Find by implicit differentiation. Then find the slope of the graph at the given point. dx 3x3y = 6, point (1, 2) %3D
Find by implicit differentiation. Then find the slope of the graph at the given point. dx 3x3y = 6, point (1, 2) %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Implicit Differentiation Tutorial**
**Problem Statement:**
Find \(\frac{dy}{dx}\) by implicit differentiation. Then find the slope of the graph at the given point.
**Equation:**
\[3x^3y = 6\]
**Given Point:**
\((1, 2)\)
### Steps for Solution:
1. **Implicit Differentiation:**
- To differentiate the given equation \(3x^3y = 6\) implicitly with respect to \(x\):
\[
\frac{d}{dx}(3x^3y) = \frac{d}{dx}(6)
\]
- Apply the product rule to the left-hand side:
\[
3x^3 \frac{dy}{dx} + y \frac{d}{dx}(3x^3) = 0
\]
- Differentiate \(3x^3\) with respect to \(x\):
\[
3x^3 \frac{dy}{dx} + y(9x^2) = 0
\]
- Rearrange to solve for \(\frac{dy}{dx}\):
\[
3x^3 \frac{dy}{dx} = -9x^2 y
\]
\[
\frac{dy}{dx} = \frac{-9x^2 y}{3x^3}
\]
\[
\frac{dy}{dx} = \frac{-9y}{3x}
\]
\[
\frac{dy}{dx} = -3 \frac{y}{x}
\]
2. **Finding the Slope at a Specific Point:**
- Substitute \(x = 1\) and \(y = 2\) into the expression for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} \bigg|_{(1, 2)} = -3 \frac{2}{1} = -6
\]
Therefore, 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%2F52a0c9f6-ada1-431f-830a-3fbdcbca3339%2Ffb56e121-be9e-4b04-bf67-d22a29d8c484%2F8aiz9fm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Implicit Differentiation Tutorial**
**Problem Statement:**
Find \(\frac{dy}{dx}\) by implicit differentiation. Then find the slope of the graph at the given point.
**Equation:**
\[3x^3y = 6\]
**Given Point:**
\((1, 2)\)
### Steps for Solution:
1. **Implicit Differentiation:**
- To differentiate the given equation \(3x^3y = 6\) implicitly with respect to \(x\):
\[
\frac{d}{dx}(3x^3y) = \frac{d}{dx}(6)
\]
- Apply the product rule to the left-hand side:
\[
3x^3 \frac{dy}{dx} + y \frac{d}{dx}(3x^3) = 0
\]
- Differentiate \(3x^3\) with respect to \(x\):
\[
3x^3 \frac{dy}{dx} + y(9x^2) = 0
\]
- Rearrange to solve for \(\frac{dy}{dx}\):
\[
3x^3 \frac{dy}{dx} = -9x^2 y
\]
\[
\frac{dy}{dx} = \frac{-9x^2 y}{3x^3}
\]
\[
\frac{dy}{dx} = \frac{-9y}{3x}
\]
\[
\frac{dy}{dx} = -3 \frac{y}{x}
\]
2. **Finding the Slope at a Specific Point:**
- Substitute \(x = 1\) and \(y = 2\) into the expression for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} \bigg|_{(1, 2)} = -3 \frac{2}{1} = -6
\]
Therefore, the slope of the graph at the point \((1, 2)\) is \(-6\).
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