Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
Hello! I need this done with work shown for it. Here's the image.
![**Problem 2: Finding Derivatives**
Given the function:
\[ f(x) = 3x^7 - 4x^5 + 11x^3 - 6x \]
Find the derivatives \( f' \), \( f'' \), and \( f''' \):
1. **First Derivative (\( f'(x) \))**:
The first derivative of the function is found by applying the power rule to each term. The power rule states that the derivative of \( ax^n \) is \( nax^{n-1} \).
\[
f'(x) = \frac{d}{dx} (3x^7) - \frac{d}{dx} (4x^5) + \frac{d}{dx} (11x^3) - \frac{d}{dx} (6x)
\]
Calculating each term separately:
\[
\frac{d}{dx} (3x^7) = 21x^6
\]
\[
\frac{d}{dx} (4x^5) = 20x^4
\]
\[
\frac{d}{dx} (11x^3) = 33x^2
\]
\[
\frac{d}{dx} (6x) = 6
\]
Therefore, the first derivative is:
\[
f'(x) = 21x^6 - 20x^4 + 33x^2 - 6
\]
2. **Second Derivative (\( f''(x) \))**:
To find the second derivative, we again apply the power rule to the first derivative.
\[
f''(x) = \frac{d}{dx} (21x^6) - \frac{d}{dx} (20x^4) + \frac{d}{dx} (33x^2) - \frac{d}{dx} (6)
\]
Calculating each term separately:
\[
\frac{d}{dx} (21x^6) = 126x^5
\]
\[
\frac{d}{dx} (20x^4) = 80x^3
\]
\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fef3b440f-f5de-4a2a-817d-48b2b1baa1e0%2Fe7816ab3-aa65-4632-93af-06fb47044d00%2Fqrve9qj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 2: Finding Derivatives**
Given the function:
\[ f(x) = 3x^7 - 4x^5 + 11x^3 - 6x \]
Find the derivatives \( f' \), \( f'' \), and \( f''' \):
1. **First Derivative (\( f'(x) \))**:
The first derivative of the function is found by applying the power rule to each term. The power rule states that the derivative of \( ax^n \) is \( nax^{n-1} \).
\[
f'(x) = \frac{d}{dx} (3x^7) - \frac{d}{dx} (4x^5) + \frac{d}{dx} (11x^3) - \frac{d}{dx} (6x)
\]
Calculating each term separately:
\[
\frac{d}{dx} (3x^7) = 21x^6
\]
\[
\frac{d}{dx} (4x^5) = 20x^4
\]
\[
\frac{d}{dx} (11x^3) = 33x^2
\]
\[
\frac{d}{dx} (6x) = 6
\]
Therefore, the first derivative is:
\[
f'(x) = 21x^6 - 20x^4 + 33x^2 - 6
\]
2. **Second Derivative (\( f''(x) \))**:
To find the second derivative, we again apply the power rule to the first derivative.
\[
f''(x) = \frac{d}{dx} (21x^6) - \frac{d}{dx} (20x^4) + \frac{d}{dx} (33x^2) - \frac{d}{dx} (6)
\]
Calculating each term separately:
\[
\frac{d}{dx} (21x^6) = 126x^5
\]
\[
\frac{d}{dx} (20x^4) = 80x^3
\]
\
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning