbook Question

As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acre-ft in 24 hours. Show that at some instant during that period the reservoir’s volume was increasing at a rate in excess of 225,000 gal/min. (An acre-foot is 43,560 ft³, the volume that would cover 1 acre to the depth of 1 ft. A cubic foot holds 7.48 gal.)

Textbook Question

Finding Functions from Derivatives

Suppose that f'(x) = 2x for all x. Find f(2) if

a. f(0) = 0

Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

y = (x² - 49) / (x² + 5x - 14)

Textbook Question

30. y = (x² - 4) / (x² - 2)

Textbook Question

Finding Functions from Derivatives

Suppose that f(−1) = 3 and that f'(x) = 0 for all x. Must f(x) = 3 for all x? Give reasons for your answer.

Textbook Question

Finding Functions from Derivatives

Suppose that f'(x) = 2x for all x. Find f(2) if

b. f(1) = 0