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...
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![**Problem Statement:**
\[
\text{[2]} \quad \text{Consider} \quad y = \frac{(x+2)^3 (2x+1)^9}{x^8 (3x+1)^4} \quad \text{where} \quad x > 0
\]
**Instructions:**
First, take the natural logarithm (Natural Log) of both sides, then find \(\frac{dy}{dx}\).
**Solution Approach:**
1. **Taking the Natural Logarithm:**
- Apply the natural logarithm to both sides of the equation for easier differentiation:
\[
\ln y = \ln \left(\frac{(x+2)^3 (2x+1)^9}{x^8 (3x+1)^4}\right)
\]
- Use logarithmic properties to simplify:
\[
\ln y = 3\ln(x+2) + 9\ln(2x+1) - 8\ln x - 4\ln(3x+1)
\]
2. **Differentiate Both Sides with Respect to \(x\):**
- The derivative of \(\ln y\) with respect to \(x\) is:
\[
\frac{1}{y} \frac{dy}{dx}
\]
- Differentiate the right side using the chain rule:
\[
3\cdot \frac{1}{x+2} \cdot 1 + 9\cdot \frac{1}{2x+1} \cdot 2 - 8\cdot \frac{1}{x} \cdot 1 - 4\frac{1}{3x+1} \cdot 3
\]
3. **Solve for \(\frac{dy}{dx}\):**
- Combine and simplify the derivatives:
\[
\frac{dy}{dx} = y \left( \frac{3}{x+2} + \frac{18}{2x+1} - \frac{8}{x} - \frac{12}{3x+1} \right)
\]
- Substitute back \(y = \frac{(x+2)^3 (2x+1)^9}{x^8 (3x+](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f45806f-9331-47bc-a32a-91feef99b6e6%2F456852bb-610f-4773-8f37-9f3f03914901%2Ftyf08uh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
\[
\text{[2]} \quad \text{Consider} \quad y = \frac{(x+2)^3 (2x+1)^9}{x^8 (3x+1)^4} \quad \text{where} \quad x > 0
\]
**Instructions:**
First, take the natural logarithm (Natural Log) of both sides, then find \(\frac{dy}{dx}\).
**Solution Approach:**
1. **Taking the Natural Logarithm:**
- Apply the natural logarithm to both sides of the equation for easier differentiation:
\[
\ln y = \ln \left(\frac{(x+2)^3 (2x+1)^9}{x^8 (3x+1)^4}\right)
\]
- Use logarithmic properties to simplify:
\[
\ln y = 3\ln(x+2) + 9\ln(2x+1) - 8\ln x - 4\ln(3x+1)
\]
2. **Differentiate Both Sides with Respect to \(x\):**
- The derivative of \(\ln y\) with respect to \(x\) is:
\[
\frac{1}{y} \frac{dy}{dx}
\]
- Differentiate the right side using the chain rule:
\[
3\cdot \frac{1}{x+2} \cdot 1 + 9\cdot \frac{1}{2x+1} \cdot 2 - 8\cdot \frac{1}{x} \cdot 1 - 4\frac{1}{3x+1} \cdot 3
\]
3. **Solve for \(\frac{dy}{dx}\):**
- Combine and simplify the derivatives:
\[
\frac{dy}{dx} = y \left( \frac{3}{x+2} + \frac{18}{2x+1} - \frac{8}{x} - \frac{12}{3x+1} \right)
\]
- Substitute back \(y = \frac{(x+2)^3 (2x+1)^9}{x^8 (3x+
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