Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
**Problem:** Evaluate \(\frac{d}{dx}(2x + 4)(3x + 2)\). **Options:** A) 6 B) \(6x^2 + 16x + 8\) C) \(12x + 16\) D) \(30x^2 + 43x + 8\) **Explanation:** To solve this problem, apply the product rule of differentiation, which states: If \(u(x) = f(x) \cdot g(x)\), then \( \frac{d}{dx}[u(x)] = f'(x)g(x) + f(x)g'(x) \). Let \(f(x) = 2x + 4\) and \(g(x) = 3x + 2\). 1. Compute \(f'(x)\): The derivative of \(2x + 4\) is 2. 2. Compute \(g'(x)\): The derivative of \(3x + 2\) is 3. 3. Apply the product rule: \[ \frac{d}{dx}[(2x + 4)(3x + 2)] = (2)(3x + 2) + (2x + 4)(3) \] Evaluate: \[ = (2)(3x) + (2)(2) + (2x)(3) + (4)(3) = 6x + 4 + 6x + 12 = 12x + 16 \] Therefore, the correct option is C) \(12x + 16\).
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Transcribed Image Text:**Problem:** Evaluate \(\frac{d}{dx}(2x + 4)(3x + 2)\). **Options:** A) 6 B) \(6x^2 + 16x + 8\) C) \(12x + 16\) D) \(30x^2 + 43x + 8\) **Explanation:** To solve this problem, apply the product rule of differentiation, which states: If \(u(x) = f(x) \cdot g(x)\), then \( \frac{d}{dx}[u(x)] = f'(x)g(x) + f(x)g'(x) \). Let \(f(x) = 2x + 4\) and \(g(x) = 3x + 2\). 1. Compute \(f'(x)\): The derivative of \(2x + 4\) is 2. 2. Compute \(g'(x)\): The derivative of \(3x + 2\) is 3. 3. Apply the product rule: \[ \frac{d}{dx}[(2x + 4)(3x + 2)] = (2)(3x + 2) + (2x + 4)(3) \] Evaluate: \[ = (2)(3x) + (2)(2) + (2x)(3) + (4)(3) = 6x + 4 + 6x + 12 = 12x + 16 \] Therefore, the correct option is C) \(12x + 16\).
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Given query is to find the Indicated derivative.

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