Homework Help is Here – Start Your Trial Now!
Math
Calculus
Multiple choice. 3 Part C. d [x³ sin x] dx (A) 3x² cos x (D) x³ cos x 3 2 (B) 3x sin x (E) None of the above. 2 3 (C) 3x² sin x + x³ cos x

Multiple choice. 3 Part C. d [x³ sin x] dx (A) 3x² cos x (D) x³ cos x 3 2 (B) 3x sin x (E) None of the above. 2 3 (C) 3x² sin x + x³ cos x

Calculus: Early Transcendentals
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...
See similar textbooks
icon
Related questions
Question
# 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: \[ \boxed{\text{(C) } 3x^2 \sin x + x^3 \cos x} \]
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: \[ \boxed{\text{(C) } 3x^2 \sin x + x^3 \cos x} \]
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS