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
100%
I am confused on this step. Apparently I am supposed to factor X5 out of the numerator, but I do not understand how you would factor this.
Can you please explain how to factor the numerator on the left into the numerator on the right?
Thank you!
![The given expression is a derivative of a function using the quotient rule. Let's break it down.
We start with the function:
\[ y' = \frac{(4 - x^5)(6x^5) - x^6(-5x^4)}{(4 - x^5)^2} \]
1. In the numerator, we first apply the product rule to the two terms \((4 - x^5)\) and \(6x^5\):
- \((4 - x^5)(6x^5)\): This term comes from the derivative of the first function multiplied by the second function.
- \(x^6(-5x^4)\): This term comes from the derivative of the second function multiplied by the first function.
Expanding these products in the numerator:
\[
(4 - x^5)(6x^5): 4 \cdot 6x^5 - x^5 \cdot 6x^5
\]
\[
= 24x^5 - 6x^{10}
\]
Similarly for the second term:
\[
- x^6(-5x^4): - (-5x^4 \cdot x^6)
\]
\[
= 5x^{10}
\]
2. Combining these simplified terms, the numerator becomes:
\[
24x^5 - 6x^{10} + 5x^{10} = 24x^5 - x^{10}
\]
3. Substituting this back into the original quotient:
\[
y' = \frac{24x^5 - x^{10}}{(4 - x^5)^2}
\]
4. Rearranging terms for readability:
\[
y' = \frac{x^5(24 - 6x^5 + 5x^5)}{(4 - x^5)^2}
\]
This derivation illustrates how to apply the quotient rule properly to a function involving powers of \(x\), demonstrating the complexity of handling polynomial derivatives and providing insight into the algebraic manipulations involved in calculus.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc438c34-6e26-4f6a-9067-f367a43a4982%2F3134eebc-bb38-4961-b1b1-cdf40cb2d259%2Fcdcahmv.png&w=3840&q=75)
Transcribed Image Text:The given expression is a derivative of a function using the quotient rule. Let's break it down.
We start with the function:
\[ y' = \frac{(4 - x^5)(6x^5) - x^6(-5x^4)}{(4 - x^5)^2} \]
1. In the numerator, we first apply the product rule to the two terms \((4 - x^5)\) and \(6x^5\):
- \((4 - x^5)(6x^5)\): This term comes from the derivative of the first function multiplied by the second function.
- \(x^6(-5x^4)\): This term comes from the derivative of the second function multiplied by the first function.
Expanding these products in the numerator:
\[
(4 - x^5)(6x^5): 4 \cdot 6x^5 - x^5 \cdot 6x^5
\]
\[
= 24x^5 - 6x^{10}
\]
Similarly for the second term:
\[
- x^6(-5x^4): - (-5x^4 \cdot x^6)
\]
\[
= 5x^{10}
\]
2. Combining these simplified terms, the numerator becomes:
\[
24x^5 - 6x^{10} + 5x^{10} = 24x^5 - x^{10}
\]
3. Substituting this back into the original quotient:
\[
y' = \frac{24x^5 - x^{10}}{(4 - x^5)^2}
\]
4. Rearranging terms for readability:
\[
y' = \frac{x^5(24 - 6x^5 + 5x^5)}{(4 - x^5)^2}
\]
This derivation illustrates how to apply the quotient rule properly to a function involving powers of \(x\), demonstrating the complexity of handling polynomial derivatives and providing insight into the algebraic manipulations involved in calculus.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.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