Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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# Calculus Multiple Choice Question

## Problem Statement

Evaluate the derivative of \( x^3 \sin x \).

\[ \frac{d}{dx} \left[ x^3 \sin x \right] \]

### Options:
- (A) \( 3x^2 \cos x \)
- (B) \( 3x^2 \sin x \)
- (C) \( 3x^2 \sin x + x^3 \cos x \)
- (D) \( x^3 \cos x \)
- (E) None of the above.

### Explanation:

To solve this, use the product rule of differentiation which states: 

\[ \frac{d}{dx} [u \cdot v] = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \]

Here, \( u = x^3 \) and \( v = \sin x \).

First, compute the derivatives of \( u \) and \( v \):
\[ \frac{du}{dx} = 3x^2 \]
\[ \frac{dv}{dx} = \cos x \]

Apply the product rule:
\[ \frac{d}{dx} [x^3 \sin x] = (3x^2)(\sin x) + (x^3)(\cos x) \]

Hence, the correct answer is:
\[ \boxed{\text{(C) } 3x^2 \sin x + x^3 \cos x} \]
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Transcribed Image Text:# Calculus Multiple Choice Question ## Problem Statement Evaluate the derivative of \( x^3 \sin x \). \[ \frac{d}{dx} \left[ x^3 \sin x \right] \] ### Options: - (A) \( 3x^2 \cos x \) - (B) \( 3x^2 \sin x \) - (C) \( 3x^2 \sin x + x^3 \cos x \) - (D) \( x^3 \cos x \) - (E) None of the above. ### Explanation: To solve this, use the product rule of differentiation which states: \[ \frac{d}{dx} [u \cdot v] = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \] Here, \( u = x^3 \) and \( v = \sin x \). First, compute the derivatives of \( u \) and \( v \): \[ \frac{du}{dx} = 3x^2 \] \[ \frac{dv}{dx} = \cos x \] Apply the product rule: \[ \frac{d}{dx} [x^3 \sin x] = (3x^2)(\sin x) + (x^3)(\cos x) \] Hence, the correct answer is: \[ \boxed{\text{(C) } 3x^2 \sin x + x^3 \cos x} \]
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