Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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![### Calculating the Derivative
**Problem Statement:**
Find \( dy \) for the following equation.
\[ y = 3x \sqrt{12 - x^3} \]
**Objective:**
Express \( \frac{dy}{dx} \) in terms of \( x \).
**Solution Steps:**
To find \( \frac{dy}{dx} \), we will use the product rule and chain rule of differentiation.
1. **Identify the Terms:**
- Let \( u = 3x \).
- Let \( v = \sqrt{12 - x^3} \).
2. **Differentiate Each Term:**
- \( \frac{du}{dx} = 3 \).
- For \( v = (12 - x^3)^{1/2} \):
- Using the chain rule, \( \frac{dv}{dx} = \frac{1}{2}(12 - x^3)^{-1/2} \cdot (-3x^2) \).
- Simplifying gives \( \frac{dv}{dx} = \frac{-3x^2}{2\sqrt{12 - x^3}} \).
3. **Apply the Product Rule:**
- The product rule states: \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).
- Substituting:
\[
\frac{dy}{dx} = 3x \cdot \frac{-3x^2}{2\sqrt{12 - x^3}} + \sqrt{12 - x^3} \cdot 3
\]
4. **Combine and Simplify:**
- \( \frac{dy}{dx} = \frac{-9x^3}{2\sqrt{12 - x^3}} + 3\sqrt{12 - x^3} \).
**Conclusion:**
Thus, \( dy = \left(\frac{-9x^3}{2\sqrt{12 - x^3}} + 3\sqrt{12 - x^3}\right) dx \).
This expression represents the derivative of \( y \) with respect to \( x \), capturing all changes in \( y \) as \( x \) varies.](https://content.bartleby.com/qna-images/question/aeadb183-1175-48ee-be1d-be4cbd36e1e9/e20c73ac-9a1d-4e7a-a891-5f478dcb1690/irc8qij_processed.jpeg)
Transcribed Image Text:### Calculating the Derivative
**Problem Statement:**
Find \( dy \) for the following equation.
\[ y = 3x \sqrt{12 - x^3} \]
**Objective:**
Express \( \frac{dy}{dx} \) in terms of \( x \).
**Solution Steps:**
To find \( \frac{dy}{dx} \), we will use the product rule and chain rule of differentiation.
1. **Identify the Terms:**
- Let \( u = 3x \).
- Let \( v = \sqrt{12 - x^3} \).
2. **Differentiate Each Term:**
- \( \frac{du}{dx} = 3 \).
- For \( v = (12 - x^3)^{1/2} \):
- Using the chain rule, \( \frac{dv}{dx} = \frac{1}{2}(12 - x^3)^{-1/2} \cdot (-3x^2) \).
- Simplifying gives \( \frac{dv}{dx} = \frac{-3x^2}{2\sqrt{12 - x^3}} \).
3. **Apply the Product Rule:**
- The product rule states: \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).
- Substituting:
\[
\frac{dy}{dx} = 3x \cdot \frac{-3x^2}{2\sqrt{12 - x^3}} + \sqrt{12 - x^3} \cdot 3
\]
4. **Combine and Simplify:**
- \( \frac{dy}{dx} = \frac{-9x^3}{2\sqrt{12 - x^3}} + 3\sqrt{12 - x^3} \).
**Conclusion:**
Thus, \( dy = \left(\frac{-9x^3}{2\sqrt{12 - x^3}} + 3\sqrt{12 - x^3}\right) dx \).
This expression represents the derivative of \( y \) with respect to \( x \), capturing all changes in \( y \) as \( x \) varies.
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