Use u(x) = (x2 - 2)³, v(x) = x2 + 3, u'(x) = 6x(x2 – 2)², and v'(x) = 2x to find the derivative of (x2 – 2)3 y = Then, rearrange the numerator in preparation for simplifying. x + 3 v(x) · u'(x) – u(x) • v'(x) y' x² +3 ). 6x(x? – 2)2 - (x? – 2)3 . 2x y' [x² + 3]? 6x(x? + 3)(x? – 2)2 -( y' = 2)3 (x² + 3)²
Use u(x) = (x2 - 2)³, v(x) = x2 + 3, u'(x) = 6x(x2 – 2)², and v'(x) = 2x to find the derivative of (x2 – 2)3 y = Then, rearrange the numerator in preparation for simplifying. x + 3 v(x) · u'(x) – u(x) • v'(x) y' x² +3 ). 6x(x? – 2)2 - (x? – 2)3 . 2x y' [x² + 3]? 6x(x? + 3)(x? – 2)2 -( y' = 2)3 (x² + 3)²
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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![Use u(x) = (x2 - 2)³, v(x) = x2 + 3, u'(x) = 6x(x2 – 2)2, and v'(x) = 2x to find the derivative of
(x2 – 2)3
y =
Then, rearrange the numerator in preparation for simplifying.
x + 3
v(x) · u'(x) – u(x) • v'(x)
y' =
² +3
). 6x(x2 – 2)2 – (x² – 2)³ · 2x
y'
[x² + 3]?
6x(x2 + 3)(x2 – 2)² |
y' =
Ja? - 23
(x² + 3)²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d73adbd-9f80-482e-aac0-f9071e3f3b13%2Ffc6752d6-84e6-452a-99d1-9570546045ee%2F60f9ud7_processed.png&w=3840&q=75)
Transcribed Image Text:Use u(x) = (x2 - 2)³, v(x) = x2 + 3, u'(x) = 6x(x2 – 2)2, and v'(x) = 2x to find the derivative of
(x2 – 2)3
y =
Then, rearrange the numerator in preparation for simplifying.
x + 3
v(x) · u'(x) – u(x) • v'(x)
y' =
² +3
). 6x(x2 – 2)2 – (x² – 2)³ · 2x
y'
[x² + 3]?
6x(x2 + 3)(x2 – 2)² |
y' =
Ja? - 23
(x² + 3)²
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