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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![### Calculus Problem: Implicit Differentiation
Given the implicit equation of a curve:
\[ x^2 + 2xy - y^2 = 2, \]
we are asked to find the derivative \(\frac{dy}{dx}\) at the point (1, 1).
**Options:**
- \(\frac{3}{2}\)
- \(0\)
- \(-\frac{3}{2}\)
- nonexistent (This option is selected)
To solve this, we typically use implicit differentiation to find \(\frac{dy}{dx}\):
1. Differentiate both sides of the given equation with respect to \(x\):
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) - \frac{d}{dx}(y^2) = \frac{d}{dx}(2)
\]
2. Apply the differentiation rules:
\[
2x + 2\left(x \frac{dy}{dx} + y \right) - 2y \frac{dy}{dx} = 0
\]
3. Simplify to find \(\frac{dy}{dx}\):
\[
2x + 2x \frac{dy}{dx} + 2y - 2y \frac{dy}{dx} = 0
\]
\[
2x + 2y + 2x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0
\]
\[
2x + 2y = 2y \frac{dy}{dx} - 2x \frac{dy}{dx}
\]
Factor out \(\frac{dy}{dx}\):
\[
2x + 2y = (2y - 2x) \frac{dy}{dx}
\]
\[
\frac{dy}{dx} = \frac{2x + 2y}{2y - 2x}
\]
4. Substitute the point (1, 1) into the derived formula for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{2(1) + 2(1)}{2(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2bf6a656-c902-430b-8919-3920c6e4552b%2Fb51f55e1-5afd-4a06-a45a-152fb91a9712%2Fx3jfc8_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Implicit Differentiation
Given the implicit equation of a curve:
\[ x^2 + 2xy - y^2 = 2, \]
we are asked to find the derivative \(\frac{dy}{dx}\) at the point (1, 1).
**Options:**
- \(\frac{3}{2}\)
- \(0\)
- \(-\frac{3}{2}\)
- nonexistent (This option is selected)
To solve this, we typically use implicit differentiation to find \(\frac{dy}{dx}\):
1. Differentiate both sides of the given equation with respect to \(x\):
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) - \frac{d}{dx}(y^2) = \frac{d}{dx}(2)
\]
2. Apply the differentiation rules:
\[
2x + 2\left(x \frac{dy}{dx} + y \right) - 2y \frac{dy}{dx} = 0
\]
3. Simplify to find \(\frac{dy}{dx}\):
\[
2x + 2x \frac{dy}{dx} + 2y - 2y \frac{dy}{dx} = 0
\]
\[
2x + 2y + 2x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0
\]
\[
2x + 2y = 2y \frac{dy}{dx} - 2x \frac{dy}{dx}
\]
Factor out \(\frac{dy}{dx}\):
\[
2x + 2y = (2y - 2x) \frac{dy}{dx}
\]
\[
\frac{dy}{dx} = \frac{2x + 2y}{2y - 2x}
\]
4. Substitute the point (1, 1) into the derived formula for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{2(1) + 2(1)}{2(
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