Textbook Question
Derivative Calculations
In Exercises 1–12, find the first and second derivatives.
y = x³/3 + x²/2 + x/4
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
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 3.9.29
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|.
<IMAGE>
f(x) = x² + 2x, x₀ = 1, dx = 0.1
Verified step by step guidance1
First, calculate the change in the function, Δf = f(x₀ + dx) − f(x₀). Substitute x₀ = 1 and dx = 0.1 into the function f(x) = x² + 2x to find f(x₀ + dx) and f(x₀).
Evaluate f(x₀ + dx) by substituting x = x₀ + dx = 1 + 0.1 = 1.1 into the function: f(1.1) = (1.1)² + 2(1.1).
Evaluate f(x₀) by substituting x = x₀ = 1 into the function: f(1) = 1² + 2(1).
Calculate the derivative of the function, fʹ(x) = d/dx (x² + 2x). This gives fʹ(x) = 2x + 2. Then, find the estimate df = fʹ(x₀) dx by substituting x₀ = 1 and dx = 0.1.
Finally, compute the approximation error |Δf − df| by taking the absolute value of the difference between the change in the function Δf and the estimate df.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Function Change Δf
The change in a function, Δf, represents the difference in the function's value as the input changes from x₀ to x₀ + dx. It is calculated by evaluating the function at the new point and subtracting the function's value at the original point: Δf = f(x₀ + dx) − f(x₀). This concept helps understand how the function behaves over small intervals.
Recommended video:
06:35
Changing Geometries
Differential Estimate df
The differential estimate, df, is an approximation of the change in the function using the derivative at a specific point. It is calculated as df = fʹ(x₀) dx, where fʹ(x₀) is the derivative of the function at x₀, and dx is the small change in x. This linear approximation is useful for estimating function changes over small intervals.
Recommended video:
05:53Finding Differentials
Approximation Error |Δf − df|
The approximation error, |Δf − df|, measures the difference between the actual change in the function and the estimated change using the derivative. It quantifies the accuracy of the linear approximation provided by the differential. A smaller error indicates a better approximation, highlighting the importance of understanding the behavior of the function and its derivative.
Recommended video:
04:57Determining Error and Relative Error
Related Practice
Textbook Question
Rates of Change
Speed of a rocket At t sec after liftoff, the height of a rocket is 3t² ft. How fast is the rocket climbing 10 sec after liftoff?
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
y = (5 − 2x)⁻³ + (1 / 8)(2 / x + 1)⁴
Textbook Question
If y = x² and dx/dt = 3, then what is dy/dt when x = –1?
Textbook Question
Derivative Calculations
In Exercises 1–12, find the first and second derivatives.
w = 3z⁷ − 7z³ + 21z²
Textbook Question
Finding Linearizations
In Exercises 1–5, find the linearization L(x) of f(x) at x = a.
f(x) = ∛x, a = −8