Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem Statement:**

Write the function in the form \( y = f(u) \) and \( u = g(x) \). Then find \(\frac{dy}{dx}\) as a function of \(x\).

\[ y = (-2x + 8)^3 \]

**Options:**

- **Option A:**  
  \( y = u^3; \, u = -2x + 8; \, \frac{dy}{dx} = -6(-2x + 8)^2 \)

- **Option B:**  
  \( y = u^3; \, u = -2x + 8; \, \frac{dy}{dx} = 3(-2x + 8)^2 \)

- **Option C:**  
  \( y = u^3; \, u = -2x + 8; \, \frac{dy}{dx} = -2(-2x + 8)^3 \)

- **Option D:**  
  \( y = 3u + 8; \, u = x^3; \, \frac{dy}{dx} = -6x^2 \)
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Transcribed Image Text:**Problem Statement:** Write the function in the form \( y = f(u) \) and \( u = g(x) \). Then find \(\frac{dy}{dx}\) as a function of \(x\). \[ y = (-2x + 8)^3 \] **Options:** - **Option A:** \( y = u^3; \, u = -2x + 8; \, \frac{dy}{dx} = -6(-2x + 8)^2 \) - **Option B:** \( y = u^3; \, u = -2x + 8; \, \frac{dy}{dx} = 3(-2x + 8)^2 \) - **Option C:** \( y = u^3; \, u = -2x + 8; \, \frac{dy}{dx} = -2(-2x + 8)^3 \) - **Option D:** \( y = 3u + 8; \, u = x^3; \, \frac{dy}{dx} = -6x^2 \)
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