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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5.3 #5
Part 1 of 2
![### Calculus: Higher Order Derivatives
**Problem Statement:**
For the function \( f(x) = 3x^5 - 4x^4 + 3x^3 + 5x^2 + 6 \), find \( f'''(x) \), the third derivative of \( f \), and \( f^{(4)}(x) \), the fourth derivative of \( f \).
**Solution:**
1. **First Derivative, \( f'(x) \):**
- Differentiate each term with respect to \( x \):
- \( 3x^5 \) becomes \( 15x^4 \)
- \( -4x^4 \) becomes \( -16x^3 \)
- \( 3x^3 \) becomes \( 9x^2 \)
- \( 5x^2 \) becomes \( 10x \)
- \( 6 \) becomes \( 0 \)
Thus,
\[
f'(x) = 15x^4 - 16x^3 + 9x^2 + 10x
\]
2. **Second Derivative, \( f''(x) \):**
- Differentiate \( f'(x) \):
- \( 15x^4 \) becomes \( 60x^3 \)
- \( -16x^3 \) becomes \( -48x^2 \)
- \( 9x^2 \) becomes \( 18x \)
- \( 10x \) becomes \( 10 \)
Thus,
\[
f''(x) = 60x^3 - 48x^2 + 18x + 10
\]
3. **Third Derivative, \( f'''(x) \):**
- Differentiate \( f''(x) \):
- \( 60x^3 \) becomes \( 180x^2 \)
- \( -48x^2 \) becomes \( -96x \)
- \( 18x \) becomes \( 18 \)
- \( 10 \) becomes \( 0 \)
Thus,
\[
f'''(x) = 180x^2 - 96x + 18
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff3158bb-c7a6-4ac4-bdef-e27086080d78%2F911ff55f-0651-4d31-83b2-0a7ef600f1c5%2Fs8hvpmq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus: Higher Order Derivatives
**Problem Statement:**
For the function \( f(x) = 3x^5 - 4x^4 + 3x^3 + 5x^2 + 6 \), find \( f'''(x) \), the third derivative of \( f \), and \( f^{(4)}(x) \), the fourth derivative of \( f \).
**Solution:**
1. **First Derivative, \( f'(x) \):**
- Differentiate each term with respect to \( x \):
- \( 3x^5 \) becomes \( 15x^4 \)
- \( -4x^4 \) becomes \( -16x^3 \)
- \( 3x^3 \) becomes \( 9x^2 \)
- \( 5x^2 \) becomes \( 10x \)
- \( 6 \) becomes \( 0 \)
Thus,
\[
f'(x) = 15x^4 - 16x^3 + 9x^2 + 10x
\]
2. **Second Derivative, \( f''(x) \):**
- Differentiate \( f'(x) \):
- \( 15x^4 \) becomes \( 60x^3 \)
- \( -16x^3 \) becomes \( -48x^2 \)
- \( 9x^2 \) becomes \( 18x \)
- \( 10x \) becomes \( 10 \)
Thus,
\[
f''(x) = 60x^3 - 48x^2 + 18x + 10
\]
3. **Third Derivative, \( f'''(x) \):**
- Differentiate \( f''(x) \):
- \( 60x^3 \) becomes \( 180x^2 \)
- \( -48x^2 \) becomes \( -96x \)
- \( 18x \) becomes \( 18 \)
- \( 10 \) becomes \( 0 \)
Thus,
\[
f'''(x) = 180x^2 - 96x + 18
\]
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