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
please explain the answer or how you got the answer.
![---
### Derivative Calculation Exercise
#### Problem Statement:
Find the derivative of \( f(x) = 8^x \ln(x) \).
Use the rules of differentiation to solve this problem. Remember, the \( ln(x) \) represents the natural logarithm of \( x \), which is a common function in calculus.
#### Solution:
1. **Identify the Components:**
- \( u(x) = 8^x \)
- \( v(x) = \ln(x) \)
2. **Apply the Product Rule**:
The product rule states that if you have two functions, \( u(x) \) and \( v(x) \), the derivative of their product is:
\[
(u(x) \cdot v(x))' = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]
3. **Differentiate Each Function:**
- Derivative of \( u(x) = 8^x \):
\[
u'(x) = 8^x \ln(8)
\]
- Derivative of \( v(x) = \ln(x) \):
\[
v'(x) = \frac{1}{x}
\]
4. **Apply the Product Rule**:
\[
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]
Substituting \( u(x) \)'s and \( v(x) \)'s derivatives:
\[
f'(x) = 8^x \ln(8) \cdot \ln(x) + 8^x \cdot \frac{1}{x}
\]
5. **Combine Like Terms**:
\[
f'(x) = 8^x \ln(8) \ln(x) + \frac{8^x}{x}
\]
So, the derivative of \( f(x) = 8^x \ln(x) \) is:
\[
f'(x) = 8^x \ln(8) \ln(x) + \frac{8^x}{x}
\]
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdaf30f56-016e-4284-9e0e-308d7a6d7bdd%2F1cebe812-d918-4a5f-b92d-d8ddc0db28ab%2F51vib3r_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Derivative Calculation Exercise
#### Problem Statement:
Find the derivative of \( f(x) = 8^x \ln(x) \).
Use the rules of differentiation to solve this problem. Remember, the \( ln(x) \) represents the natural logarithm of \( x \), which is a common function in calculus.
#### Solution:
1. **Identify the Components:**
- \( u(x) = 8^x \)
- \( v(x) = \ln(x) \)
2. **Apply the Product Rule**:
The product rule states that if you have two functions, \( u(x) \) and \( v(x) \), the derivative of their product is:
\[
(u(x) \cdot v(x))' = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]
3. **Differentiate Each Function:**
- Derivative of \( u(x) = 8^x \):
\[
u'(x) = 8^x \ln(8)
\]
- Derivative of \( v(x) = \ln(x) \):
\[
v'(x) = \frac{1}{x}
\]
4. **Apply the Product Rule**:
\[
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]
Substituting \( u(x) \)'s and \( v(x) \)'s derivatives:
\[
f'(x) = 8^x \ln(8) \cdot \ln(x) + 8^x \cdot \frac{1}{x}
\]
5. **Combine Like Terms**:
\[
f'(x) = 8^x \ln(8) \ln(x) + \frac{8^x}{x}
\]
So, the derivative of \( f(x) = 8^x \ln(x) \) is:
\[
f'(x) = 8^x \ln(8) \ln(x) + \frac{8^x}{x}
\]
---
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 2 steps with 2 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