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Let x³ + y5 + 7x + 6y = 1. Compute dy at (1,-1). dx

Let x³ + y5 + 7x + 6y = 1. Compute dy at (1,-1). dx

Calculus: Early Transcendentals
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 (
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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