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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Question
![**Problem Statement:**
If \( f(x) = 2\sqrt{x} \ln(x) \), find \( f'(x) \).
---
**Solution Steps:**
1. **Identify and Apply the Product Rule:**
- The function \( f(x) = 2\sqrt{x} \ln(x) \) is a product of two functions: \( u(x) = 2\sqrt{x} \) and \( v(x) = \ln(x) \).
- Use the product rule: \( (uv)' = u'v + uv' \).
2. **Find Derivatives:**
- \( u(x) = 2\sqrt{x} = 2x^{1/2} \)
- \( u'(x) = \frac{d}{dx}[2x^{1/2}] = x^{-1/2} = \frac{1}{\sqrt{x}} \).
- \( v(x) = \ln(x) \)
- \( v'(x) = \frac{1}{x} \).
3. **Apply the Product Rule:**
- \( f'(x) = u'(x) v(x) + u(x) v'(x) \)
- \( f'(x) = \left(\frac{1}{\sqrt{x}}\right) \ln(x) + \left(2\sqrt{x}\right) \frac{1}{x} \).
4. **Simplify:**
- \( f'(x) = \frac{\ln(x)}{\sqrt{x}} + \frac{2}{\sqrt{x}} \)
- \( f'(x) = \frac{\ln(x) + 2}{\sqrt{x}} \).
---
**Final Expression:**
\[ f'(x) = \frac{\ln(x) + 2}{\sqrt{x}} \]
---
**Evaluate \( f'(1) \):**
- Substitute \( x = 1 \) into \( f'(x) \):
- \( f'(1) = \frac{\ln(1) + 2}{\sqrt{1}} \).
- Since \( \ln(1) = 0 \), \( f'(1) = \frac{0 + 2}{1} = 2 \).
---
**Result:**
\[ f'(1) = 2 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe345ae0b-fd83-4d2e-a467-23dbb110ff23%2Fdd11c51c-06bf-4faa-8a7b-6adf4d8f8867%2Fwoav175_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
If \( f(x) = 2\sqrt{x} \ln(x) \), find \( f'(x) \).
---
**Solution Steps:**
1. **Identify and Apply the Product Rule:**
- The function \( f(x) = 2\sqrt{x} \ln(x) \) is a product of two functions: \( u(x) = 2\sqrt{x} \) and \( v(x) = \ln(x) \).
- Use the product rule: \( (uv)' = u'v + uv' \).
2. **Find Derivatives:**
- \( u(x) = 2\sqrt{x} = 2x^{1/2} \)
- \( u'(x) = \frac{d}{dx}[2x^{1/2}] = x^{-1/2} = \frac{1}{\sqrt{x}} \).
- \( v(x) = \ln(x) \)
- \( v'(x) = \frac{1}{x} \).
3. **Apply the Product Rule:**
- \( f'(x) = u'(x) v(x) + u(x) v'(x) \)
- \( f'(x) = \left(\frac{1}{\sqrt{x}}\right) \ln(x) + \left(2\sqrt{x}\right) \frac{1}{x} \).
4. **Simplify:**
- \( f'(x) = \frac{\ln(x)}{\sqrt{x}} + \frac{2}{\sqrt{x}} \)
- \( f'(x) = \frac{\ln(x) + 2}{\sqrt{x}} \).
---
**Final Expression:**
\[ f'(x) = \frac{\ln(x) + 2}{\sqrt{x}} \]
---
**Evaluate \( f'(1) \):**
- Substitute \( x = 1 \) into \( f'(x) \):
- \( f'(1) = \frac{\ln(1) + 2}{\sqrt{1}} \).
- Since \( \ln(1) = 0 \), \( f'(1) = \frac{0 + 2}{1} = 2 \).
---
**Result:**
\[ f'(1) = 2 \]
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