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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![**Implicit Differentiation Example**
To find \( y'' \) (the second derivative) using implicit differentiation, start with the given equation:
\[ 2x^3 + 3y^3 = 2 \]
**Steps to Solve:**
1. **Differentiate both sides with respect to \( x \):**
- For \( 2x^3 \), use the power rule: \( \frac{d}{dx}(2x^3) = 6x^2 \).
- For \( 3y^3 \), use the chain rule: \( \frac{d}{dx}(3y^3) = 9y^2 \frac{dy}{dx} \).
2. **Resulting Derivative Equation:**
Combine these results:
\[ 6x^2 + 9y^2 \frac{dy}{dx} = 0 \]
3. **Solve for \( \frac{dy}{dx} \):**
Isolate \( \frac{dy}{dx} \):
\[ 9y^2 \frac{dy}{dx} = -6x^2 \]
\[ \frac{dy}{dx} = \frac{-6x^2}{9y^2} \]
4. **Differentiate Again for \( y'' \):**
Differentiate \( \frac{dy}{dx} = \frac{-6x^2}{9y^2} \) using the quotient and chain rules to find \( y'' \).
**Note:** The detailed steps for finding \( y'' \) involve applying implicit differentiation again to \( \frac{dy}{dx} \).
---
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Transcribed Image Text:**Implicit Differentiation Example**
To find \( y'' \) (the second derivative) using implicit differentiation, start with the given equation:
\[ 2x^3 + 3y^3 = 2 \]
**Steps to Solve:**
1. **Differentiate both sides with respect to \( x \):**
- For \( 2x^3 \), use the power rule: \( \frac{d}{dx}(2x^3) = 6x^2 \).
- For \( 3y^3 \), use the chain rule: \( \frac{d}{dx}(3y^3) = 9y^2 \frac{dy}{dx} \).
2. **Resulting Derivative Equation:**
Combine these results:
\[ 6x^2 + 9y^2 \frac{dy}{dx} = 0 \]
3. **Solve for \( \frac{dy}{dx} \):**
Isolate \( \frac{dy}{dx} \):
\[ 9y^2 \frac{dy}{dx} = -6x^2 \]
\[ \frac{dy}{dx} = \frac{-6x^2}{9y^2} \]
4. **Differentiate Again for \( y'' \):**
Differentiate \( \frac{dy}{dx} = \frac{-6x^2}{9y^2} \) using the quotient and chain rules to find \( y'' \).
**Note:** The detailed steps for finding \( y'' \) involve applying implicit differentiation again to \( \frac{dy}{dx} \).
---
**Need Help?**
Click "Read It" or "Watch It" for additional resources and guidance on implicit differentiation.
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