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Use the product rule to find the derivative of j(x) = (4x³) (2x² − 3). If we set f(x) ƒ'(x) g'(x) Thus j'(x): = = = = 4x³ and g(x) 2x² (2x² − 3) + (4x³). Simplifying, we have j'(x) = = 3, then and

Use the product rule to find the derivative of j(x) = (4x³) (2x² − 3). If we set f(x) ƒ'(x) g'(x) Thus j'(x): = = = = 4x³ and g(x) 2x² (2x² − 3) + (4x³). Simplifying, we have j'(x) = = 3, then and

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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Use the product rule to find the derivative of j(x) = (4x³) (2x² −– 3). If we set f(x) ƒ'(x) = g'(x) Thus j'(x) = = = = = 4x³ and g(x) 2x² · (2x² − 3) + (4x³). Simplifying, we have j'(x) = = 3, then and
Transcribed Image Text:Use the product rule to find the derivative of j(x) = (4x³) (2x² −– 3). If we set f(x) ƒ'(x) = g'(x) Thus j'(x) = = = = = 4x³ and g(x) 2x² · (2x² − 3) + (4x³). Simplifying, we have j'(x) = = 3, then and
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