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:**
Use implicit differentiation to find \( y' \) at the point \( (3, 0) \) for the given equation:
\[ 2x^3 + e^{3xy} = 55 \]
**Solution Steps:**
1. **Equation:**
- Start with the equation \( 2x^3 + e^{3xy} = 55 \).
2. **Differentiate Implicitly:**
- Apply implicit differentiation to both sides of the equation with respect to \( x \).
3. **Chain Rule Application:**
- Differentiate \( 2x^3 \) to get \( 6x^2 \).
- For \( e^{3xy} \), apply the chain rule:
- Differentiate \( e^{3xy} \) which gives \( e^{3xy} \cdot \frac{d}{dx}(3xy) \).
- The derivative of \( 3xy \) using the product rule is \( 3y + 3x \cdot y' \).
4. **Equate Derivatives:**
- Set the derivatives equal to zero since the derivative of the constant 55 is zero:
\[
6x^2 + e^{3xy} \cdot (3y + 3x \cdot y') = 0
\]
5. **Substitute \( (3, 0) \):**
- Substitute \( x = 3 \) and \( y = 0 \) into the differentiated equation.
- Simplify to solve for \( y' \).
6. **Solve for \( y' \):**
- Rearrange the equation to find \( y' \).
**Final Answer:**
Enter your computed value of \( y'(3, 0) \) in the box provided, and hit "Preview" to verify.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F718d0bec-2150-4df1-8dd3-9a5e25eabd78%2F57016ac1-0962-4271-8369-031f52d6d248%2Fsks5r4r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Use implicit differentiation to find \( y' \) at the point \( (3, 0) \) for the given equation:
\[ 2x^3 + e^{3xy} = 55 \]
**Solution Steps:**
1. **Equation:**
- Start with the equation \( 2x^3 + e^{3xy} = 55 \).
2. **Differentiate Implicitly:**
- Apply implicit differentiation to both sides of the equation with respect to \( x \).
3. **Chain Rule Application:**
- Differentiate \( 2x^3 \) to get \( 6x^2 \).
- For \( e^{3xy} \), apply the chain rule:
- Differentiate \( e^{3xy} \) which gives \( e^{3xy} \cdot \frac{d}{dx}(3xy) \).
- The derivative of \( 3xy \) using the product rule is \( 3y + 3x \cdot y' \).
4. **Equate Derivatives:**
- Set the derivatives equal to zero since the derivative of the constant 55 is zero:
\[
6x^2 + e^{3xy} \cdot (3y + 3x \cdot y') = 0
\]
5. **Substitute \( (3, 0) \):**
- Substitute \( x = 3 \) and \( y = 0 \) into the differentiated equation.
- Simplify to solve for \( y' \).
6. **Solve for \( y' \):**
- Rearrange the equation to find \( y' \).
**Final Answer:**
Enter your computed value of \( y'(3, 0) \) in the box provided, and hit "Preview" to verify.
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