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 Statement:**
Differentiate \(\frac{dy}{dx}\) by applying the chain rule repeatedly. Show work.
Given:
\[ y = \left( \sqrt{x} + \frac{x}{x-2} \right)^7 \]
**Explanation:**
The given function \( y \) is a composite function where the outer function is \((u)^7\) and the inner function \(u\) is \( \sqrt{x} + \frac{x}{x-2} \).
To differentiate using the chain rule, follow these steps:
1. **Differentiate the outer function:**
Let \( y = u^7 \).
Then, \(\frac{dy}{du} = 7u^6 \).
2. **Differentiate the inner function \(u\):**
Let \( u = \sqrt{x} + \frac{x}{x-2} \).
To differentiate \(u\), split it into two parts:
\[
u = \sqrt{x} + \frac{x}{x-2}
\]
- Differentiate \(\sqrt{x}\):
\[ \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \]
- Differentiate \(\frac{x}{x-2}\):
Use the quotient rule where \( f(x) = x \) and \( g(x) = x - 2 \):
\[
\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}
\]
\[
= \frac{(x - 2)(1) - x(1)}{(x - 2)^2}
\]
\[
= \frac{x - 2 - x}{(x - 2)^2}
\]
\[
= -\frac{2}{(x - 2)^2}
\]
Therefore,
\[
\frac{du}{dx} = \frac{1}{2\sqrt{x}} - \frac{2}{(x - 2)^2}
\]
3. **Combine the results using the chain rule:**
\[
\frac{dy](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63c12736-f43f-4777-9d94-03ebf9ed40f6%2F8d764745-361e-4483-ba1e-3c6d87013591%2F5xsmry_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Differentiate \(\frac{dy}{dx}\) by applying the chain rule repeatedly. Show work.
Given:
\[ y = \left( \sqrt{x} + \frac{x}{x-2} \right)^7 \]
**Explanation:**
The given function \( y \) is a composite function where the outer function is \((u)^7\) and the inner function \(u\) is \( \sqrt{x} + \frac{x}{x-2} \).
To differentiate using the chain rule, follow these steps:
1. **Differentiate the outer function:**
Let \( y = u^7 \).
Then, \(\frac{dy}{du} = 7u^6 \).
2. **Differentiate the inner function \(u\):**
Let \( u = \sqrt{x} + \frac{x}{x-2} \).
To differentiate \(u\), split it into two parts:
\[
u = \sqrt{x} + \frac{x}{x-2}
\]
- Differentiate \(\sqrt{x}\):
\[ \frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \]
- Differentiate \(\frac{x}{x-2}\):
Use the quotient rule where \( f(x) = x \) and \( g(x) = x - 2 \):
\[
\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}
\]
\[
= \frac{(x - 2)(1) - x(1)}{(x - 2)^2}
\]
\[
= \frac{x - 2 - x}{(x - 2)^2}
\]
\[
= -\frac{2}{(x - 2)^2}
\]
Therefore,
\[
\frac{du}{dx} = \frac{1}{2\sqrt{x}} - \frac{2}{(x - 2)^2}
\]
3. **Combine the results using the chain rule:**
\[
\frac{dy
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