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**
Given the equation \( x^3 + y^5 + 7x + 6y = 1 \), compute \(\frac{dy}{dx}\) at the point (1, -1).
**Solution Approach**
To find \(\frac{dy}{dx}\), we need to implicitly differentiate the given equation with respect to \( x \).
Start by differentiating each term of the equation with respect to \( x \):
\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(y^5) + \frac{d}{dx}(7x) + \frac{d}{dx}(6y) = \frac{d}{dx}(1) \]
**Step-by-step Derivation:**
1. Differentiate \( x^3 \):
\[ \frac{d}{dx}(x^3) = 3x^2 \]
2. Differentiate \( y^5 \) using the chain rule:
\[ \frac{d}{dx}(y^5) = 5y^4 \frac{dy}{dx} \]
3. Differentiate \( 7x \):
\[ \frac{d}{dx}(7x) = 7 \]
4. Differentiate \( 6y \) using the chain rule:
\[ \frac{d}{dx}(6y) = 6 \frac{dy}{dx} \]
5. The derivative of 1, a constant, is 0:
\[ \frac{d}{dx}(1) = 0 \]
**Combining All Terms:**
\[ 3x^2 + 5y^4 \frac{dy}{dx} + 7 + 6 \frac{dy}{dx} = 0 \]
**Solve for \(\frac{dy}{dx}\):**
Rearrange to isolate \( \frac{dy}{dx} \):
\[ 5y^4 \frac{dy}{dx} + 6 \frac{dy}{dx} = -3x^2 - 7 \]
\[ \left(5y^4 + 6 \right) \frac{dy}{dx} = -3x^2 - 7 \]
\[ \frac{dy}{dx} = \frac{-3x^2 - 7}{5y^4 + 6} \]
**Substitute the Point (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09277b3e-bfbb-4704-b58b-5d83ab5b8c9a%2F896123b0-c9c9-44f0-a511-12cc668e1585%2Fkq73syg_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Given the equation \( x^3 + y^5 + 7x + 6y = 1 \), compute \(\frac{dy}{dx}\) at the point (1, -1).
**Solution Approach**
To find \(\frac{dy}{dx}\), we need to implicitly differentiate the given equation with respect to \( x \).
Start by differentiating each term of the equation with respect to \( x \):
\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(y^5) + \frac{d}{dx}(7x) + \frac{d}{dx}(6y) = \frac{d}{dx}(1) \]
**Step-by-step Derivation:**
1. Differentiate \( x^3 \):
\[ \frac{d}{dx}(x^3) = 3x^2 \]
2. Differentiate \( y^5 \) using the chain rule:
\[ \frac{d}{dx}(y^5) = 5y^4 \frac{dy}{dx} \]
3. Differentiate \( 7x \):
\[ \frac{d}{dx}(7x) = 7 \]
4. Differentiate \( 6y \) using the chain rule:
\[ \frac{d}{dx}(6y) = 6 \frac{dy}{dx} \]
5. The derivative of 1, a constant, is 0:
\[ \frac{d}{dx}(1) = 0 \]
**Combining All Terms:**
\[ 3x^2 + 5y^4 \frac{dy}{dx} + 7 + 6 \frac{dy}{dx} = 0 \]
**Solve for \(\frac{dy}{dx}\):**
Rearrange to isolate \( \frac{dy}{dx} \):
\[ 5y^4 \frac{dy}{dx} + 6 \frac{dy}{dx} = -3x^2 - 7 \]
\[ \left(5y^4 + 6 \right) \frac{dy}{dx} = -3x^2 - 7 \]
\[ \frac{dy}{dx} = \frac{-3x^2 - 7}{5y^4 + 6} \]
**Substitute the Point (
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