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
**Problem Statement:**
Find \(\frac{dy}{dx}\) by implicit differentiation. Then find the slope of the graph at the given point.
Given equation:
\[ 3x^3 y = 6 \]
Given point:
\[ (1, 2) \]
**Explanation:**
To find \(\frac{dy}{dx}\) using implicit differentiation, we will differentiate both sides of the given equation with respect to \(x\).
1. Start with the given equation:
\[ 3x^3 y = 6 \]
2. Differentiate both sides with respect to \(x\), applying the product rule on the left-hand side:
\[ \frac{d}{dx}(3x^3 y) = \frac{d}{dx}(6) \]
\[ 3 \frac{d}{dx}(x^3) y + 3x^3 \frac{d}{dx}(y) = 0 \]
3. Calculate the derivatives:
\[ 3 \cdot 3x^2 y + 3x^3 \frac{dy}{dx} = 0 \]
\[ 9x^2 y + 3x^3 \frac{dy}{dx} = 0 \]
4. Solve for \(\frac{dy}{dx}\):
\[ 3x^3 \frac{dy}{dx} = -9x^2 y \]
\[ \frac{dy}{dx} = \frac{-9x^2 y}{3x^3} \]
\[ \frac{dy}{dx} = \frac{-9y}{3x} \]
\[ \frac{dy}{dx} = -\frac{3y}{x} \]
5. Substitute the given point \((1, 2)\) into the derivative to find the slope at that specific point:
\[ \left. \frac{dy}{dx} \right|_{(1,2)} = -\frac{3 \cdot 2}{1} \]
\[ \left. \frac{dy}{dx} \right|_{(1,2)} = -6 \]
So, the slope of the graph at the point \((1, 2)\) is \(-6\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb62ce946-c165-4529-a471-ed49759bc0da%2F04204bff-4ca3-4a33-a30a-fee05ea60744%2Fwmmtmpj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Implicit Differentiation Problem
**Problem Statement:**
Find \(\frac{dy}{dx}\) by implicit differentiation. Then find the slope of the graph at the given point.
Given equation:
\[ 3x^3 y = 6 \]
Given point:
\[ (1, 2) \]
**Explanation:**
To find \(\frac{dy}{dx}\) using implicit differentiation, we will differentiate both sides of the given equation with respect to \(x\).
1. Start with the given equation:
\[ 3x^3 y = 6 \]
2. Differentiate both sides with respect to \(x\), applying the product rule on the left-hand side:
\[ \frac{d}{dx}(3x^3 y) = \frac{d}{dx}(6) \]
\[ 3 \frac{d}{dx}(x^3) y + 3x^3 \frac{d}{dx}(y) = 0 \]
3. Calculate the derivatives:
\[ 3 \cdot 3x^2 y + 3x^3 \frac{dy}{dx} = 0 \]
\[ 9x^2 y + 3x^3 \frac{dy}{dx} = 0 \]
4. Solve for \(\frac{dy}{dx}\):
\[ 3x^3 \frac{dy}{dx} = -9x^2 y \]
\[ \frac{dy}{dx} = \frac{-9x^2 y}{3x^3} \]
\[ \frac{dy}{dx} = \frac{-9y}{3x} \]
\[ \frac{dy}{dx} = -\frac{3y}{x} \]
5. Substitute the given point \((1, 2)\) into the derivative to find the slope at that specific point:
\[ \left. \frac{dy}{dx} \right|_{(1,2)} = -\frac{3 \cdot 2}{1} \]
\[ \left. \frac{dy}{dx} \right|_{(1,2)} = -6 \]
So, the slope of the graph at the point \((1, 2)\) is \(-6\).
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