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...
Related questions
Question
2. What is the process for finding dy/dx when y is an implicit function of x? Illustrate with an
example.
![**Finding the Derivative \( \frac{dy}{dx} \) for Implicit Functions**
**Question 2:** What is the process for finding \( \frac{dy}{dx} \) when \( y \) is an implicit function of \( x \)? Illustrate with an example.
**Explanation:**
To find \( \frac{dy}{dx} \) for an implicit function, follow these steps:
1. **Differentiate both sides of the equation with respect to \( x \):** Use the chain rule for terms involving \( y \), treating \( y \) as a function of \( x \).
2. **Collect all terms involving \( \frac{dy}{dx} \) on one side of the equation.**
3. **Solve for \( \frac{dy}{dx} \):** Isolate \( \frac{dy}{dx} \) to find the derivative.
**Example:**
Consider the implicit equation \( x^2 + y^2 = 25 \).
1. Differentiate both sides with respect to \( x \):
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)
\]
2. Using the chain rule:
\[
2x + 2y \frac{dy}{dx} = 0
\]
3. Solve for \( \frac{dy}{dx} \):
\[
2y \frac{dy}{dx} = -2x
\]
\[
\frac{dy}{dx} = -\frac{x}{y}
\]
This process illustrates how to find the derivative of \( y \) with respect to \( x \) when \( y \) is defined implicitly.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa71aa627-012c-4819-b4d9-6da9d6e2c8f5%2Fbdb239be-f16a-425f-9ed4-a26e260e3e76%2F056ddlf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding the Derivative \( \frac{dy}{dx} \) for Implicit Functions**
**Question 2:** What is the process for finding \( \frac{dy}{dx} \) when \( y \) is an implicit function of \( x \)? Illustrate with an example.
**Explanation:**
To find \( \frac{dy}{dx} \) for an implicit function, follow these steps:
1. **Differentiate both sides of the equation with respect to \( x \):** Use the chain rule for terms involving \( y \), treating \( y \) as a function of \( x \).
2. **Collect all terms involving \( \frac{dy}{dx} \) on one side of the equation.**
3. **Solve for \( \frac{dy}{dx} \):** Isolate \( \frac{dy}{dx} \) to find the derivative.
**Example:**
Consider the implicit equation \( x^2 + y^2 = 25 \).
1. Differentiate both sides with respect to \( x \):
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)
\]
2. Using the chain rule:
\[
2x + 2y \frac{dy}{dx} = 0
\]
3. Solve for \( \frac{dy}{dx} \):
\[
2y \frac{dy}{dx} = -2x
\]
\[
\frac{dy}{dx} = -\frac{x}{y}
\]
This process illustrates how to find the derivative of \( y \) with respect to \( x \) when \( y \) is defined implicitly.
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