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:**
Given the function \( f(x) = \sqrt{8x + 3} \), compute the first, second, and third derivatives of \( f(x) \).
**Solution:**
1. **First Derivative (\( f'(x) \))**:
- To find \( f'(x) \), we use the chain rule.
- Let \( u = 8x + 3 \).
- Then \( f(x) = \sqrt{u} = u^{1/2} \).
- The derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2}u^{-1/2} \).
- The derivative of \( u \) with respect to \( x \) is \( u' = 8 \).
- Therefore, \( f'(x) \) becomes:
\[ f'(x) = \frac{1}{2}(8x + 3)^{-1/2} \cdot 8 = \frac{4}{\sqrt{8x + 3}} \]
2. **Second Derivative (\( f''(x) \))**:
- To find \( f''(x) \), we take the derivative of \( f'(x) \).
- Let \( v = 8x + 3 \).
- We have \( f'(x) = \frac{4}{\sqrt{v}} \).
- Rewrite \( f'(x) = 4v^{-1/2} \).
- The derivative of \( 4v^{-1/2} \) is \( 4 \cdot (-\frac{1}{2})v^{-3/2} \cdot v' = 4 \cdot (-\frac{1}{2})v^{-3/2} \cdot 8 \).
- Therefore, \( f''(x) \) becomes:
\[ f''(x) = -\frac{16}{(8x + 3)^{3/2}} \]
3. **Third Derivative (\( f'''(x) \))**:
- To find \( f'''(x) \), we take the derivative of \( f''(x) \).
- Let \( w = 8x + 3 \).
- We](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09277b3e-bfbb-4704-b58b-5d83ab5b8c9a%2F132d1bf3-b057-4613-bdbf-3c0e1657a918%2Fwpuetsq_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Given the function \( f(x) = \sqrt{8x + 3} \), compute the first, second, and third derivatives of \( f(x) \).
**Solution:**
1. **First Derivative (\( f'(x) \))**:
- To find \( f'(x) \), we use the chain rule.
- Let \( u = 8x + 3 \).
- Then \( f(x) = \sqrt{u} = u^{1/2} \).
- The derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2}u^{-1/2} \).
- The derivative of \( u \) with respect to \( x \) is \( u' = 8 \).
- Therefore, \( f'(x) \) becomes:
\[ f'(x) = \frac{1}{2}(8x + 3)^{-1/2} \cdot 8 = \frac{4}{\sqrt{8x + 3}} \]
2. **Second Derivative (\( f''(x) \))**:
- To find \( f''(x) \), we take the derivative of \( f'(x) \).
- Let \( v = 8x + 3 \).
- We have \( f'(x) = \frac{4}{\sqrt{v}} \).
- Rewrite \( f'(x) = 4v^{-1/2} \).
- The derivative of \( 4v^{-1/2} \) is \( 4 \cdot (-\frac{1}{2})v^{-3/2} \cdot v' = 4 \cdot (-\frac{1}{2})v^{-3/2} \cdot 8 \).
- Therefore, \( f''(x) \) becomes:
\[ f''(x) = -\frac{16}{(8x + 3)^{3/2}} \]
3. **Third Derivative (\( f'''(x) \))**:
- To find \( f'''(x) \), we take the derivative of \( f''(x) \).
- Let \( w = 8x + 3 \).
- We
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