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
![**Problem Statement:**
Find the derivative of the function.
\[ y = \ln(3x^2 - x) + 8x \]
**Solution:**
To find the derivative of the function \( y \) with respect to \( x \), use the following steps:
1. **Differentiate the logarithmic term:**
The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
Here, \( u = 3x^2 - x \).
- The derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = 6x - 1 \).
Therefore, the derivative of \( \ln(3x^2 - x) \) is:
\[
\frac{1}{3x^2 - x} \cdot (6x - 1)
\]
2. **Differentiate the linear term:**
The derivative of \( 8x \) with respect to \( x \) is \( 8 \).
**Combine the derivatives:**
The derivative of the entire function is:
\[
y' = \frac{6x - 1}{3x^2 - x} + 8
\]
**Conclusion:**
The derivative of the function \( y = \ln(3x^2 - x) + 8x \) is:
\[
y' = \frac{6x - 1}{3x^2 - x} + 8
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc7aab756-0f3b-421c-a400-c3f015f6942d%2F9f7ddd7f-cd28-451e-a278-cfda37deb920%2Fsu4bt4e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the derivative of the function.
\[ y = \ln(3x^2 - x) + 8x \]
**Solution:**
To find the derivative of the function \( y \) with respect to \( x \), use the following steps:
1. **Differentiate the logarithmic term:**
The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
Here, \( u = 3x^2 - x \).
- The derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = 6x - 1 \).
Therefore, the derivative of \( \ln(3x^2 - x) \) is:
\[
\frac{1}{3x^2 - x} \cdot (6x - 1)
\]
2. **Differentiate the linear term:**
The derivative of \( 8x \) with respect to \( x \) is \( 8 \).
**Combine the derivatives:**
The derivative of the entire function is:
\[
y' = \frac{6x - 1}{3x^2 - x} + 8
\]
**Conclusion:**
The derivative of the function \( y = \ln(3x^2 - x) + 8x \) is:
\[
y' = \frac{6x - 1}{3x^2 - x} + 8
\]
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