Use implicit differentiation to find dy/dx and d'y/dx? x? + y? = 9 dy O A. x d²y x2 +y? y' dx? dx dy O B. x d?y x+ y? dx y' dx? y3 dy x dy x2 - Oc. dx y dx2 y2 dy OD. x d'y y' dx2 x² + y? y? dx
Use implicit differentiation to find dy/dx and d'y/dx? x? + y? = 9 dy O A. x d²y x2 +y? y' dx? dx dy O B. x d?y x+ y? dx y' dx? y3 dy x dy x2 - Oc. dx y dx2 y2 dy OD. x d'y y' dx2 x² + y? y? dx
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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![**Implicit Differentiation Problem**
Use implicit differentiation to find \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \).
Given the equation:
\[ x^2 + y^2 = 9 \]
Determine which of the following options correctly represents the first and second derivatives.
**Options:**
- **A.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^3}
\]
- **B.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x + y^2}{y^3}
\]
- **C.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x^2 - y}{y^2}
\]
- **D.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^2}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65e9754d-e7bf-4951-a913-18b478a6d2d4%2Fb7655fa7-7e35-449d-a427-d2cb3cc93a3b%2F5yfdl9b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Implicit Differentiation Problem**
Use implicit differentiation to find \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \).
Given the equation:
\[ x^2 + y^2 = 9 \]
Determine which of the following options correctly represents the first and second derivatives.
**Options:**
- **A.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^3}
\]
- **B.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x + y^2}{y^3}
\]
- **C.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x^2 - y}{y^2}
\]
- **D.**
\[
\frac{dy}{dx} = -\frac{x}{y}, \quad \frac{d^2y}{dx^2} = -\frac{x^2 + y^2}{y^2}
\]
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