Question 2 Differentiate y= (3 cos x – 2 sin (2r))² A y' = [-6 cos x + 4 sin (2x)][8 cos (2x) + 6 sin x] %3D B y' = - 2[3 cos x – 2 sin (2x)] c y'= [6 cos x – 4 sin (2x)][4cos (2x) – 3 sin x] D y' = - 2[3 cos x – 2 sin (2x)][3 sin x + 4 cos (2x)]
Question 2 Differentiate y= (3 cos x – 2 sin (2r))² A y' = [-6 cos x + 4 sin (2x)][8 cos (2x) + 6 sin x] %3D B y' = - 2[3 cos x – 2 sin (2x)] c y'= [6 cos x – 4 sin (2x)][4cos (2x) – 3 sin x] D y' = - 2[3 cos x – 2 sin (2x)][3 sin x + 4 cos (2x)]
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...
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![Question 2
Differentiate y= (3 cos x – 2 sin (2x))2
(A
y' = [-6 cos x + 4 sin (2x)][8 cos (2x) + 6 sin x]
В
y' = - 2[3 cos x – 2 sin (2x)]
y'= [6 cos x – 4 sin (2x)][4cos (2x) – 3 sin x]
y' = - 2[3 cos x
2 sin (2x)][3 sin x + 4 cos (2x)]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F271325e3-200b-44fa-b6ff-6e61ad07120a%2Fb009661d-a78b-45b5-96f9-538523d49b2d%2Fs0cucmu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 2
Differentiate y= (3 cos x – 2 sin (2x))2
(A
y' = [-6 cos x + 4 sin (2x)][8 cos (2x) + 6 sin x]
В
y' = - 2[3 cos x – 2 sin (2x)]
y'= [6 cos x – 4 sin (2x)][4cos (2x) – 3 sin x]
y' = - 2[3 cos x
2 sin (2x)][3 sin x + 4 cos (2x)]
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