Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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Question
### Problem Statement Find \( f'(x) \). Given: \[ f(x) = 9x^3 (x^2 - 9) \] ### Solution To find the derivative \( f'(x) \), apply the product rule and chain rule. #### Steps: 1. **Identify the functions**: - Let \( u(x) = 9x^3 \) - Let \( v(x) = (x^2 - 9) \) 2. **Find the derivatives**: - \( u'(x) = \frac{d}{dx}(9x^3) = 27x^2 \) - \( v'(x) = \frac{d}{dx}(x^2 - 9) = 2x \) 3. **Apply the product rule**: - The product rule states that if \( f(x) = u(x) \cdot v(x) \), then: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] 4. **Calculate \( f'(x) \)**: \[ f'(x) = 27x^2 \cdot (x^2 - 9) + 9x^3 \cdot 2x \] \[ f'(x) = 27x^2 \cdot x^2 - 27x^2 \cdot 9 + 18x^4 \] \[ f'(x) = 27x^4 - 243x^2 + 18x^4 \] \[ f'(x) = 45x^4 - 243x^2 \] Therefore: \[ f'(x) = 45x^4 - 243x^2 \] The derivative is found and the blue box signifies where this answer should be entered.
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Transcribed Image Text:### Problem Statement Find \( f'(x) \). Given: \[ f(x) = 9x^3 (x^2 - 9) \] ### Solution To find the derivative \( f'(x) \), apply the product rule and chain rule. #### Steps: 1. **Identify the functions**: - Let \( u(x) = 9x^3 \) - Let \( v(x) = (x^2 - 9) \) 2. **Find the derivatives**: - \( u'(x) = \frac{d}{dx}(9x^3) = 27x^2 \) - \( v'(x) = \frac{d}{dx}(x^2 - 9) = 2x \) 3. **Apply the product rule**: - The product rule states that if \( f(x) = u(x) \cdot v(x) \), then: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] 4. **Calculate \( f'(x) \)**: \[ f'(x) = 27x^2 \cdot (x^2 - 9) + 9x^3 \cdot 2x \] \[ f'(x) = 27x^2 \cdot x^2 - 27x^2 \cdot 9 + 18x^4 \] \[ f'(x) = 27x^4 - 243x^2 + 18x^4 \] \[ f'(x) = 45x^4 - 243x^2 \] Therefore: \[ f'(x) = 45x^4 - 243x^2 \] The derivative is found and the blue box signifies where this answer should be entered.
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