Textbook Question
In Exercises 41–58, find dy/dt.
y = ((3t − 4) / (5t + 2))⁻⁵
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.2.27
Graphs
Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d).
Verified step by step guidance1
Step 1: Analyze the graph of the function y = f₁(x). The graph is a parabola opening upwards, which suggests that the function is quadratic, such as y = x².
Step 2: Recall that the derivative of a quadratic function y = x² is linear, specifically y' = 2x. This means the derivative graph will be a straight line passing through the origin.
Step 3: Compare the derivative graphs (a)-(d) provided. The graph labeled (b) is a straight line passing through the origin, which matches the derivative of y = x².
Step 4: Match the function y = f₁(x) with its derivative graph (b), as the characteristics align: quadratic function leads to a linear derivative.
Step 5: Verify the match by considering the slope of the derivative graph (b). The slope corresponds to the rate of change of the original function, confirming the relationship between y = f₁(x) and its derivative.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative and Slope
The derivative of a function at a point is the slope of the tangent line to the graph of the function at that point. It provides the rate of change of the function's value with respect to changes in the input. Understanding how the slope changes across the graph helps in matching functions to their derivatives, as the derivative graph represents these slopes.
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05:13
Slopes of Tangent Lines
Critical Points and Derivative Sign
Critical points occur where the derivative is zero or undefined, indicating potential maxima, minima, or inflection points. The sign of the derivative (positive or negative) indicates whether the function is increasing or decreasing. Analyzing these aspects helps in identifying the corresponding derivative graph, as changes in sign reflect changes in the function's behavior.
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04:50Critical Points
Graphical Interpretation of Derivatives
Graphically, the derivative of a function is represented as a new graph that shows the slope of the original function at each point. For example, a linear function's derivative is constant, while a quadratic function's derivative is linear. Recognizing these patterns is crucial for matching a function to its derivative graph, as each function type has a distinct derivative shape.
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01:47Graphical Applications of Exponential & Logarithmic Derivatives: Example 8
Related Practice
Textbook Question
In Exercises 5–10, find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together.
y = (1 / x³), (−2, −1/8)
Textbook Question
Differential Estimates of Change
In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.
The change in the volume V = x³ of a cube when the edge lengths change from x₀ to x₀ + dx
Textbook Question
Normal lines parallel to a line Find the normal lines to the curve xy + 2x – y = 0 that are parallel to the line 2x + y = 0.
Textbook Question
In Exercises 19–22, find the slope of the curve at the point indicated.
y = (x − 1) / (x + 1), x = 0
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Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
xy = cot(xy)