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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Question
![**Finding the Slope of the Tangent Line Using Implicit Differentiation**
**Problem Statement:**
Use implicit differentiation to find the slope of the tangent line to the curve \( 9xy^3 + 5xy = 14 \) at the point \( (1, 1) \).
**Explanation:**
To find the slope of the tangent line to the given curve at the specified point, follow these steps:
1. **Differentiate Both Sides with Respect to \( x \):**
- Implicitly differentiate both sides of the equation \( 9xy^3 + 5xy = 14 \). Remember to use the product rule for terms involving both \( x \) and \( y \), and the chain rule for derivative of \( y \) with respect to \( x \).
2. **Solve for \( \frac{dy}{dx} \):**
- After differentiating, solve for \( \frac{dy}{dx} \), which represents the slope of the tangent line at any point on the curve.
3. **Substitute the Point \( (1, 1) \):**
- Plug in \( x = 1 \) and \( y = 1 \) into the derivative to find the specific slope at the point \( (1, 1) \).
**Detailed Solution:**
1. Differentiate the equation \( 9xy^3 + 5xy = 14 \) with respect to \( x \):
\[
\frac{d}{dx}(9xy^3) + \frac{d}{dx}(5xy) = \frac{d}{dx}(14)
\]
Using the product rule for \( 9xy^3 \):
\[
9 \left( x \cdot 3y^2 \cdot \frac{dy}{dx} + y^3 \cdot 1 \right) + 5 \left( x \cdot \frac{dy}{dx} + y \cdot 1 \right) = 0
\]
Simplify:
\[
27xy^2 \frac{dy}{dx} + 9y^3 + 5x \frac{dy}{dx} + 5y = 0
\]
Combine like terms:
\[
(27xy^2 + 5x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d7d1679-9649-4c1d-b706-e07713ee238e%2F7e31e9b0-9ed1-49cc-a64c-cb3c64cdc8e2%2Fjmt7oii_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Slope of the Tangent Line Using Implicit Differentiation**
**Problem Statement:**
Use implicit differentiation to find the slope of the tangent line to the curve \( 9xy^3 + 5xy = 14 \) at the point \( (1, 1) \).
**Explanation:**
To find the slope of the tangent line to the given curve at the specified point, follow these steps:
1. **Differentiate Both Sides with Respect to \( x \):**
- Implicitly differentiate both sides of the equation \( 9xy^3 + 5xy = 14 \). Remember to use the product rule for terms involving both \( x \) and \( y \), and the chain rule for derivative of \( y \) with respect to \( x \).
2. **Solve for \( \frac{dy}{dx} \):**
- After differentiating, solve for \( \frac{dy}{dx} \), which represents the slope of the tangent line at any point on the curve.
3. **Substitute the Point \( (1, 1) \):**
- Plug in \( x = 1 \) and \( y = 1 \) into the derivative to find the specific slope at the point \( (1, 1) \).
**Detailed Solution:**
1. Differentiate the equation \( 9xy^3 + 5xy = 14 \) with respect to \( x \):
\[
\frac{d}{dx}(9xy^3) + \frac{d}{dx}(5xy) = \frac{d}{dx}(14)
\]
Using the product rule for \( 9xy^3 \):
\[
9 \left( x \cdot 3y^2 \cdot \frac{dy}{dx} + y^3 \cdot 1 \right) + 5 \left( x \cdot \frac{dy}{dx} + y \cdot 1 \right) = 0
\]
Simplify:
\[
27xy^2 \frac{dy}{dx} + 9y^3 + 5x \frac{dy}{dx} + 5y = 0
\]
Combine like terms:
\[
(27xy^2 + 5x
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