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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Question
![**Problem:**
Find \(\frac{dy}{dx}\) implicitly for the equation:
\[ x^2 e^{-x} + 3y^2 - xy = 0 \]
**Solution:**
Differentiate both sides of the equation with respect to \(x\). Use the product rule and chain rule where necessary.
1. Differentiate \(x^2 e^{-x}\):
- Apply the product rule: \((u \cdot v)' = u'v + uv'\)
- Let \(u = x^2\) and \(v = e^{-x}\)
- \(u' = 2x\) and \(v' = -e^{-x}\)
\[
\frac{d}{dx}(x^2 e^{-x}) = 2x e^{-x} + x^2(-e^{-x}) = 2x e^{-x} - x^2 e^{-x}
\]
2. Differentiate \(3y^2\):
- Use the chain rule: \(\frac{d}{dx}(3y^2) = 6y \frac{dy}{dx}\)
3. Differentiate \(-xy\):
- Use the product rule: \(u = x\), \(v = y\)
- \(u' = 1\), \(v' = \frac{dy}{dx}\)
\[
\frac{d}{dx}(-xy) = -\left(x \frac{dy}{dx} + y \right) = -x \frac{dy}{dx} - y
\]
Now substitute these derivatives into the equation:
\[
2x e^{-x} - x^2 e^{-x} + 6y \frac{dy}{dx} - x \frac{dy}{dx} - y = 0
\]
Combine terms with \(\frac{dy}{dx}\):
\[
(6y - x) \frac{dy}{dx} = y - 2x e^{-x} + x^2 e^{-x}
\]
Solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{y - 2x e^{-x} + x^2 e^{-x}}{6y - x}
\]
**Answer:**
\[
\frac{dy}{dx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66c6604b-039c-4afd-872a-b167ca11814a%2F1e2b4b36-7c4c-4c00-9348-97e94f2afb0c%2Fhiwd5ea_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find \(\frac{dy}{dx}\) implicitly for the equation:
\[ x^2 e^{-x} + 3y^2 - xy = 0 \]
**Solution:**
Differentiate both sides of the equation with respect to \(x\). Use the product rule and chain rule where necessary.
1. Differentiate \(x^2 e^{-x}\):
- Apply the product rule: \((u \cdot v)' = u'v + uv'\)
- Let \(u = x^2\) and \(v = e^{-x}\)
- \(u' = 2x\) and \(v' = -e^{-x}\)
\[
\frac{d}{dx}(x^2 e^{-x}) = 2x e^{-x} + x^2(-e^{-x}) = 2x e^{-x} - x^2 e^{-x}
\]
2. Differentiate \(3y^2\):
- Use the chain rule: \(\frac{d}{dx}(3y^2) = 6y \frac{dy}{dx}\)
3. Differentiate \(-xy\):
- Use the product rule: \(u = x\), \(v = y\)
- \(u' = 1\), \(v' = \frac{dy}{dx}\)
\[
\frac{d}{dx}(-xy) = -\left(x \frac{dy}{dx} + y \right) = -x \frac{dy}{dx} - y
\]
Now substitute these derivatives into the equation:
\[
2x e^{-x} - x^2 e^{-x} + 6y \frac{dy}{dx} - x \frac{dy}{dx} - y = 0
\]
Combine terms with \(\frac{dy}{dx}\):
\[
(6y - x) \frac{dy}{dx} = y - 2x e^{-x} + x^2 e^{-x}
\]
Solve for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{y - 2x e^{-x} + x^2 e^{-x}}{6y - x}
\]
**Answer:**
\[
\frac{dy}{dx
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