Book cover for Thomas Calculus

Thomas Calculus

George B. Thomas, Jr.

ISBN #9780321878960

13th Edition

6,812 Questions

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127,035 Students Helped

Homework Questions

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This section introduces the derivative as the fundamental concept that unifies the ideas of slope, tangent line, and instantaneous rate of change of a function. By defining the derivative through the limit of the difference quotient, the text explains how to compute the slope of a curve at any given point. Through examples such as the reciprocal function and the falling rock problem, it showcases the utility of derivatives in solving real-world problems across various disciplines.

Learning Objectives

1

Define the derivative of a function at a point using the limit of the difference quotient.

2

Explain how the derivative represents the slope of the tangent line and the instantaneous rate of change.

3

Apply the definition of the derivative to compute slopes for both specific points and as a function.

4

Utilize the derivative concept to solve practical problems in physics, economics, and the sciences.

Key Concepts

CONCEPT

DEFINITION

Derivative

The derivative of a function ƒ at a point x0 is defined as ƒ′(x0) = lim (h→0) [ƒ(x0 + h) - ƒ(x0)]/h, representing the slope of the tangent line or the instantaneous rate of change.

Difference Quotient

The expression [ƒ(x0 + h) - ƒ(x0)]/h, where h ≠ 0, which when its limit as h approaches 0 exists, defines the derivative at x0.

Tangent Line

A line that touches a curve at a single point with a slope equal to the derivative of the function at that point.

Instantaneous Rate of Change

The derivative at a point, indicating the rate at which a function is changing at that precise moment.

Differentiability

A function is differentiable at a point if the derivative exists at that point; if it exists for every point in the domain, the function is called differentiable.

Example Problems

Example 1

In Exercises $1-4,$ use the grid and a straight edge to make a rough estimate of the slope of the curve (in $y$ -units per $x$ -unit) at the points $P_{1}$ and $P_{2}$ .

Example 2

Use the grid and a straight edge to make a rough estimate of the slope of the curve (in $y$ -units per $x$ -unit) at the points $P_{1}$ and $P_{2}$ .

Example 3

Use the grid and a straight edge to make a rough estimate of the slope of the curve (in $y$ -units per $x$ -unit) at the points $P_{1}$ and $P_{2}$ .

Example 4

Use the grid and a straight edge to make a rough estimate of the slope of the curve (in $y$ -units per $x$ -unit) at the points $P_{1}$ and $P_{2}$ .

Example 5

In Exercises $5-10,$ find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. \begin{equation} y=4-x^{2}, \quad(-1,3) \end{equation}
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Step-by-Step Explanations

QUESTION

Using the limit definition of the derivative, how do you find the derivative of ƒ(x) = 1/x at a point x = a (with a ≠ 0)?

STEP-BY-STEP ANSWER:

Step 1: Write the difference quotient: (Æ’(a+h) - Æ’(a)) / h = (1/(a+h) - 1/a) / h.
Step 2: Combine the fractions in the numerator: (1/(a+h) - 1/a) = (a - (a+h)) / [a(a+h)] = (-h) / [a(a+h)].
Step 3: Substitute back into the difference quotient: ((-h)/[a(a+h)])/h.
Step 4: Simplify by canceling h (assuming h ≠ 0): -1/[a(a+h)].
Step 5: Take the limit as h approaches 0: lim (h→0) -1/[a(a+h)] = -1/a².
Final Answer: The derivative of ƒ(x) = 1/x at x = a is -1/a².

Derivative of 1/x at x = a

QUESTION

How can the derivative be interpreted as an instantaneous rate of change? Use a practical example such as a falling object.

STEP-BY-STEP ANSWER:

Step 1: Consider the function representing the height of a free falling object: ƒ(t) = 16t² (with appropriate units).
Step 2: Write the difference quotient: (Æ’(t+h) - Æ’(t)) / h.
Step 3: Calculate the expression: (16(t+h)² - 16t²)/h.
Step 4: Expand and simplify the numerator to 16(2t*h + h²) and simplify to 16(2t + h).
Step 5: Take the limit as h → 0, yielding ƒ′(t) = 32t.
Step 6: At t = 1, the instantaneous speed is 32 ft/sec.
Final Answer: The derivative gives the instantaneous rate at which the height changes, yielding 32 ft/sec at t = 1 sec.

Derivative as an Instantaneous Rate of Change

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Common Mistakes

  • Misapplying the limit procedure by not properly simplifying the difference quotient before taking the limit.
  • Confusing the derivative with the average rate of change over an interval, rather than the instantaneous rate.
  • Overlooking domain restrictions (e.g., a ? 0 for 1/x) when computing derivatives.
  • Failing to correctly cancel terms, particularly forgetting that the cancelation of h is valid only when h is not zero.
  • Mixing up the interpretation of the derivative as both the slope of the tangent and the instantaneous rate of change.