Ch. 3 - Derivatives

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

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.5.61

Chapter 3, Problem 3.5.61

Slopes on the graph of the tangent function Graph y = tan x and its derivative together on (−π/2, π/2). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers.

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Graph the function \( y = \tan(x) \) on the interval \( (-\pi/2, \pi/2) \). Note that the tangent function has vertical asymptotes at \( x = -\pi/2 \) and \( x = \pi/2 \), and it is undefined at these points.

Find the derivative of \( y = \tan(x) \). The derivative is \( y' = \sec^2(x) \), which represents the slope of the tangent line to the graph of \( \tan(x) \) at any point where it is defined.

Analyze the behavior of \( y' = \sec^2(x) \) on the interval \( (-\pi/2, \pi/2) \). Since \( \sec^2(x) \) is always positive and increases without bound as \( x \) approaches \( \pm\pi/2 \), the slope of the tangent function becomes arbitrarily large near these points.

Determine whether the slope is ever negative. Since \( \sec^2(x) > 0 \) for all \( x \) in \( (-\pi/2, \pi/2) \), the slope of the tangent function is never negative on this interval.

Conclude that the graph of the tangent function does not have a smallest or largest slope because \( \sec^2(x) \) increases without bound as \( x \) approaches \( \pm\pi/2 \). The slope is always positive and becomes arbitrarily large near the vertical asymptotes.

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Key Concepts

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

Tangent Function Behavior

The tangent function, defined as y = tan(x), has a periodic nature with vertical asymptotes at odd multiples of π/2. Within the interval (−π/2, π/2), the function increases from negative infinity to positive infinity, which affects the slopes of its tangent lines. Understanding this behavior is crucial for analyzing the slopes of the tangent function.

Derivative of the Tangent Function

The derivative of the tangent function, given by y' = sec²(x), indicates the slope of the tangent line at any point on the graph. Since sec²(x) is always positive for x in (−π/2, π/2), the slope of the tangent function is never negative in this interval. This concept is essential for determining the nature of slopes on the graph.

Limits and Asymptotic Behavior

As x approaches the vertical asymptotes of the tangent function at −π/2 and π/2, the function's values approach negative and positive infinity, respectively. This behavior implies that the slopes of the tangent function also approach infinity near these asymptotes. Understanding limits helps in identifying the largest and smallest slopes within the specified interval.

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Related Practice

Textbook Question

Approximation Error

In Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Find

a. the change Δf = f(x₀ + dx) − f(x₀);

b. the value of the estimate df = fʹ(x₀) dx; and

c. the approximation error |Δf − df|.

f(x) = x⁻¹, x₀ = 0.5, dx = 0.1

Textbook Question

Differentiating Implicitly

Use implicit differentiation to find dy/dx in Exercises 1–14.

x²(x – y)² = x² – y²

Textbook Question

In Exercises 29 and 30, find the slope of the curve at the given points.

y² + x² = y⁴ – 2x at (–2,1) and (–2,–1)

Textbook Question

For Exercises 55 and 56, evaluate each limit by first converting each to a derivative at a particular x-value.

lim (x → −1) (x²/⁹ − 1) / (x + 1)

Textbook Question

Normal lines to a parabola Show that if it is possible to draw three normal lines from the point (a, 0) to the parabola x = y² shown in the accompanying diagram, then a must be greater than 1/2. One of the normal lines is the x-axis. For what value of a are the other two normal lines perpendicular?

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Textbook Question

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.

x³/² + 2y³/² = 17, (1, 4)