Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
The task is to write the given function in the form \( y = f(u) \) where \( u = g(x) \), and then find \(\frac{dy}{dx}\) as a function of \(x\).

The given function is:

\[ y = \left( \frac{x^2}{8} + 5x - \frac{3}{x} \right)^4 \]

The problem asks for the selection of the correct form from the multiple choices below:

**Options:**

A. \( y = f(u) = u^4 \) and \( u = g(x) = \left( \frac{x^2}{8} + 5x - \frac{3}{x} \right) \)

B. \( y = f(u) = u \) and \( u = g(x) = \frac{x^2}{8} + 5x \)

C. \( y = f(u) = u^4 \) and \( u = g(x) = \frac{x^2}{8} + 5x - \frac{3}{x} \)

D. \( y = f(u) = u \) and \( u = g(x) = \frac{x^2}{8} + 5x - \frac{3}{x} \)

---

**Explanation:** 

The task involves selecting the correct expressions for \( y \) as a function of \( u \) and \( u \) as a function of \( x \) that match the initial given function structure. The correct answer should accurately represent the original function's form through these substitutions.
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Transcribed Image Text:The task is to write the given function in the form \( y = f(u) \) where \( u = g(x) \), and then find \(\frac{dy}{dx}\) as a function of \(x\). The given function is: \[ y = \left( \frac{x^2}{8} + 5x - \frac{3}{x} \right)^4 \] The problem asks for the selection of the correct form from the multiple choices below: **Options:** A. \( y = f(u) = u^4 \) and \( u = g(x) = \left( \frac{x^2}{8} + 5x - \frac{3}{x} \right) \) B. \( y = f(u) = u \) and \( u = g(x) = \frac{x^2}{8} + 5x \) C. \( y = f(u) = u^4 \) and \( u = g(x) = \frac{x^2}{8} + 5x - \frac{3}{x} \) D. \( y = f(u) = u \) and \( u = g(x) = \frac{x^2}{8} + 5x - \frac{3}{x} \) --- **Explanation:** The task involves selecting the correct expressions for \( y \) as a function of \( u \) and \( u \) as a function of \( x \) that match the initial given function structure. The correct answer should accurately represent the original function's form through these substitutions.
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