fay = 3x Ln(x²1) 3x = 3Ln(x²-1) + 3× · 3x BLn (x²-) +

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...
icon
Related questions
icon
Concept explainers
Question

I'm looking to see where the mistake was made and what the correct approach is.

**Calculus: Differentiating a Function Involving Natural Logarithms**

Given the function:
\[ f(x) = 3x \ln(x^3 - 1) \]

We want to find the derivative, \( f'(x) \).

### Steps:

1. **Apply the Product Rule**: The function is a product of two functions, \( u = 3x \) and \( v = \ln(x^3 - 1) \). Therefore, we use the product rule for differentiation which states:
   \[ (uv)' = u'v + uv' \]

2. **Differentiate Each Component**:
   - The derivative of \( u = 3x \) is \( u' = 3 \).
   - To find \( v' \) where \( v = \ln(x^3 - 1) \), we use the chain rule:
     \[ v' = \frac{1}{x^3 - 1} \cdot (3x^2) = \frac{3x^2}{x^3 - 1} \]

3. **Apply the Product Rule**:
   \[ f'(x) = (3x)' \ln(x^3 - 1) + 3x \cdot \left( \frac{3x^2}{x^3 - 1} \right) \]
   \[ f'(x) = 3 \ln(x^3 - 1) + \frac{3x \cdot 3x^2}{x^3 - 1} \]

4. **Simplify the Expression**:
   \[ f'(x) = 3 \ln(x^3 - 1) + \frac{9x^3}{x^3 - 1} \]

5. **Final Result**:
   \[ f'(x) = 3 \ln(x^3 - 1) + \frac{3x}{x^3 - 1} \]

This completes the differentiation process, allowing students to recognize the techniques and rules applied for finding derivatives involving products and natural logarithms.
Transcribed Image Text:**Calculus: Differentiating a Function Involving Natural Logarithms** Given the function: \[ f(x) = 3x \ln(x^3 - 1) \] We want to find the derivative, \( f'(x) \). ### Steps: 1. **Apply the Product Rule**: The function is a product of two functions, \( u = 3x \) and \( v = \ln(x^3 - 1) \). Therefore, we use the product rule for differentiation which states: \[ (uv)' = u'v + uv' \] 2. **Differentiate Each Component**: - The derivative of \( u = 3x \) is \( u' = 3 \). - To find \( v' \) where \( v = \ln(x^3 - 1) \), we use the chain rule: \[ v' = \frac{1}{x^3 - 1} \cdot (3x^2) = \frac{3x^2}{x^3 - 1} \] 3. **Apply the Product Rule**: \[ f'(x) = (3x)' \ln(x^3 - 1) + 3x \cdot \left( \frac{3x^2}{x^3 - 1} \right) \] \[ f'(x) = 3 \ln(x^3 - 1) + \frac{3x \cdot 3x^2}{x^3 - 1} \] 4. **Simplify the Expression**: \[ f'(x) = 3 \ln(x^3 - 1) + \frac{9x^3}{x^3 - 1} \] 5. **Final Result**: \[ f'(x) = 3 \ln(x^3 - 1) + \frac{3x}{x^3 - 1} \] This completes the differentiation process, allowing students to recognize the techniques and rules applied for finding derivatives involving products and natural logarithms.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps with 4 images

Blurred answer
Knowledge Booster
Points, Lines and Planes
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.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
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: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning