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Function as, y (x) = (x² – 2) (3x + 2) %3D

Function as, y (x) = (x² – 2) (3x + 2) %3D

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...
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How do we get u'(x)=2x and v'(x)=3 ? Can you show this step by step? I can follow the rest of the problem, thanks

Given: Function as, y (x) = (x² – 2) (3x + 2) %3D Formula used: Product rule as: y' (x) = u (x) · v (x) + u' (x) · v (x). %3D Calculation: Let u = (x2- 2) and v = (3x + 2). Therefore, u' (x) = 2x and v (x) = 3 %3D Using the product rule, y' (x) = u (x) v (x) + u' (x) · v (x) 3D(x²-2) (3) + (2x) (3x + 2) %3D = 3x2-6+6x2 + 4x 3D9X² +4x -6 Thus, the value of y' (x) is 9x? + 4x - 6.
Transcribed Image Text:Given: Function as, y (x) = (x² – 2) (3x + 2) %3D Formula used: Product rule as: y' (x) = u (x) · v (x) + u' (x) · v (x). %3D Calculation: Let u = (x2- 2) and v = (3x + 2). Therefore, u' (x) = 2x and v (x) = 3 %3D Using the product rule, y' (x) = u (x) v (x) + u' (x) · v (x) 3D(x²-2) (3) + (2x) (3x + 2) %3D = 3x2-6+6x2 + 4x 3D9X² +4x -6 Thus, the value of y' (x) is 9x? + 4x - 6.
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