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:**
1. Use implicit differentiation to find dy/dx when \( x = 1 \) and \( y = 1 \).
\( y^2 + xy = 2. \)
**Solution Explanation:**
Implicit differentiation is a technique used to find the derivative of a function in terms of both \( x \) and \( y \) without explicitly solving for \( y \).
Given the equation \( y^2 + xy = 2 \), we want to find \( \frac{dy}{dx} \) when \( x = 1 \) and \( y = 1 \).
**Steps:**
1. Differentiate both sides of the equation with respect to \( x \):
- The derivative of \( y^2 \) is \( 2y \frac{dy}{dx} \) (using the chain rule).
- The derivative of \( xy \) is \( x \frac{dy}{dx} + y \) (using the product rule).
2. Therefore, differentiate the entire equation:
\[
2y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0
\]
3. Rearrange the terms to isolate \( \frac{dy}{dx} \):
\[
(2y + x) \frac{dy}{dx} = -y
\]
4. Solve for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{-y}{2y + x}
\]
5. Substitute \( x = 1 \) and \( y = 1 \) into the equation:
\[
\frac{dy}{dx} = \frac{-1}{2(1) + 1} = \frac{-1}{3}
\]
**Conclusion:**
The value of \( \frac{dy}{dx} \) when \( x = 1 \) and \( y = 1 \) is \( -\frac{1}{3} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f5b0569-83e9-4f3d-93a8-6d46166ffdc9%2Fd4b33628-acc6-4108-8914-e0480fbbdd81%2Fakftegg_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
1. Use implicit differentiation to find dy/dx when \( x = 1 \) and \( y = 1 \).
\( y^2 + xy = 2. \)
**Solution Explanation:**
Implicit differentiation is a technique used to find the derivative of a function in terms of both \( x \) and \( y \) without explicitly solving for \( y \).
Given the equation \( y^2 + xy = 2 \), we want to find \( \frac{dy}{dx} \) when \( x = 1 \) and \( y = 1 \).
**Steps:**
1. Differentiate both sides of the equation with respect to \( x \):
- The derivative of \( y^2 \) is \( 2y \frac{dy}{dx} \) (using the chain rule).
- The derivative of \( xy \) is \( x \frac{dy}{dx} + y \) (using the product rule).
2. Therefore, differentiate the entire equation:
\[
2y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0
\]
3. Rearrange the terms to isolate \( \frac{dy}{dx} \):
\[
(2y + x) \frac{dy}{dx} = -y
\]
4. Solve for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{-y}{2y + x}
\]
5. Substitute \( x = 1 \) and \( y = 1 \) into the equation:
\[
\frac{dy}{dx} = \frac{-1}{2(1) + 1} = \frac{-1}{3}
\]
**Conclusion:**
The value of \( \frac{dy}{dx} \) when \( x = 1 \) and \( y = 1 \) is \( -\frac{1}{3} \).
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