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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![**Problem Statement:**
Determine \(\frac{dy}{dx}\) when \(y = \ln (2x^2 - 4x)\).
**Solution:**
To find the derivative of \(y\) with respect to \(x\), use the chain rule for differentiation.
1. Start with the given function \(y = \ln (2x^2 - 4x)\).
2. Let \(u = 2x^2 - 4x\). Using the chain rule, \(y = \ln u\).
3. According to the chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
4. Find \(\frac{dy}{du}\): Since \(y = \ln u\),
\[
\frac{dy}{du} = \frac{1}{u}
\]
5. Find \(\frac{du}{dx}\): Given \(u = 2x^2 - 4x\),
\[
\frac{du}{dx} = \frac{d}{dx} (2x^2 - 4x) = 4x - 4
\]
6. Plug in \(\frac{du}{dx}\) and \(\frac{dy}{du}\) into the chain rule:
\[
\frac{dy}{dx} = \frac{1}{u} \cdot (4x - 4)
\]
7. Substitute back \(u = 2x^2 - 4x\) into the equation:
\[
\frac{dy}{dx} = \frac{4x - 4}{2x^2 - 4x}
\]
Simplify the expression if possible:
\[
\frac{dy}{dx} = \frac{4(x - 1)}{2x(x - 2)} = \frac{2(x - 1)}{x(x - 2)}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec8465cd-fdc1-4278-9e3f-7f6c30db5e12%2F250de52b-6c37-4d39-82f9-34bb776281ba%2Fjhjaavn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Determine \(\frac{dy}{dx}\) when \(y = \ln (2x^2 - 4x)\).
**Solution:**
To find the derivative of \(y\) with respect to \(x\), use the chain rule for differentiation.
1. Start with the given function \(y = \ln (2x^2 - 4x)\).
2. Let \(u = 2x^2 - 4x\). Using the chain rule, \(y = \ln u\).
3. According to the chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
4. Find \(\frac{dy}{du}\): Since \(y = \ln u\),
\[
\frac{dy}{du} = \frac{1}{u}
\]
5. Find \(\frac{du}{dx}\): Given \(u = 2x^2 - 4x\),
\[
\frac{du}{dx} = \frac{d}{dx} (2x^2 - 4x) = 4x - 4
\]
6. Plug in \(\frac{du}{dx}\) and \(\frac{dy}{du}\) into the chain rule:
\[
\frac{dy}{dx} = \frac{1}{u} \cdot (4x - 4)
\]
7. Substitute back \(u = 2x^2 - 4x\) into the equation:
\[
\frac{dy}{dx} = \frac{4x - 4}{2x^2 - 4x}
\]
Simplify the expression if possible:
\[
\frac{dy}{dx} = \frac{4(x - 1)}{2x(x - 2)} = \frac{2(x - 1)}{x(x - 2)}
\]
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